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Jul 20, 2024

# WBCHSE Class 11 Chemistry: Some Basic Concepts of Chemistry – Chhaya Publication

Unit-1 (Some Basic Concepts of Chemisiry) of Chhaya Chemistry (xi) Part-1

Q1. The ratio of molecular masses of two substances $$A$$ and $$B$$ is $$2: 3$$. If equal masses of these two substances be vaporised, the ratio of their volumes will be-
(A) $$2: 1$$
(B) $$1: 2$$
(C) $$2: 3$$
(D) $$3: 2$$

(D) $$3: 2$$

The molecular mass ratio is given as $$M_A : M_B = 2:3$$.
Let, the molecular mass of substance $$A$$ and substance $$B$$ be $$2x$$ unit and $$3x$$ unit respectively (where $$x > 0$$ ).
Given equal masses $$m_A = m_B = m (say)$$.
Using the ideal gas equation $$PV = nRT$$, where $$n = \frac{m}{M}$$,
Therefore, $$n_A = \frac{m}{2x}$$
$$n_B = \frac{m}{3x}$$
Since $$V \propto n$$ at constant $$P$$ and $$T$$,
$$\frac{V_A}{V_B} = \frac{n_A}{n_B} = \frac{\frac{m}{2x}}{\frac{m}{3x}} = \frac{3}{2}$$
Hence the ratio of volumes $$V_A : V_B = 3:2$$.

Q2. In natural chlorine gas the ratio of the number of moles of $${ }^{35} \mathrm{Cl}$$ and $${ }^{37} \mathrm{Cl}$$ is-
(A) $$2: 3$$
(B) $$3: 1$$
(C) $$2: 1$$
(D) $$4: 3$$

(B) $$3: 1$$

Let the percentage of $$^{35}Cl$$ be $$x$$, and that of $$^{37}Cl$$ be $$(100-x)$$.

The average atomic mass of chlorine (Cl) that we have 35.5. Therefore-
$$35.5 = \frac{35x+37(100-x)}{100}$$
$$\Rightarrow 3550 = 35x+3700-37x$$
$$\Rightarrow -150 = -2x$$
$$∴ x = 75$$  and  $$100-x = 25$$

Hence, the ratio of number of moles of $$^{35}Cl$$ and  $$^{37}Cl$$ is:
$$= \frac{75}{35.5} : \frac{25}{35.5} = 3:2$$ ….. (B)

Q3. Naturally occurring boron (average molecular mass = 10.8 u ) has two isotopes $${ }^{10} \mathrm{~B}$$ and $${ }^{11} \mathrm{~B}$$. The percentage of $${ }^{10} \mathrm{~B}$$ in the naturally occurring boron is-
(A) 20
(B) 23
(C) 31
(D) 27

(A) 20

Let the percentage of $$^{10}B$$ be $$x$$, and that of $$^{11}B$$ be $$(100-x)$$.

The average atomic mass is given by:
$$10.8 = \frac{10x+11(100-x)}{100}$$
$$\Rightarrow 1080 = 10x+1100-11x$$
$$\Rightarrow -20 = -x$$
$$∴ x = 20$$ ……. (A)

Q4. 2.0 g of a mixture of $$\mathrm{Na}_{2} \mathrm{CO}_{3}$$ and $$\mathrm{NaHCO}_{3}$$ is heated strongly at $$300^{\circ} \mathrm{C}$$ till it attains constant mass. On cooling, the mixture attains a mass of 1.752 g . The percentage of $$\mathrm{Na}_{2} \mathrm{CO}_{3}$$ in the initial mixture is-
(A) 50.2
(B) 66.4
(C) 56.8
(D) 43.8

(B) 66.4

Given:
Initial mass of the mixture: 2.0 g
Final mass after heating: 1.752 g

The reaction is:
$$2 \mathrm{NaHCO_3} \rightarrow \mathrm{Na_2CO_3} + \mathrm{H_2O} + \mathrm{CO_2}$$

Let $$x$$ grams be the mass of $$\mathrm{Na_2CO_3}$$ in the initial mixture, and $$y$$ grams be the mass of $$\mathrm{NaHCO_3}$$ in the initial mixture.

The initial total mass is:
$$x + y = 2.0 \, \text{g}$$

After heating, the $$\mathrm{NaHCO_3}$$ decomposes into $$\mathrm{Na_2CO_3}$$, $$\mathrm{H_2O}$$, and $$\mathrm{CO_2}$$. The $$\mathrm{Na_2CO_3}$$ from the $$\mathrm{NaHCO_3}$$ decomposition and the initial $$\mathrm{Na_2CO_3}$$ remain:

The decomposition reaction shows that 2 moles of $$\mathrm{NaHCO_3}$$ (168 g) produce 1 mole of $$\mathrm{Na_2CO_3}$$ (106 g).

So, the retained mass of $$\mathrm{Na_2CO_3}$$ after decomposition is:
$$\frac{106}{168} \times y$$

The total final mass is the sum of the retained $$\mathrm{Na_2CO_3}$$ from the decomposition and the initial $$\mathrm{Na_2CO_3}$$:
$$x + \frac{106}{168} \times y = 1.752$$

We now have two equations:
1. $$x + y = 2.0$$
2. $$x + \frac{106}{168} \times y = 1.752$$

Substituting $$y = 2.0-x$$ into the second equation:
$$\Rightarrow x + \frac{106}{168} \times (2.0-x) = 1.752$$
$$\Rightarrow x + \frac{106}{168} \times 2.0- \frac{106}{168} \times x = 1.752$$
$$\Rightarrow x + \frac{212}{168}- \frac{106}{168} \times x = 1.752$$
$$\Rightarrow x + 1.262-0.631 x = 1.752$$
$$\Rightarrow 0.369 x = 0.49$$
$$\Rightarrow x = \frac{0.49}{0.369}$$
$$∴ x \approx 1.328 \, \text{g}$$

So, the mass of $$\mathrm{Na_2CO_3}$$ in the initial mixture is approximately 1.328 g.

The percentage of $$\mathrm{Na_2CO_3}$$ in the initial mixture is:
$$\frac{1.328}{2.0} \times 100 \approx 66.4\%$$

Therefore, the correct answer is: (B) 66.4

Q5. 1.38 g of silver carbonate is heated strongly. On cooling, the mass of the residue becomes-
(A) 1.38 g
(B) 1.82 g
(C) 1.08 g
(D) 1.42 g

(C) 1.08 g

Let’s find the mass of the residue, which is silver $$Ag$$ after the decomposition.

Silver carbonate decomposes as follows:
$$2Ag_2CO_3 \rightarrow 4Ag + CO_2 + 2O_2$$

Molar masses:
$$Ag_2CO_3 = 108 \times 2 + 12 + 16 \times 3 = 276 \ \text{g/mol}$$
$$Ag = 108 \ \text{g/mol}$$

The mass ratio:
$$\frac{4 \times 108}{2 \times 276} = \frac{mass \ of \ residue}{1.38}$$
$$\Rightarrow \frac{432}{552} \times 1.38 = 1.08 g$$

Therefore, the correct answer is: (C) 1.08 g

Q6. The equivalent weight of a metal is 9 . If the chloride of the metal has a vapour density of 66.8 , the atomic mass of the metal will be-
(A) 16.4
(B) 27
(C) 26.3
(D) 32.5

(B) 27

Given equivalent weight of metal = 9
Vapour density of metal chloride = 66.8
∴ Molecular weight of metal chloride = $$2 \times \text{VD} = 2 \times 66.8 = 133.6$$

Valency of metal:
$$= \frac{\text { molecular weight of metal chloride }}{\text { equivalent weight of metal }+35.5}$$
$$= \frac{133.6}{9+35.5}=3$$

Therefore, atomic weight of the metal = equivalent weight $$\times$$ valency $$=9 \times 3=27$$

Q7. 3.04 g of a metal hydroxide is heated strongly till it decomposes completely to the corresponding metal oxide having a mass of 2.0 g . The equivalent weight of the metal is-
(A) 12.1
(B) 16.2
(C) 9.3
(D) 20.0

(C) 9.3

Here, $$18 x$$ is equivalent to $$1.04$$
$$18 x \approx 1.04$$
$$\Rightarrow x\approx \frac{{1.04}}{{18}}=0.05778$$

Again, $$\quad 2 m+16 x \approx 2$$
$$\Rightarrow 2m\approx 2-16\times (0.05778)\\\Rightarrow m\approx \frac{{1.07552}}{2}=0.53776$$

$$\therefore$$ Equivalent weight of metal in metal oxide $$\left(\mathrm{M}_{2} \mathrm{O}_{x}\right)$$
$$=\frac{{2m}}{{2m+16x}}\times 8\\=\frac{{16m}}{{2m+16x}}\\=\frac{{16\times 0.53776}}{2}=9.3\ (As\ 2m+16x\approx 2)$$

Q8. The percentage of nitrogen in an organic compound containing $$\mathrm{C}, \mathrm{H}$$ and N is 20 . Lowest possible molecular mass of the compound will be-
(A) 80
(B) 70
(C) 60
(D) 48

(B) 70

Let’s assume the molecular mass of the compound is $$M$$.

The mass of nitrogen in the compound is $$0.20M$$.

Given that the molecular mass of nitrogen (N) is 14 g/mol, we can set up the following equation:
$$14z = 0.20M$$

Where $$z$$ is the number of nitrogen atoms in the compound.

Solving for $$M$$:
$$M = \frac{14z}{0.20}$$
$$M = 70z$$

To find the lowest possible molecular mass, we take the smallest possible integer value for $$z$$, which is 1 (since $$z$$ must be at least 1 to have nitrogen in the compound):
$$M = 70 \times 1$$
$$M = 70$$

Thus, the lowest possible molecular mass of the compound is 70.

The correct answer is (B) 70.

Q9. A crystalline salt $$\left(\mathrm{M}_{2} \mathrm{SO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}\right)$$ on heating loses $$55.9 \%$$ of its mass. If the atomic mass of $$M$$ be $$23$$ , the value of $$x$$ will be-
(A) 7
(B) 8
(C) 9
(D) 10

(D) 10

First, let’s write down the chemical reaction:

$M_2SO_4 \cdot xH_2O \rightarrow M_2SO_4 + xH_2O$

Given:
Atomic mass of $$M$$ is 23.
The compound loses 55.9% of its mass (which is the mass of water).

Let’s calculate the molar mass of the anhydrous salt ($$M_2SO_4$$):

$$\text{Molar mass of } M_2SO_4 = 2 \times 23 + 32 + 4 \times 16$$
$$= 46 + 32 + 64$$
$$= 142 \text{ g/mol}$$

Next, let’s calculate the molar mass of $$xH_2O$$:

$$\text{Molar mass of } xH_2O = x \times (2 \times 1 + 16)$$
$$= 18x \text{ g/mol}$$

The total molar mass of the hydrated salt $$M_2SO_4 \cdot xH_2O$$ is:

$$(142 + 18x) \text{ g/mol}$$

The compound loses 55.9% of its mass as water, so:

$$\frac{18x}{142+18x} = 0.559$$

Solving for $$x$$:

$$18x = 0.559 (142+18x)$$
$$\Rightarrow 18x = 79.378+10.062x$$
$$\Rightarrow 18x-10.062x = 79.378$$
$$\Rightarrow 7.938x = 79.378$$
$$\Rightarrow x = \frac{79.378}{7.938}$$
$$∴ x \approx 10$$

Thus, the value of $$x$$ is 10.

The correct answer is (D) 10.

Q10. Equivalent weight of A in the compound $$\mathrm{A}_{x} \mathrm{O}_{y}$$ is ‘ $$E$$. If atomic mass of $$A$$ be ‘ $$a$$ ‘ and $$y=1.5 x$$, then which of the following relations will hold good-
(A) $$a=1.5 E$$
(B) $$E=0.2 \times a$$
(C) $$a=3 \times E$$
(D) $$E=0.4 \times a$$

(C) $$a=3 \times E$$

From the formula of the compound $$\mathrm{A}_x \mathrm{O}_y$$, we can say, 16 grams of $$\mathrm{O}_2$$ combine with $$a$$ grams of $$A$$,

Therefore, equivalent weight of element A:
$$E = \frac{a \times x}{16 \times y} \times 8$$.

Given that $$y = 1.5x$$, let’s use this information to derive the correct relationship:

Substitute $$y = 1.5x$$:
$$E = \frac{a \times x}{16 \times 1.5x} \times 8$$
$$\Rightarrow E = \frac{a \times x}{24x} \times 8$$
$$\Rightarrow E = \frac{a}{24} \times 8$$
$$\Rightarrow E = \frac{8a}{24}$$
$$∴ E = \frac{a}{3}$$

Thus, the correct relationship is: (C) $$a = 3E$$

Q11. Which of the following contains highest number of constituent particles-
(A) 0.36 g diamond
(B) $$2.24 \mathrm{~L} \mathrm{CO}_{2}$$ gas at STP
(C) 0.54 g ice
(D) $$5 \mathrm{~L} \mathrm{H}_{2}$$ gas at 1 atm pressure and $$27^{\circ} \mathrm{C}$$ temperature

(D) $$5 \mathrm{~L} \mathrm{H}_{2}$$ gas at 1 atm pressure and $$27^{\circ} \mathrm{C}$$ temperature

$$\begin{array}{l}\text{(A)}\ \text{0}\text{.36 g of Diamond (Pure Carbon, C):}\\\text{Molar mass of carbon:}C=12\,\text{g/mol}\\\text{Number of moles of carbon: }\\\text{=}\frac{{0.36\,\text{g}}}{{12\,\text{g/mol}}}=0.03\,\text{mol}\\\text{Number of carbon atoms:}\\\text{Number of atoms}=0.03\,\text{mol}\times 6.022\times {{10}^{{23}}}\,\text{atoms/mol}\\=1.8066\times {{10}^{{22}}}\,\text{atoms}\end{array}$$

$$\begin{array}{l}\text{(B)}\ \text{2}\text{.24L of C}{{\text{O}}_{\text{2}}}\text{ }\!\!~\!\!\text{ gas at STP:}\\\text{At STP (Standard Temperature and Pressure) 1 mole of any gas occupies 22}\text{.4L}\text{.}\\\text{Number of moles of C}{{\text{O}}_{\text{2}}}\text{:}\\\text{moles of }C{{O}_{2}}=\frac{{2.24\,\text{L}}}{{22.4\,\text{L/mol}}}=0.1\text{ mol}\\\text{Number of }\!\!~\!\!\text{ C}{{\text{O}}_{\text{2}}}\text{ }\!\!~\!\!\text{ molecules:}\\=0.1\,\text{mol}\times 6.022\times {{10}^{{23}}}\,\text{molecules/mol}\\=6.022\times {{10}^{{22}}}\,\text{molecules}\end{array}$$

$$\begin{array}{l}\text{(C)0}\text{.54 g of Ice (Water, }{{\text{H}}_{\text{2}}}\text{O):}\\\text{Molar mass of water }{{\text{H}}_{\text{2}}}\text{O=18}\,\text{g/mol}\\\text{Number of moles of water:}\\\text{moles of }{{H}_{2}}O=\frac{{0.54\,\text{g}}}{{18\,\text{g/mol}}}=0.03\,\text{mol}\\\text{Number of molecules:}\\=0.03\,\text{mol}\times 6.022\times {{10}^{{23}}}\,\text{molecules/mol}\\=1.8066\times {{10}^{{22}}}\,\text{molecules}\end{array}$$

$$\begin{array}{l}\text{(D) 5L of }{{\text{H}}_{\text{2}}}\text{ gas at 1 atm pressure and 27 }\!\!{}^\circ\!\!\text{ C temperature:}\\\text{Use the ideal gas law PV=nRT}\\P=1\,\text{atm},\\V=5\,\text{L},\\R=0.0821\,\text{L atm / (mol K)},\\T=27{}^\circ C=300\,\text{K}\\\text{Number of moles of }{{\text{H}}_{\text{2}}}\text{:}\\n=\frac{{PV}}{{RT}}=\frac{{1\times 5}}{{0.0821\times 300}}\approx 0.203\,\text{mol}\\\text{Number of molecules:}\\=0.203\,\text{mol}\times 6.022\times {{10}^{{23}}}\,\text{molecules/mol}\\\approx 1.2235\times {{10}^{{23}}}\,\text{molecules}\end{array}$$

The highest number of constituent particles is in 5L $$H_2$$ gas at 1 atm pressure and 27°C temperature.

(D) 5L $$H_2$$ gas at 1 atm pressure and 27°C temperature

Q12. The volume of 1.0 g of a gaseous substance (vapour density $$=14$$ ) at $$27^{\circ} \mathrm{C}$$ and 1 atm pressure is-
(A) 1 L
(B) 0.88 L
(C) 0.64 L
(D) 0.52 L

(B) 0.88 L

To find the volume of 1.0 g of a gaseous substance at 27°C and 1 atm pressure, we first need to determine the molar mass of the gas using its vapor density.

Vapor density (VD) is defined as the mass of a certain volume of the gas compared to the mass of the same volume of hydrogen. The formula for vapor density is:
$$\text{VD} = \frac{\text{molar mass of the gas}}{2}$$

Given that the vapor density is 14:
$$\text{molar mass of the gas} = 2 \times \text{VD} = 2 \times 14 = 28 \, \text{g/mol}$$

Now, we have 1.0 g of the gas. To find the number of moles ($$n$$):
$$n = \frac{\text{mass}}{\text{molar mass}} = \frac{1.0 \, \text{g}}{28 \, \text{g/mol}} = \frac{1}{28} \, \text{mol}$$

Next, we use the ideal gas law to find the volume ($$V$$). The ideal gas law is:
$$PV = nRT$$

Where:
– $$P$$ is the pressure (1 atm)
– $$V$$ is the volume (L)
– $$n$$ is the number of moles ($$\frac{1}{28} \, \text{mol}$$)
– $$R$$ is the ideal gas constant ($$0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}$$)
– $$T$$ is the temperature in Kelvin (27°C = 27 + 273 = 300 K)

Substitute the values into the ideal gas law:
$$\Rightarrow V = \frac{nRT}{P} = \frac{\left(\frac{1}{28}\right) \times 0.0821 \times 300}{1}$$
$$\Rightarrow V = \frac{0.0821 \times 300}{28}$$
$$\Rightarrow V = \frac{24.63}{28}$$
$$∴ V \approx 0.88 \, \text{L}$$

Thus, the volume of 1.0 g of the gaseous substance at 27°C and 1 atm pressure is:
(B) 0.88L

Q13. A metal (equivalent weight $$=65.6$$ ) has a specific heat of $$0.13 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}$$. The atomic mass of the metal is-
(A) 492.0
(B) 459.2
(C) 196.8
(D) 216.9

(C) 196.8

Specific heat in cal $$\mathrm{g}^{-1} \mathrm{~K}^{-1}$$ unit
$$\displaystyle =\frac{{0.13\times 1.987}}{{8.314}}=0.031\ \text{cal }{{\text{g}}^{{\text{-1}}}}\ {{\text{K}}^{{\text{-1}}}}$$

Approximate atomic mass of the metal:
$$= \frac{6.4}{0.031} = 206.45$$

Valency of the metal $$=\frac{206.45}{65.6} \approx 3$$
$$\therefore$$ Atomic mass of the metal $$=3 \times 65.6=196.8$$

Q14. 3.39 g of a mixture of $$\mathrm{CaCl}_{2}$$ and $$NaCl$$ is treated with excess of sodium carbonate solution and the entire amount of $$\mathrm{Ca}^{2+}$$ ions is precipitated as $$\mathrm{CaCO}_{3}$$. This $$\mathrm{CaCO}_{3}$$, on being heated strongly produces 1.12 g of $$CaO$$ . The percentage of $$NaCl$$ in the initial mixture is-
(A) 30.49
(B) 31.27
(C) 32.74
(D) 34.51

(D) 34.51

$$\begin{array}{l} \underset{100 \mathrm{~g}}{\mathrm{CaCO}_{3}} \rightarrow \underset{56 \mathrm{~g}}{\mathrm{CaO}}+\mathrm{CO}_{2} \\ 56 \mathrm{~g} \mathrm{CaO} \equiv 100 \mathrm{~g} \mathrm{CaCO}_{3} \\ \therefore 1.12 \mathrm{~g} \mathrm{CaO} \equiv 2 \mathrm{~g} \mathrm{CaCO}_{3} \\ \begin{array}{l} \mathrm{CaCl}_{2} \rightarrow \\ 111 \mathrm{~g} \\ 2.22 \mathrm{~g} \end{array} \begin{array}{c} \mathrm{CaCO}_{3} \\ 100 \mathrm{~g} \\ 2 \mathrm{~g} \end{array} \\ \end{array} \\ \therefore \text{Percentage of NaCl in the initial mixture:}\\ =\frac{3.39-2.22}{3.39} \times 100=34.51$$

Q15. A mixture of 224 L . (STP) of $$\mathrm{H}_{2}$$ gas and 44.8 L (STP) of $$\mathrm{O}_{2}$$ gas is exploded in a closed steel vessel. The amount of water produced is-
(A) 18 g
(B) 36 g
(C) 72 g
(b) 90 g

(C) 72 g

1. Balanced Chemical Equation

The reaction between hydrogen gas (H₂) and oxygen gas (O₂) to produce water (H₂O) is:

2 H₂(g) + O₂(g) → 2 H₂O(l)

2. Determine the Limiting Reactant

• Mole Ratios: From the balanced equation, 2 moles of H₂ react with 1 mole of O₂.

• Moles at STP: Remember that 1 mole of any ideal gas occupies 22.4 L at STP.

• Moles of H₂ = $$\frac{224 \ L}{22.4 \ L/mol} = 10 \ moles$$

• Moles of O₂ = $$\frac{44.8 \ L}{22.4 \ L/mol} = 2 \ moles$$

• Comparison: Since we need 2 moles of H₂ for every 1 mole of O₂, and we have an excess of H₂, oxygen (O₂) is the limiting reactant.

3. Calculate Moles of Water Produced

• From the balanced equation, 1 mole of O₂ produces 2 moles of H₂O.

• Since O₂ is the limiting reactant, 2 moles of O₂ will produce 2 moles O₂ * (2 moles H₂O / 1 mole O₂) = 4 moles of H₂O.

4. Calculate the Mass of Water

• Molar mass of H₂O = 18 g/mol

• Mass of water = Moles of water * Molar mass of water

• Mass of water = 4 moles * 18 g/mol = 72 g

Answer: The amount of water produced is 72 g (option C).

Q16. 100 mL x(\mathrm{M}) \mathrm{Ba}(\mathrm{OH})_{2} \) solution requires 500 mL of $$0.82(\mathrm{M}) \mathrm{HCl}$$ solution for its neutralisation. The amount of $$\mathrm{BaCO}_{3}$$ (molecular mass $$=197.34$$ ) that will be precipitated if excess of $$\mathrm{CO}_{2}$$ gas is passed through the above $$\mathrm{Ba}(\mathrm{OH})_{2}$$ solution is-
(A) 80.90 g
(B) 40.45 g
(C) 101.27 g
(D) 112.34 g

(C) 101.27 g

For the neutralization reaction:
$$Ba(OH)_2 + 2HCl \rightarrow BaCl_2 + 2H_2O$$
$$100 \ mL \ of \ Ba(OH)_2 = 500 \ mL \ of \ 0.82M \ HCl$$
$$Moles \ of \ HCl = 500 \times 0.82 = 410 \ mL$$
$$Moles \ of \ Ba(OH)_2 = 100 \times 2 = 200 \ mL$$
$$Moles \ of \ BaCO_3 = 100 \times 0.82 = 82 \ moles$$
The amount of $$BaCO_3$$:
$$\frac{82}{1000} \times 197.34 = 101.27g$$

Q17. 1 L of air at STP contains $$9.375 \times 10^{-3} \mathrm{~mol}$$ of $$\mathrm{O}_{2}$$. So percentage of oxygen $$(V / V)$$ in the air at STP is-
(A) 20
(B) 21
(C) 22
(D) 22.1

(B) 21

1 mole of any gas at STP occupies 22.4L.
$$\text{Volume of } 9.375 \times 10^{-3} \text{ mol of } O_2 = 9.375 \times 10^{-3} \times 22.4 = 0.21L$$
$$\text{Percentage of } O_2 \text{ (V/V) in the air} = \frac{0.21}{1} \times 100 = 21\%$$

Q18. 11.81 g of a metal nitrate when heated strongly produces 0.5 g of the corresponding metal oxide. Equivalent weight of the metal is-
(A) 24
(B) 27
(C) 31.6
(D) 34

Something wrong in the question.

Q19. At a fixed temperature and pressure, 2.5 L of a sample of $$\mathrm{NH}_{3}$$ is mixed with 1.2 L of HCl gas. The volume of the resultant mixture at the same temperature and pressure will be-
(A) 3.7 L
(B) 1.85 L
(C) 1.3 L
(D) 1.6 L

(C) 1.3 L

To solve this problem, we need to consider the chemical reaction between $$NH_3$$ and $$HCl$$ and then determine the resultant volume of the gases at the same temperature and pressure.

Step 1: Write the balanced chemical reaction

The balanced chemical reaction between ammonia ($$NH_3$$) and hydrogen chloride ($$HCl$$) is:
$$NH_3 (g) + HCl (g) \rightarrow NH_4Cl (s)$$

Step 2: Determine the limiting reactant

We start with 2.5 L of $$NH_3$$ and 1.2 L of $$HCl$$. According to the balanced equation, 1 mole of $$NH_3$$ reacts with 1 mole of $$HCl$$.

Step 3: Identify the limiting reactant

Since the volumes of gases are proportional to their moles at the same temperature and pressure, we can compare their volumes directly:
– We have 2.5 L of $$NH_3$$
– We have 1.2 L of $$HCl$$

The reaction will proceed until one of the reactants is completely consumed. The limiting reactant is $$HCl$$ because we have less volume of it.

Step 4: Determine the volume of gases after the reaction

When 1.2 L of $$HCl$$ reacts with 1.2 L of $$NH_3$$, they will produce $$NH_4Cl$$ (solid) and leave an excess of $$NH_3$$.

Remaining $$NH_3$$:
$$2.5 \, \text{L} – 1.2 \, \text{L} = 1.3 \, \text{L}$$

The $$NH_4Cl$$ produced is a solid and does not contribute to the gas volume. The volume of the resultant mixture (gas phase) will be the remaining volume of $$NH_3$$, which is 1.3 L.

Thus, the volume of the resultant mixture at the same temperature and pressure is:
(C) 1.3L

Q20. $$\mathrm{CO}_{2}$$ produced by complete decomposition of 10 g of $$\mathrm{CaCO}_{3}$$ is passed through a sodium carbonate solution containing $$1 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{CO}_{3}$$ dissolved in it. Amount of $$\mathrm{NaHCO}_{3}$$ produced will be-
(A) 8.4 g
(B) 10.2 g
(C) 12.8 g
(D) 16.8 g

(D) 16.8 g

To solve this problem, we need to follow these steps:

1. Determine the amount of $$CO_2$$ produced from the decomposition of $$CaCO_3$$.
2. Use stoichiometry to find out how much $$NaHCO_3$$ is produced when this $$CO_2$$ reacts with $$Na_2CO_3$$.

Step 1: Decomposition of $$CaCO_3$$

The reaction for the decomposition of calcium carbonate ($$CaCO_3$$) is:
$$CaCO_3 (s) \rightarrow CaO (s) + CO_2 (g)$$

We need to find the moles of $$CaCO_3$$ in 10 grams.

Molar mass of $$CaCO_3$$:
$$\text{Ca} = 40 \, \text{g/mol}$$
$$\text{C} = 12 \, \text{g/mol}$$
$$\text{O}_3 = 16 \times 3 = 48 \, \text{g/mol}$$
$$\text{Molar mass of } CaCO_3 = 40 + 12 + 48 = 100 \, \text{g/mol}$$

So, moles of $$CaCO_3$$ in 10 grams:
$$\text{Moles of } CaCO_3 = \frac{10 \, \text{g}}{100 \, \text{g/mol}} = 0.1 \, \text{mol}$$

From the decomposition reaction, 1 mole of $$CaCO_3$$ produces 1 mole of $$CO_2$$. Therefore, 0.1 moles of $$CaCO_3$$ will produce 0.1 moles of $$CO_2$$.

Step 2: Reaction of $$CO_2$$ with $$Na_2CO_3$$

The reaction between $$CO_2$$ and sodium carbonate ($$Na_2CO_3$$) to produce sodium bicarbonate ($$NaHCO_3$$) is:
$$CO_2 (g) + Na_2CO_3 (aq) + H_2O (l) \rightarrow 2 NaHCO_3 (aq)$$

From the balanced equation, 1 mole of $$CO_2$$ reacts with 1 mole of $$Na_2CO_3$$ to produce 2 moles of $$NaHCO_3$$.

Since we have 0.1 moles of $$CO_2$$, it will react with 0.1 moles of $$Na_2CO_3$$ to produce:
$$\text{Moles of } NaHCO_3 = 2 \times 0.1 = 0.2 \, \text{moles}$$

Molar mass of $$NaHCO_3$$:
$$\text{Na} = 23 \, \text{g/mol}$$
$$\text{H} = 1 \, \text{g/mol}$$
$$\text{C} = 12 \, \text{g/mol}$$
$$\text{O}_3 = 16 \times 3 = 48 \, \text{g/mol}$$
$$\text{Molar mass of } NaHCO_3 = 23 + 1 + 12 + 48 = 84 \, \text{g/mol}$$

Therefore, the mass of $$NaHCO_3$$ produced:
$$\text{Mass of } NaHCO_3 = 0.2 \, \text{moles} \times 84 \, \text{g/mol} = 16.8 \, \text{g}$$

Thus, the amount of $$NaHCO_3$$ produced is:
(D) 16.8g

Q21. Under a given set of condition 20 L dihydrogen and 15 L dioxygen are mixed together and allowed to react with each other to produce $$\mathrm{H}_{2} \mathrm{O}$$ (steam). If the extent of reaction is only $$50 \%$$ of the theoretical value, the total volume of gas mixture obtained at the end of the reaction is-
(A) 15 L
(B) 20 L
(C) 25 L
(D) 30 L

(D) 30 L

To solve this problem, let’s start with the balanced chemical reaction between dihydrogen ($$H_2$$) and dioxygen ($$O_2$$) to form water vapor ($$H_2O$$):

$$2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$$

Step 1: Determine the theoretical reaction volumes

From the balanced equation:
– 2 volumes of $$H_2$$ react with 1 volume of $$O_2$$ to produce 2 volumes of $$H_2O$$.

Given:
– 20 L of $$H_2$$
– 15 L of $$O_2$$

According to the stoichiometric ratio, to completely react with 15 L of $$O_2$$, we would need:
$$2 \times 15 \, \text{L} = 30 \, \text{L} \, \text{of} \, H_2$$

Since we only have 20 L of $$H_2$$, $$H_2$$ is the limiting reactant.

Step 2: Theoretical reaction considering limiting reactant

20 L of $$H_2$$ would theoretically react with:
$$\frac{20 \, \text{L} \, H_2}{2} = 10 \, \text{L} \, O_2$$

This theoretical reaction would produce:
$$20 \, \text{L} \, H_2O$$

Step 3: Extent of reaction is only 50%

If the extent of the reaction is only 50%, then only 50% of the limiting reactant $$H_2$$ will react:
$$20 \, \text{L} \times 0.5 = 10 \, \text{L} \, H_2 \text{ will react}$$

This will consume:
$$\frac{10 \, \text{L} \, H_2}{2} = 5 \, \text{L} \, O_2$$

And produce:
$$10 \, \text{L} \, H_2O$$

Step 4: Remaining volumes of gases

– Initial volumes: $$H_2 = 20 \, \text{L}, O_2 = 15 \, \text{L}$$
– Volumes reacted: $$H_2 = 10 \, \text{L}, O_2 = 5 \, \text{L}$$
– Volumes remaining: $$H_2 = 20 – 10 = 10 \, \text{L}$$, $$O_2 = 15 – 5 = 10 \, \text{L}$$

Step 5: Total volume of gas mixture at the end

The total volume of gas mixture obtained at the end of the reaction includes the remaining unreacted gases and the product:
$$10 \, \text{L} \, H_2 + 10 \, \text{L} \, O_2 + 10 \, \text{L} \, H_2O = 30 \, \text{L}$$

Thus, the total volume of the gas mixture at the end of the reaction is:
(D) 30L

Q22. 250 mL of an aqueous solution of oxalic acid requires 1.6 g NaOH for its complete neutralisation. Molar strength of the acid solution is-
(A) 0.05
(B) 0.07
(C) 0.08
(D) 0.1

(C) 0.08

1. Balanced Chemical Equation

Oxalic acid (H₂C₂O₄) is a diprotic acid, meaning it has two acidic protons. The balanced neutralization reaction with sodium hydroxide (NaOH) is:

H₂C₂O₄(aq) + 2 NaOH(aq) → Na₂C₂O₄(aq) + 2 H₂O(l)

2. Moles of NaOH

• Molar mass of NaOH = 40 g/mol

• Moles of NaOH = $$\frac{ \text{mass of NaOH}}{ \text{molar mass of NaOH}}$$

• Moles of NaOH = $$\frac{{1.6\ \text{g}}}{{40\ \text{g/mol}}}\text{ }\!\!~\!\!\text{ }=0.04\ \text{mol}$$

3. Moles of Oxalic Acid

• From the balanced equation, 1 mole of H₂C₂O₄ reacts with 2 moles of NaOH.

• Moles of H₂C₂O₄ = $$\frac{ \text{moles of NaOH}}{2}$$

• Moles of H₂C₂O₄ = $$\frac{0.04 \text{ mol}}{2} = 0.02 \ mol$$

4. Molar Strength (Molarity)

• Molarity (M) = $$\frac{ \text{moles of solute}}{volume of solution in liters}$$

• Volume of solution = 250 mL = 0.25 L

• Molarity of H₂C₂O₄ = $$\frac{0.02 \ mol}{0.25 \ L} = 0.08 \ M$$

Answer: The molar strength of the oxalic acid solution is 0.08 M (option C).

Q23. $$100 \mathrm{~mL} 0.1(\mathrm{M})$$ solution of oxalic acid reacts completely with 20 mL of a solution of $$\mathrm{KMnO}_{4}$$. Molarity of $$\mathrm{KMnO}_{4}$$ solution $$\left(\mathrm{mol} \mathrm{L}^{-1}\right.$$ ) is-
(A) 0.1
(B) 0.15
(C) 0.18
(D) 0.2

(D) 0.2

To solve this problem, we need to consider the balanced chemical reaction between oxalic acid ($$\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4$$) and potassium permanganate ($$\mathrm{KMnO}_4$$) in acidic medium. The balanced equation for the reaction is:

$$2 \mathrm{KMnO}_4 + 5 \mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 + 6 \mathrm{H}_2\mathrm{SO}_4 \rightarrow 2 \mathrm{MnSO}_4 + 10 \mathrm{CO}_2 + 8 \mathrm{H}_2\mathrm{O} + K_2\mathrm{SO}_4$$

From the equation, we see that 2 moles of $$\mathrm{KMnO}_4$$ react with 5 moles of oxalic acid.

Step 1: Calculate moles of oxalic acid

Given:
– Volume of oxalic acid solution = 100 mL = 0.1 L
– Molarity of oxalic acid solution = 0.1 M

Moles of oxalic acid:
$$\text{Moles of oxalic acid} = \text{Molarity} \times \text{Volume (in L)}$$
$$\text{Moles of oxalic acid} = 0.1 \, \text{M} \times 0.1 \, \text{L}$$
$$\text{Moles of oxalic acid} = 0.01 \, \text{moles}$$

Step 2: Calculate moles of $$\mathrm{KMnO}_4$$ required

From the balanced equation:
– 2 moles of $$\mathrm{KMnO}_4$$ react with 5 moles of oxalic acid.

Moles of $$\mathrm{KMnO}_4$$ required for 0.01 moles of oxalic acid:
$$\text{Moles of } \mathrm{KMnO}_4 = \frac{2}{5} \times 0.01$$
$$\text{Moles of } \mathrm{KMnO}_4 = 0.004 \, \text{moles}$$

Step 3: Calculate the molarity of the $$\mathrm{KMnO}_4$$ solution

Given:
– Volume of $$\mathrm{KMnO}_4$$ solution = 20 mL = 0.02 L

Molarity of $$\mathrm{KMnO}_4$$ solution:
$$\text{Molarity} = \frac{\text{Moles}}{\text{Volume (in L)}}$$
$$\text{Molarity} = \frac{0.004 \, \text{moles}}{0.02 \, \text{L}}$$
$$\text{Molarity} = 0.2 \, \text{M}$$

Thus, the molarity of the $$\mathrm{KMnO}_4$$ solution is:
(D) 0.2

Q24. 0.38 of an organic compound $$(M=60)$$ containing $$\mathrm{C}, \mathrm{H}$$ and $$O$$ on combustion produces $$0.44 \mathrm{~g} \mathrm{CO}_{2}$$ and $$0.18 \ g$$  $$\mathrm{H}_{2} \mathrm{O}$$. The molecular formula of the compound is-
(A) $$\mathrm{CH}_{2} \mathrm{O}$$
(B) $$\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}$$
(C) $$\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}$$
(D) $$\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$$

Somothing wrong.

Q25. 0.05 mol of a mixture of $$BaO$$  and  $$CaO$$ requires 50 mL . of an aqueous solution of  $$HCl$$ for its complete neutralisation. The molar strength of the acid solution is-
(A) 1 (M)
(B) 2 (M)
(C) 2.5 (M)
(D) 3.0 (M)

(B) 2 (M)

To solve this problem, we need to consider the neutralization reactions of $$\text{BaO}$$ and $$\text{CaO}$$ with $$\text{HCl}$$. Both reactions will neutralize $$\text{HCl}$$ in a 1:2 molar ratio because both $$\text{BaO}$$ and $$\text{CaO}$$ are basic oxides and react with $$\text{HCl}$$ to form their respective chlorides and water.

The balanced equations are:

$$\text{BaO} + 2\text{HCl} \rightarrow \text{BaCl}_2 + \text{H}_2\text{O}$$
$$\text{CaO} + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O}$$

Step 1: Determine the total moles of $$\text{HCl}$$ required for neutralization

Given:
– 0.05 mol of a mixture of $$\text{BaO}$$ and $$\text{CaO}$$
– Total moles of $$\text{BaO}$$ and $$\text{CaO}$$ combined = 0.05 mol

Each mole of $$\text{BaO}$$ or $$\text{CaO}$$ requires 2 moles of $$\text{HCl}$$ for neutralization. Therefore, 0.05 mol of the mixture will require:
$$0.05 \text{ mol} \times 2 = 0.1 \text{ mol of } \text{HCl}$$

Step 2: Determine the molarity of the $$\text{HCl}$$ solution

Given:
– Volume of $$\text{HCl}$$ solution = 50 mL = 0.05 L

We need to find the molarity (M) of the $$\text{HCl}$$ solution using the moles of $$\text{HCl}$$ and the volume of the solution.

$$\text{Molarity} = \frac{\text{Moles of } \text{HCl}}{\text{Volume of solution in L}}$$
$$\text{Molarity} = \frac{0.1 \text{ mol}}{0.05 \text{ L}}$$
$$\text{Molarity} = 2 \text{ M}$$

Therefore, the molarity of the $$\text{HCl}$$ solution is:
(B) 2 (M)

Q26. In order to prepare 250 mL 2.0 (M) $$\mathrm{HNO}_{3}$$ solution, 45 g of a sample of conc. $$\mathrm{HNO}_{3}$$ is required. Percentage of $$\mathrm{HNO}_{3}(\mathrm{~W} / \mathrm{W})$$ in the sample of concentrated nitric acid is-
(A) 55
(B) 65
(C) 70
(D) 75

(C) 70

To solve this problem, we need to determine the percentage of $$\mathrm{HNO}_3$$ (weight/weight) in the sample of concentrated nitric acid. The steps are as follows:

Step 1: Calculate the number of moles of $$\mathrm{HNO}_3$$ required

Given:
– Desired molarity of the $$\mathrm{HNO}_3$$ solution = 2.0 M
– Volume of the $$\mathrm{HNO}_3$$ solution = 250 mL = 0.25 L

The number of moles of $$\mathrm{HNO}_3$$ required is:
$$\text{Moles of } \mathrm{HNO}_3 = \text{Molarity} \times \text{Volume (in L)}$$
$$\text{Moles of } \mathrm{HNO}_3 = 2.0 \, \text{M} \times 0.25 \, \text{L}$$
$$\text{Moles of } \mathrm{HNO}_3 = 0.5 \, \text{moles}$$

Step 2: Calculate the mass of $$\mathrm{HNO}_3$$ required

The molar mass of $$\mathrm{HNO}_3$$ is approximately:
$$\text{Molar mass of } \mathrm{HNO}_3 = 1 \, (\text{H}) + 14 \, (\text{N}) + 3 \times 16 \, (\text{O})$$
$$\text{Molar mass of } \mathrm{HNO}_3 = 1 + 14 + 48$$
$$\text{Molar mass of } \mathrm{HNO}_3 = 63 \, \text{g/mol}$$

The mass of $$\mathrm{HNO}_3$$ required is:
$$\text{Mass of } \mathrm{HNO}_3 = \text{Moles} \times \text{Molar mass}$$
$$\text{Mass of } \mathrm{HNO}_3 = 0.5 \, \text{moles} \times 63 \, \text{g/mol}$$
$$\text{Mass of } \mathrm{HNO}_3 = 31.5 \, \text{g}$$

Step 3: Determine the percentage of $$\mathrm{HNO}_3$$ in the sample

Given:
– Mass of the sample of concentrated $$\mathrm{HNO}_3$$ required = 45 g
– This sample provides 31.5 g of $$\mathrm{HNO}_3$$

The percentage of $$\mathrm{HNO}_3$$ in the sample is:
$$\text{Percentage of } \mathrm{HNO}_3 (\text{W/W}) = \left( \frac{\text{Mass of } \mathrm{HNO}_3}{\text{Total mass of sample}} \right) \times 100$$
$$\text{Percentage of } \mathrm{HNO}_3 = \left( \frac{31.5 \, \text{g}}{45 \, \text{g}} \right) \times 100$$
$$\text{Percentage of } \mathrm{HNO}_3 = \left( \frac{31.5}{45} \right) \times 100$$
$$\text{Percentage of } \mathrm{HNO}_3 = 0.7 \times 100$$
$$\text{Percentage of } \mathrm{HNO}_3 = 70\%$$

Therefore, the percentage of $$\mathrm{HNO}_3$$ in the sample of concentrated nitric acid is:
(C) 70

Q27. Mole fraction of water in an aqueous solution of glucose is 0.9 . Molality of the solution is-
(A) $$4.38 \mathrm{~mol} \mathrm{~kg}^{-1}$$
(B) $$6.17 \mathrm{~mol} \mathrm{~kg}^{-1}$$
(C) $$5.06 \mathrm{~mol} \mathrm{~kg}^{-1}$$
(D) $$5.29 \mathrm{~mol} \mathrm{~kg}^{-1}$$

(B) $$6.17 \mathrm{~mol} \mathrm{~kg}^{-1}$$

To find the molality of the solution, we first need to understand the relationship between mole fraction and molality.

The mole fraction of water ($$\chi_{\text{H}_2\text{O}}$$) is given as 0.9. Therefore, the mole fraction of glucose ($$\chi_{\text{glucose}}$$) will be:
$$\chi_{\text{glucose}} = 1 – \chi_{\text{H}_2\text{O}} = 1 – 0.9 = 0.1$$

Let’s assume we have 1 mole of solution. This means:
– Moles of water ($$n_{\text{H}_2\text{O}}$$) = 0.9
– Moles of glucose ($$n_{\text{glucose}}$$) = 0.1

To find molality, we need the moles of solute (glucose) per kilogram of solvent (water).

1. Calculate the mass of water:
The molar mass of water ($$\text{H}_2\text{O}$$) is approximately 18 g/mol.

$$\text{Mass of water} = n_{\text{H}_2\text{O}} \times \text{molar mass of water} = 0.9 \times 18 \, \text{g} = 16.2 \, \text{g} = 0.0162 \, \text{kg}$$

2. Calculate the molality (m):
$$\text{Molality} (m) = \frac{n_{\text{glucose}}}{\text{mass of water in kg}} = \frac{0.1}{0.0162 \, \text{kg}} \approx 6.17 \, \text{mol/kg}$$

So, the molality of the solution is:
$$\boxed{6.17 \, \text{mol/kg}}$$

The correct answer is (B) $$6.17 \mathrm{~mol} \mathrm{~kg}^{-1}$$.

Q28. The molarity and molality of an aqueous solution of urea are $$2.05 \mathrm{~mol} \mathrm{~L}^{-1}$$  and  $$2.0 \mathrm{~mol} \mathrm{~kg}^{-1}$$ respectively. The density of the solution is-
(A) 1.00
(B) 1.09
(C) 1.15
(D) 1.21

(C) 1.15

1. Given Data:
– Molarity ($$M$$) = $$2.05 \mathrm{~mol} \mathrm{~L}^{-1}$$
– Molality ($$m$$) = $$2.0 \mathrm{~mol} \mathrm{~kg}^{-1}$$
– Molar mass of urea ($$\text{NH}_2\text{CONH}_2$$) = 60 g/mol

2. Definitions:
– Molarity ($$M$$): Moles of solute per liter of solution.
– Molality ($$m$$): Moles of solute per kilogram of solvent.

3. Assume 1 liter of solution:
– Moles of urea in 1 liter of solution = $$2.05 \mathrm{~mol}$$
– Mass of urea in 1 liter of solution = $$2.05 \mathrm{~mol} \times 60 \mathrm{~g/mol} = 123 \mathrm{~g} = 0.123 \mathrm{~kg}$$

4. Molality relation:
– Molality ($$m$$) = $$\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$$
– Given $$m = 2.0 \mathrm{~mol} \mathrm{~kg}^{-1}$$, and we have $$2.05 \mathrm{~mol}$$ of urea,
– Mass of solvent (water) in kg = $$\frac{\text{Moles of solute}}{m} = \frac{2.05 \mathrm{~mol}}{2.0 \mathrm{~mol} \mathrm{~kg}^{-1}} = 1.025 \mathrm{~kg}$$

5. Total mass of the solution:
– Mass of solution = Mass of solute (urea) + Mass of solvent (water)
– Mass of solution = $$0.123 \mathrm{~kg} + 1.025 \mathrm{~kg} = 1.148 \mathrm{~kg}$$

6. Density of the solution:
– Density ($$\rho$$) = $$\frac{\text{Mass of solution}}{\text{Volume of solution}} = \frac{1.148 \mathrm{~kg}}{1 \mathrm{~L}} = 1.148 \mathrm{~kg/L} \approx 1.15 \mathrm{~g/mL}$$

So, the density of the solution is:
$$\boxed{1.15 \mathrm{~g/mL}}$$

The correct answer is (C) $$1.15$$.

Q29. The density of pure ethyl alcohol is $$1.15 \mathrm{~g} \mathrm{~mL}^{-1}$$. In order of prepare 100 mL of $$0.5(\mathrm{M})$$ aqueous solution of ethyl alcohol, what volume of water be mixed with requisite amount of pure ethyl alcohol-
(A) 100 mL
(B) 98 mL
(C) 95 mL
(D) 90 mL

(B) 98 mL

To prepare a 100 mL of 0.5 M aqueous solution of ethyl alcohol, we need to determine the volume of pure ethyl alcohol required and then calculate the volume of water to be mixed with it.

1. Molarity (M) definition:
$$M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}$$

2. Calculate moles of ethyl alcohol needed:
Given:
$$M = 0.5 \mathrm{~mol/L}$$
$$\text{Volume of solution} = 100 \mathrm{~mL} = 0.1 \mathrm{~L}$$

$$\text{Moles of ethyl alcohol} = M \times \text{Volume} = 0.5 \mathrm{~mol/L} \times 0.1 \mathrm{~L} = 0.05 \mathrm{~mol}$$

3. Molar mass of ethyl alcohol ($$\text{C}_2\text{H}_5\text{OH}$$):
$$\text{Molar mass} = 2 \times 12 \, (\text{C}) + 6 \times 1 \, (\text{H}) + 16 \, (\text{O}) = 24 + 6 + 16 = 46 \mathrm{~g/mol}$$

4. Calculate the mass of ethyl alcohol:
$$\text{Mass of ethyl alcohol} = \text{moles} \times \text{molar mass} = 0.05 \mathrm{~mol} \times 46 \mathrm{~g/mol} = 2.3 \mathrm{~g}$$

5. Calculate the volume of ethyl alcohol:
Given the density of pure ethyl alcohol is $$1.15 \mathrm{~g/mL}$$,
$$\text{Volume of ethyl alcohol} = \frac{\text{mass}}{\text{density}} = \frac{2.3 \mathrm{~g}}{1.15 \mathrm{~g/mL}} = 2 \mathrm{~mL}$$

6. Calculate the volume of water to be mixed:
To prepare 100 mL of the solution,
$$\text{Volume of water} = \text{Total volume} – \text{Volume of ethyl alcohol} = 100 \mathrm{~mL} – 2 \mathrm{~mL} = 98 \mathrm{~mL}$$

Therefore, the volume of water to be mixed with the requisite amount of pure ethyl alcohol is:
$$\boxed{98 \mathrm{~mL}}$$

The correct answer is (B) 98 mL.

Q30. ‘ $$A$$ ‘ and ‘ $$B$$ ‘ are two different aqueous solutions of urea having percentage strength $$(W / W)$$ of 10 and 20 respectively. If 100 g of solution ‘ $$A$$ ‘ be mixed with 200 g of solution ‘ $$B$$ ‘, the percentage strength $$(W / W)$$ of the resulting mixture will be-
(A) 12.34
(B) 14.39
(C) 16.66
(D) 18.17

(C) 16.66

To determine the percentage strength (w/w) of the resulting mixture, we need to find the total mass of urea and the total mass of the solution after mixing.

1. Calculate the mass of urea in each solution:
– Solution $$A$$ has 10% urea by mass.
$$\text{Mass of urea in solution } A = 10\% \times 100 \text{ g} = 0.10 \times 100 \text{ g} = 10 \text{ g}$$

– Solution $$B$$ has 20% urea by mass.
$$\text{Mass of urea in solution } B = 20\% \times 200 \text{ g} = 0.20 \times 200 \text{ g} = 40 \text{ g}$$

2. Calculate the total mass of urea in the mixture:
$$\text{Total mass of urea} = \text{Mass of urea in } A + \text{Mass of urea in } B = 10 \text{ g} + 40 \text{ g} = 50 \text{ g}$$

3. Calculate the total mass of the mixture:
$$\text{Total mass of mixture} = \text{Mass of solution } A + \text{Mass of solution } B = 100 \text{ g} + 200 \text{ g} = 300 \text{ g}$$

4. Calculate the percentage strength (w/w) of the resulting mixture:
$\text{Percentage strength (w/w)} = \left( \frac{\text{Total mass of urea}}{\text{Total mass of mixture}} \right) \times 100\% = \left( \frac{50 \text{ g}}{300 \text{ g}} \right) \times 100\% = \frac{50}{300} \times 100\% = \frac{1}{6} \times 100\% \approx 16.66\%$

Therefore, the percentage strength (w/w) of the resulting mixture is:
$$\boxed{16.66}$$

The correct answer is (C) 16.66.

Q31. A and B are two different oxides of the metal M. Each gram of the oxide A contains 0.888 g metal and that of the oxide B contains 0.798 g metal. If the formula of the oxide A is $$\mathrm{M}_{2} \mathrm{O}$$, the formula of the oxide B will be-
(A) $$\mathrm{MO}_{2}$$
(B) MO
(C) $$\mathrm{M}_{2} \mathrm{O}_{3}$$
(D) $$\mathrm{M}_{2} \mathrm{O}$$

(B) MO

To find the formula of oxide B, we can use the given information about the mass of the metal in each oxide and the formula of oxide A.

1. Oxide A ($$\mathrm{M}_2\mathrm{O}$$) Analysis:
– Each gram of oxide A contains 0.888 g of metal M.
– The remaining mass in 1 g of oxide A is oxygen:
$$\text{Mass of oxygen in 1 g of oxide A} = 1 – 0.888 = 0.112 \text{ g}$$

2. Determine the molar mass ratio for oxide A:
– Molar mass of metal M is $$M$$.
– Molar mass of oxygen is 16 g/mol.
– Formula of oxide A is $$\mathrm{M}_2\mathrm{O}$$, so its molar mass is $$2M + 16$$.

For 1 g of $$\mathrm{M}_2\mathrm{O}$$:
$$\text{Mass of metal M in 1 g of oxide A} = 0.888 \text{ g}$$
$$\text{Mass of oxygen in 1 g of oxide A} = 0.112 \text{ g}$$

The ratio of masses in oxide A (M to O) is:
$$\frac{0.888}{2M} = \frac{0.112}{16}$$

Solving for $$M$$:
$$\frac{0.888}{2M} = \frac{0.112}{16}$$
$$\frac{0.888}{2M} = 0.007$$
$$0.888 = 0.014M$$
$$M = \frac{0.888}{0.014}$$
$$M = 63.43 \text{ g/mol}$$

3. Oxide B Analysis:
– Each gram of oxide B contains 0.798 g of metal M.
– The remaining mass in 1 g of oxide B is oxygen:
$$\text{Mass of oxygen in 1 g of oxide B} = 1 – 0.798 = 0.202 \text{ g}$$

4. Determine the formula for oxide B:
– Let the formula of oxide B be $$\mathrm{M}_x\mathrm{O}_y$$.
– The molar masses of M and O must fit the ratio of masses provided.

The ratio of metal to oxygen in oxide B is:
$$\frac{0.798}{xM} = \frac{0.202}{y \cdot 16}$$
Substitute $$M = 63.43 \text{ g/mol}$$:
$$\frac{0.798}{x \cdot 63.43} = \frac{0.202}{y \cdot 16}$$

Simplify:
$$\frac{0.798}{63.43x} = \frac{0.202}{16y}$$
$$\Rightarrow \frac{0.798}{63.43x} = \frac{0.202}{16y}$$
$$\Rightarrow \frac{0.798}{0.202} = \frac{63.43x}{16y}$$
$$\Rightarrow 3.95 = \frac{63.43x}{16y}$$
$$\Rightarrow 3.95 \cdot 16y = 63.43x$$
$$\Rightarrow 63.2y = 63.43x$$
$$∴ y \approx x$$

Since the mass ratio matches that of $$\mathrm{MO}$$:
$$x = y = 1$$

So the formula of oxide B is:
$$\mathrm{MO}$$

The correct answer is (B) MO.

Q32. The molecules of a compound contains $$24 \%$$ of sulphur as one of the elements. If each molecule of the compound contains three sulphur atoms, the molar mass of the compound will be-
(A) 240
(B) 320
(C) 360
(D) 400

(D) 400

To determine the molar mass of the compound, we can use the given percentage of sulfur and the fact that each molecule contains three sulfur atoms.

1. Given data:
– The compound contains $$24\%$$ sulfur by mass.
– Each molecule contains three sulfur atoms.

2. Calculate the mass of sulfur in the compound:
– Let the molar mass of the compound be $$M$$.
– The molar mass of sulfur ($$\text{S}$$) is $$32 \mathrm{~g/mol}$$.

Since each molecule contains three sulfur atoms, the total mass of sulfur in one mole of the compound is:

$\text{Mass of sulfur in one mole of the compound} = 3 \times 32 \mathrm{~g} = 96 \mathrm{~g}$

3. Determine the percentage of sulfur:
The percentage of sulfur by mass in the compound is given as $$24\%$$. This means:
$$\frac{96 \mathrm{~g}}{M} \times 100\% = 24\%$$

4. Solve for the molar mass $$M$$:
$$\frac{96}{M} \times 100 = 24 \\ \Rightarrow \frac{96 \times 100}{24} = M \\ \Rightarrow M = \frac{9600}{24} \\ \Rightarrow M = 400 \mathrm{~g/mol}$$

Thus, the molar mass of the compound is:
$$\boxed{400 \mathrm{~g/mol}}$$

The correct answer is (D) 400.

Q33. The empirical formula of a compound is $$\mathrm{CH}_{2} \mathrm{O}$$. Its vapour density is 2.813 times that of oxygen. Molecular formula of the compound is-
(A) $$\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{4}$$
(B) $$\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}$$
(C) $$\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$$
(D) $$\mathrm{CH}_{2} \mathrm{O}$$

(B) $$\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}$$

To determine the molecular formula, we need to find the molecular mass of the compound. Given that its empirical formula is $$\mathrm{CH}_2 \mathrm{O}$$, we first calculate the molar mass of the empirical formula.

The molar mass of $$\mathrm{CH}_2 \mathrm{O}$$ is:
$$\text{C} = 12 \, \text{g/mol}$$
$$\text{H}_2 = 2 \times 1 \, \text{g/mol} = 2 \, \text{g/mol}$$
$$\text{O} = 16 \, \text{g/mol}$$
$$\text{Total molar mass of} \ \mathrm{CH}_2 \mathrm{O} = 12 + 2 + 16 = 30 \, \text{g/mol}$$

The vapor density (VD) is given as 2.813 times that of oxygen. The molar mass of oxygen (O₂) is 32 g/mol. Vapor density is defined as half of the molar mass of the gas:
$$\text{VD} = \frac{\text{Molecular mass}}{2}$$

So, if the vapor density is 2.813 times that of oxygen, we can find the molecular mass (M) of the compound:
$$\text{VD} = 2.813 \times 16 = 45.008$$
$$\text{Molecular mass} = 2 \times 45.008 = 90.016 \, \text{g/mol}$$

The molecular formula is a whole number multiple (n) of the empirical formula:
$$n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{90.016}{30} = 3$$

Thus, the molecular formula is:
$$(\mathrm{CH}_2 \mathrm{O})_3 = \mathrm{C}_3 \mathrm{H}_6 \mathrm{O}_3$$

Therefore, the correct answer is (B) $$\mathrm{C}_3 \mathrm{H}_6 \mathrm{O}_3$$.

Q34. 100 mL . of a mixture of CO and $$\mathrm{CH}_{4}$$ requires 80 mL of $$\mathrm{O}_{2}$$ for its complete oxidation (all the volumes are measured at same temperature and pressure). The percentage of $$\mathrm{CH}_{4}(\mathrm{~V} / \mathrm{V}$$ ) in the gas mixture is-
(A) 20
(B) 40
(C) 60
(D) 80

(A) 20

To determine the percentage of $$\mathrm{CH}_4$$ in the gas mixture, we need to analyze the reactions involved and the stoichiometry.

1. The oxidation reaction of carbon monoxide ($$\mathrm{CO}$$):
$$\mathrm{2CO} + \mathrm{O_2} \rightarrow \mathrm{2CO_2}$$
This indicates that 2 volumes of $$\mathrm{CO}$$ require 1 volume of $$\mathrm{O_2}$$ for complete combustion.

2. The oxidation reaction of methane ($$\mathrm{CH_4}$$):
$$\mathrm{CH_4} + 2\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\mathrm{H_2O}$$
This indicates that 1 volume of $$\mathrm{CH_4}$$ requires 2 volumes of $$\mathrm{O_2}$$ for complete combustion.

Let $$x$$ be the volume of $$\mathrm{CH_4}$$ in the 100 mL mixture, and $$(100 – x)$$ be the volume of $$\mathrm{CO}$$.

From the reactions:
– The volume of $$\mathrm{O_2}$$ required for $$x$$ mL of $$\mathrm{CH_4}$$ is $$2x$$ mL.
– The volume of $$\mathrm{O_2}$$ required for $$(100 – x)$$ mL of $$\mathrm{CO}$$ is $$\frac{(100 – x)}{2}$$ mL.

The total volume of $$\mathrm{O_2}$$ required is 80 mL, so:
$$2x + \frac{(100 – x)}{2} = 80$$

Solving this equation:
$$2x + 50 – \frac{x}{2} = 80$$
$$\Rightarrow \frac{4x – x}{2} = 30$$
$$\Rightarrow 3x = 60$$
$$∴ x = 20$$

So, the volume of $$\mathrm{CH_4}$$ in the mixture is 20 mL.

The percentage of $$\mathrm{CH_4}$$ in the mixture is:
$$\frac{20}{100} \times 100\% = 20\%$$

Therefore, the correct answer is (A) 20.

Q35. 90 mL of a mixture of nitrogen and nitric oxide is passed over heated copper and the volume of the resulting gas reduces to 60 mL . (All the volumes are measured at same temperatue and pressure). The percentage of nitric oxide $$(V / V)$$ in the initial gas mixture is-
(A) 58.32
(B) 66.66
(C) 68.39
(D) 72.05

(B) 66.66

To determine the percentage of nitric oxide ($$\mathrm{NO}$$) in the initial gas mixture, we need to understand the reaction between nitric oxide and heated copper.

The relevant chemical reaction is:
$$\mathrm{2NO} + \mathrm{2Cu} \rightarrow \mathrm{N_2} + \mathrm{2CuO}$$

From this reaction, it is evident that 2 volumes of $$\mathrm{NO}$$ react to produce 1 volume of $$\mathrm{N_2}$$, with the remaining nitrogen ($$\mathrm{N_2}$$) being inert in this process.

Let’s denote the volume of $$\mathrm{NO}$$ in the initial mixture as $$x$$ mL and the volume of $$\mathrm{N_2}$$ as $$(90 – x)$$ mL.

When $$x$$ mL of $$\mathrm{NO}$$ reacts:
– The volume of gas reduced is $$x$$ mL, since $$x$$ mL of $$\mathrm{NO}$$ produces $$\frac{x}{2}$$ mL of $$\mathrm{N_2}$$, which effectively means a reduction in volume of $$x – \frac{x}{2} = \frac{x}{2}$$ mL.
– The initial nitrogen ($$90 – x$$ mL) remains unchanged.

According to the problem, the total volume of the resulting gas after passing over heated copper is 60 mL:
$$(90 – x) + \frac{x}{2} = 60$$

Solving for $$x$$:
$$90 – \frac{x}{2} = 60$$
$$\Rightarrow 90 – 60 = \frac{x}{2}$$
$$\Rightarrow 30 = \frac{x}{2}$$
$$∴ x = 60$$

So, the volume of $$\mathrm{NO}$$ in the initial mixture is 60 mL.

The percentage of $$\mathrm{NO}$$ in the initial mixture is:
$$\frac{60}{90} \times 100\% = \frac{2}{3} \times 100\% = 66.66\%$$

Therefore, the correct answer is (B) 66.66.

Q36. 100 mL of $$\mathrm{CO}_{2}$$ is passed over red hot coke. The volume of the resulting gas mixture reduces to 60 mL . (at same temperature and pressure). Percentage of $$\mathrm{CO}_{2}(\mathrm{~V} / \mathrm{V})$$ in the resulting gas mixture is-
(A) 35
(B) 75
(C) 25
(D) 65

Something wrong

Q37. 50 mL . of propane gas is exploded with excess of oxygen. On cooling the volume of the gas mixture becomes 170 mL . Volume of oxygen present in the initial gas mixture is-
(A) 250 mL
(B) 260 mL
(C) 268 mL
(D) 270 mL

(D) 270 mL

1. Balanced Chemical Equation

The combustion of propane (C₃H₈) with oxygen (O₂) produces carbon dioxide (CO₂) and water (H₂O):

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

2. Use the Volume Ratios

• From the balanced equation, 1 volume of C₃H₈ reacts with 5 volumes of O₂ to produce 3 volumes of CO₂.

• Since the volume of a gas is directly proportional to moles at constant temperature and pressure, we can use these volume ratios directly.

3. Calculate Oxygen Consumed and Remaining

• 50 mL of C₃H₈ will react with 5 × 50 mL = 250 mL of O₂.

• 50 mL of C₃H₈ will produce 3 × 50 mL = 150 mL of CO₂.

• Since the final volume is 170 mL and 150 mL of that is CO₂, the remaining 20 mL must be unreacted O₂ (170 mL – 150 mL = 20 mL).

4. Calculate the Initial Volume of Oxygen

• The initial volume of O₂ is the amount consumed plus the amount remaining: 250 mL + 20 mL = 270 mL

Answer: The volume of oxygen present in the initial gas mixture is 270 mL (option D).

38. 1 mol of a gaseous hydrocarbon $$\left(\mathrm{C}_{x} \mathrm{H}_{y}\right)$$ requires 6 mol of $$\mathrm{O}_{2}$$ for its complete combustion and produces 4 mol $$\mathrm{CO}_{2}$$. Thus the values of $$x$$ and $$y$$ are-
(A) $$x=2, y=4$$
(B) $$x=4, y=8$$
(C) $$x=3, y=6$$
(D) $$x=2, y=2$$

(B) $$x=4, y=8$$

1. Write the Balanced Chemical Equation

The general combustion reaction for a hydrocarbon is:

CₓHᵧ + zO₂ → xCO₂ + (y/2)H₂O

We need to find the values of x, y, and z.

2. Use the Given Mole Ratios

• 1 mol CₓHᵧ : 6 mol O₂ : 4 mol CO₂

• From CO₂: The ratio tells us that 1 mole of the hydrocarbon produces 4 moles of CO₂. This means x (the number of carbon atoms in the hydrocarbon) must be 4.

• From O₂: We now know the hydrocarbon is C₄Hᵧ. Let’s rewrite the balanced equation:

C₄Hᵧ + zO₂ → 4CO₂ + (y/2)H₂O

To produce 4 moles of CO₂, we need 4 moles of oxygen from the O₂. The remaining oxygen (6 moles total – 4 moles for CO₂ = 2 moles) must be used to form water.

• From H₂O: 2 moles of oxygen atoms are used to form water. Since each water molecule (H₂O) contains one oxygen atom, this means 2 moles of water are produced. Therefore, (y/2) = 2, and y = 4.

3. The Molecular Formula

With x = 4 and y = 4, the molecular formula of the hydrocarbon is C₄H₈.

Answer: The correct option is (B) x = 4, y = 8.

Q39. When 1.56 g of a metal hydroxide $$\mathrm{M}(\mathrm{OH})_{x}$$ is heated strongly it produces 1.02 g of the corresponding metal oxide. Thus equivalent weight of M is-
(A) 9
(B) 12
(C) 23
(D) 31

(A) 9

1. Understand the Reaction

When a metal hydroxide (M(OH)ₓ) is heated, it decomposes to form the corresponding metal oxide (M₂Oₓ) and water (H₂O):

2 M(OH)ₓ → M₂Oₓ + xH₂O

2. Determine the Mass of Water Lost

• Mass of metal hydroxide (M(OH)ₓ) = 1.56 g

• Mass of metal oxide (M₂Oₓ) = 1.02 g

• Mass of water lost (xH₂O) = 1.56 g – 1.02 g = 0.54 g

3. Calculate the Mass of Oxygen in Water

• Since the molar mass of water (H₂O) is 18 g/mol , and oxygen (O) has a molar mass of 16 g/mol , the mass ratio of oxygen in water is 16/18.

• Mass of oxygen in water = $$\frac{0.54 \text{ g H₂O}\times 16\text{ g O}}{18\text{ g H₂O}} = 0.48\text{ g O}$$

4. Determine the Equivalent Weight of the Metal (M)

• Concept: The equivalent weight of an element is the mass that combines with or displaces 8 g of oxygen.

• Calculation:

• We know that 0.48 g of oxygen combines with (1.02 g metal oxide – 0.48 g oxygen) = 0.54 g of the metal (M).

• Set up a proportion to find the equivalent weight (x) of the metal:
$$\frac{{0.48\text{ g O}}}{{8\text{ g O}}}\text{ }=\text{ }\frac{{0.54\text{ g M}}}{{x\text{ g M}}}$$

• Solving for $$x: x$$ = $$\frac{0.54\text{ g M}\times 8\text{ g O}}{0.48\text{ g O}} = 9\text{g M}$$

Answer: The equivalent weight of the metal (M) is 9 (option A).

Q40. The vapour density of a gas with respect to air is 1.528 . If the vapour density of air be 14.4 , the mass of 2 L of the gas under consideration is-
(A) 2.56 g
(B) 3.02 g
(C) 3.92 g
(D) 4.27 g

(C) 3.92 g

1. Understand Vapor Density

• Definition: Vapor density is the density of a vapor or gas in relation to the density of a reference gas, usually air or hydrogen.

• Formula: Vapor Density (VD) = $$\frac{ \text{ Molar mass of gas}}{ \text{ Molar mass of reference gas}}$$

2. Calculate the Molar Mass of the Gas

• Given: VD (gas with respect to air) = 1.528

• Given: VD (air) = 14.4 (This is assumed to be the molar mass of air, which is approximately 29 g/mol , since the units are not specified).

• Using the formula: 1.528 = ​$$\frac{{\text{ Molar mass of gas}}}{{29\text{ g/mol}}}$$

• Molar mass of gas = ​$$1.528\times 29\text{ g/mol}≈44.31\text{ g/mol}$$

3. Calculate the Mass of 2 L of the Gas

• Moles of gas: We know that 1 mole of any gas occupies 22.4 L at STP (Standard Temperature and Pressure). Assuming STP conditions, we can calculate the moles of gas in 2 L:

• Moles of gas = $$\frac{2\text{ L}}{22.4\text{ L/mol}} ≈ 0.0893\text{ mol}$$

• Mass of gas: Mass = $$\frac{ \text{ moles}}{ \text{ times (molar mass)}}$$

• Mass of gas ≈ ​$$0.0893\text{ mol}\times 44.31\text{ g/mol }\approx \text{ }3.96\text{ g}$$

Answer: The mass of 2 L of the gas is approximately 3.96 g. This is closest to option (C) 3.92 g.

Q41. The nitride of a metal $$M$$ has the formula $$\mathrm{M}_{3} \mathrm{~N}_{x}$$. It contains $$28 \%$$ nitrogen ( $$W / W$$ ). Equivalent weight of M is-
(A) 9
(B) 12
(C) 14
(D) 18

(B) 12

1. Understand the Concept of Equivalent Weight

• The equivalent weight of an element is the mass of that element that combines with or displaces:

• 8 grams of oxygen (O)

• 1 gram of hydrogen (H)

• 35.5 grams of chlorine (Cl)

2. Set up the Calculation

• We know that the nitride (M₃Nx) contains 28% nitrogen by weight. This means that 100 g of the compound contains 28 g of nitrogen and 72 g of metal (M).

• Since the formula is M₃Nx, this means 72 g of the metal (M) combines with 28 g of nitrogen (N).

• We need to find the mass of metal (M) that combines with 14 g of nitrogen (the equivalent weight of nitrogen is 14, as it combines with 3 g of hydrogen in ammonia, NH₃).

3. Proportionality

Let the equivalent weight of metal M be ‘E’. We can set up a proportion:

$$\frac{ \text{ Equivalent weight of M}}{ \text{Equivalent weight of N}}= \frac{ \text{ Mass of M}}{ \text{ Mass of N}}$$

$$∴ \frac{E}{14} = \frac{72}{28}$$

4. Solve for the Equivalent Weight (E)

Cross-multiply and solve for E:

$$E × 28 = 14 × 72$$
$$⇒ E = \frac{14 × 72}{28}$$
$$∴ E = 36$$

5. However…

There’s a nuance. The equivalent weight we calculated (36) assumes nitrogen has a valency of 3 (like in NH₃). In nitrides, nitrogen typically has a valency of 1 (forming N³⁻).

To correct for this: Divide the calculated equivalent weight by 3:

Answer: The equivalent weight of the metal (M) is 12 (option B).

Q42. 1 mol of an element ‘ $$X$$ ‘ reacis with $$1 \mathrm{~mol} \mathrm{O}_{2}$$ gas to form 1 mol of a gaseous oxide. If vapour density of the oxide be 32 , the element ‘ X ‘ is-
(A) N
(B) P
(C) S
(D) C

(C) S

1. Determine the Molar Mass of the Oxide

• Vapor density is defined as the mass of a certain volume of gas divided by the mass of an equal volume of hydrogen at the same temperature and pressure.

• Since the vapor density of hydrogen gas is 1, the molar mass of a gas is twice its vapor density.

• Molar mass of the oxide = 2 × Vapor density = 2 × 32 = 64 g/mol

2. Determine the Molar Mass of ‘X’

• The oxide has the formula XO₂ (1 mole of X reacts with 1 mole of O₂).

• Molar mass of oxygen (O₂) = 32 g/mol

• Molar mass of ‘X’ = Molar mass of oxide – Molar mass of O₂ = 64 g/mol – 32 g/mol = 32 g/mol

3. Identify the Element

• The element with a molar mass of approximately 32 g/mol is sulfur (S).

Answer: The element ‘X’ is (C) S (sulfur).

Q43. Anhydrous sulphate of a metal M contains $$15.8 \%$$ of the metal. If specific heat of the metal be $$0.237 \mathrm{cal} \mathrm{g}^{-1} \mathrm{~K}^{-1}$$, the formula of the salt is-
(A) $$\mathrm{MSO}_{4}$$
(B) $$\mathrm{M}_{2} \mathrm{SO}_{4}$$
(C) $$\mathrm{M}_{2}\left(\mathrm{SO}_{4}\right)_{3}$$
(D) cannot be predicted

(C) $$\mathrm{M}_{2}\left(\mathrm{SO}_{4}\right)_{3}$$

1. Determine the approximate atomic mass of the metal (M):

• Dulong-Petit Law: This law states that the product of an element’s specific heat and its atomic mass is approximately equal to 6.4 cal/gK.

• We are given the specific heat of the metal (0.237 cal/gK).

• Approximate atomic mass (M) = ​$$\frac{6.4\text{ cal/gK }}{0.237\text{ cal/gK}} = 27\text{ g/mol}$$

2. Calculate the mass of sulfate (SO₄) in the compound:

• Since the metal constitutes 15.8% of the compound, the sulfate part makes up 100% – 15.8% = 84.2%.

3. Assume a 100g sample of the compound:

• Mass of metal (M) = 15.8 g

• Mass of sulfate (SO₄) = 84.2 g

4. Calculate the moles of metal (M) and sulfate (SO₄):

• Moles of metal (M) = ​$$\frac{15.8\text{ g}}{27\text{g/mol}}≈ 0.585\text{ mol}$$

• Moles of sulfate (SO₄) = ​$$\frac{84.2\text{ g}}{96\text{ g/mol (molar mass of SO₄)}} ≈ 0.877 \text{mol}$$

5. Determine the mole ratio of metal (M) to sulfate (SO₄):

• Divide the moles of each by the smallest number of moles:

• Ratio of M = $$\frac{0.0.585\text{ mol}}{0.585\text{ mol}}≈ 1$$

• Ratio of SO₄ = $$\frac{0.877\text{ mol}}{0.585\text{ mol}}≈ 1.5$$

• Since we need whole numbers, multiply both ratios by 2:

• Ratio of M = 1 × 2 = 2

• Ratio of SO₄ = 1.5 × 2 = 3

6. Write the formula:

• The ratio of M to SO₄ is 2:3, giving us the formula M₂(SO₄)₃.

Therefore, the answer is (C) M₂(SO₄)₃.

Q44. The vapour density of a gas mixture consisting of $$\mathrm{NO}_{2}$$ and $$\mathrm{N}_{2} \mathrm{O}_{4}$$ is 38.3 . Number of moles of $$\mathrm{NO}_{2}$$ in 100 mol mixture is-
(A) 31.26
(B) 33.48
(C) 36.42
(D) 41.73

(B) 33.48

1. Define the relationship between vapor density and molar mass:

Since the molar mass of hydrogen (H₂) is 2 g/mol , we can relate vapor density to molar mass (M) as follows:

VD = M / 2

2. Calculate the average molar mass of the gas mixture:

Given the vapor density (VD) of the mixture is 38.3, we can find the average molar mass (M) using the relationship above:

M = VD × 2 = 38.3 × 2 = 76.6 g/mol

3. Set up an equation to represent the mole fractions and molar masses:

Let x be the mole fraction of NO₂ in the mixture. Since the mixture contains only NO₂ and N₂O₄, the mole fraction of N₂O₄ is (1 – x). We can now express the average molar mass of the mixture as a weighted average of the molar masses of NO₂ (46 g/mol ) and N₂O₄ (92 g/mol ):

$$76.6 = (x × 46) + ((1 – x) × 92)$$

4. Solve for x, the mole fraction of NO₂:

Expanding the equation:

$$76.6 = 46x + 92-92x$$

$$\Rightarrow 46x = 15.4$$

$$\therefore x ≈ 0.335$$

5. Calculate the number of moles of NO₂ in 100 mol of the mixture:

Since the mole fraction of NO₂ (x) is approximately 0.335, in a 100 mol mixture, the number of moles of NO₂ is:

Moles of NO₂ = 0.335 × 100 mol ≈ 33.5 mol

Therefore, the answer is (B) 33.48.

Q45. How many mL of $$9.52(\mathrm{M}) \mathrm{HNO}_{3}$$ will be required to dissolve 3.975 g of cupric oxide-
(A) 25.8
(B) 20.2
(C) 14.5
(D) 10.5

(D) 10.5

1. Write the balanced chemical equation for the reaction:

Cupric oxide (CuO) reacts with nitric acid (HNO₃) to produce copper(II) nitrate (Cu(NO₃)₂) and water (H₂O):

$$CuO(s) + 2HNO_3(aq) \longrightarrow Cu(NO_3)_2(aq) + H_2O(l)$$

2. Calculate the moles of cupric oxide (CuO):

Moles of CuO = ​$$\frac{\text{mass of CuO}}{\text{molar mass of CuO}}$$

Moles of CuO = ​$$\frac{3.975 \text{ g}}{79.55 \text{ g/mol }} \approx 0.05 \text{ mol}$$

3. Determine the moles of nitric acid (HNO₃) required:

From the balanced equation, 1 mole of CuO reacts with 2 moles of HNO₃. Therefore:

Moles of HNO₃ = 2 × Moles of CuO = 2 × 0.05 mol = 0.1 mol

4. Calculate the volume of the nitric acid solution needed:

Molarity (M) is defined as moles of solute per liter of solution:

Molarity (M) = ​$$\frac{\text{moles of solute}}{\text{volume of solution (L)}}$$

Rearranging the equation to solve for volume:

Volume of solution (L) = ​$$\frac{\text{moles of solute}}{\text{Molarity (M)}}$$

Volume of HNO₃ solution = ​$$\frac{0.1 \text{ mol}}{9.52 \text{ M}} \approx 0.0105 \text{ L}$$

5. Convert the volume from liters to milliliters:

Volume of HNO₃ solution (mL) = 0.0105 L × 1000 mL/L = 10.5 mL

Therefore, the answer is (D) 10.5.

Q46. 7.75 mL of a certain HCl acid solution reacts completely with 3.76 g of $$\mathrm{CaCO}_{3}$$. If density of the HCl acid solution be $$1.18 \mathrm{~g} \mathrm{~mL}^{-1}$$, the percentage strength ( $$W$$ / W) of this solution will be-
(A) 25.1
(B) 30.0
(C) 36.5
(D) 41.7

(B) 30.0

1. Write the balanced chemical equation:

$$CaCO_3(s) + 2HCl(aq) \longrightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$$

2. Calculate the moles of CaCO₃:

Moles of CaCO₃ = ​$$\frac{ \text{ mass of CaCO₃}}{ \text{ molar mass of CaCO₃}}$$

Moles of CaCO₃ =​$$\frac{3.76 \text{ g}}{100.09 \text{ g/mol }} \approx 0.0376 \text{ mol}$$

3. Determine the moles of HCl reacted:

From the balanced equation, 1 mole of CaCO₃ reacts with 2 moles of HCl.

Moles of HCl = 2 × moles of CaCO₃ = 2 × 0.0376 mol = 0.0752 mol

4. Calculate the mass of HCl:

Mass of HCl = (moles of HCl) × (molar mass of HCl)

Mass of HCl = 0.0752 mol × 36.46 g/mol ≈ 2.74 g

5. Calculate the mass of the HCl solution:

Mass of HCl solution = (volume of solution) × (density of solution)

Mass of HCl solution = 7.75 mL × 1.18 g/mL ≈ 9.145 g

6. Determine the percentage strength (w/w) of the HCl solution:

Percentage strength (w/w) = ​$$\frac{\text{mass of HCl}}{\text{mass of HCl solution}} \times$$​ 100%

Percentage strength (w/w) = ​​$$\frac{2.74 \text{ g}}{9.145 \text{ g}} \times 100 \approx 30$$​%

Therefore, the percentage strength (w/w) of the HCl solution is approximately (B) 30.0.

Q47. The percentage strength ( $$W / W$$ ) of a solution of HCl is 25. $$W$$ gram of this solution reacts with excess of $$\mathrm{CaCO}_{3}$$ to produce 2.24 L of $$\mathrm{CO}_{2}$$ (at STP). So the value of $$W$$ is-
(A) 25
(B) 28.3
(C) 29.2
(D) 32.7

(C) 29.2

1. Calculate the moles of CO₂ produced:

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L.

Moles of CO₂ = $$\frac{ \text{ Volume of CO₂}}{ \text{ Molar Volume at STP}} = \frac{2.24 \ L}{22.4 \ L/mol} = 0.1 \ mol$$

2. Determine the moles of HCl reacted:

From the balanced chemical equation for the reaction between HCl and CaCO₃:

$$CaCO_3(s) + 2HCl(aq) \longrightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$$

We see that 1 mole of CO₂ is produced from 2 moles of HCl.

Moles of HCl = 2 × moles of CO₂ = 2 × 0.1 mol = 0.2 mol

3. Calculate the mass of HCl:

Mass of HCl = (moles of HCl) × (molar mass of HCl) = 0.2 mol × 36.46 g/mol = 7.292 g

4. Determine the mass of the HCl solution (W):

The percentage strength (w/w) of the HCl solution is 25%. This means that 25 g of HCl is present in 100 g of the solution.

Since we have 7.292 g of HCl:

Mass of HCl solution (W) = $$\frac{7.292 \text{ g HCl}}{25 \text{ g HCl}} × 100 \text{ g solution}≈ 29.17 \text{ g}$$

Therefore, the value of W is approximately (C) 29.2.

Q48. Number of moles of $$\mathrm{O}_{2}$$ required for complete combustion of 1 mol of the alkyne $$\mathrm{C}_{n} \mathrm{H}_{2 n-2}$$ is-
(A) $$\frac{3}{2} n$$
(B) $$\frac{3}{2} n+1$$
(C) $$\left(\frac{3 n-1}{2}\right)$$
(D) $$\frac{3}{2} n+2$$

(C) $$\left(\frac{3 n-1}{2}\right)$$

1. Write the balanced chemical equation for the combustion of an alkyne:

The general formula for an alkyne is CnH2n-2. Complete combustion involves reacting with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).

The balanced chemical equation is:

$${{C}_{n}}{{H}_{{2n-2}}}+\text{ }\frac{{\left( {3n-1} \right)}}{2}{{O}_{2}}\to \text{ }n\text{ }C{{O}_{2}}+\text{ }\left( {n-1} \right)\text{ }{{H}_{2}}O$$

2. Determine the mole ratio between the alkyne and oxygen:

From the balanced equation, we see that:

• 1 mole of CnH2n-2 reacts with (3n-1)/2 moles of O2

The problem asks for the number of moles of O2 required for 1 mole of CnH2n-2. As determined in step 2, this is (3n-1)/2 moles of O2.

Therefore, the correct answer is (C) ​$$\left(\frac{3 n-1}{2}\right)$$

Q49. $$\mathrm{MnO}_{2}$$ reacts with HCl to form $$\mathrm{MnCl}_{2}, \mathrm{H}_{2} \mathrm{O}$$ and $$\mathrm{Cl}_{2}$$. Thus the amount of $$\mathrm{Cl}_{2}$$ formed when $$1 \mathrm{~g} \mathrm{MnO}_{2}$$ reacts with 1 g HCl is given by-
(A) 0.6257 g
(B) 0.5819 g
(C) 0.5369 g
(D) 0.4863 g

(D) 0.4863 g

1. Write the balanced chemical equation:

The reaction between manganese dioxide (MnO2) and hydrochloric acid (HCl) is:

MnO2 + 4HCl → MnCl2 + 2H2O + Cl2

​2. Calculate the number of moles of each reactant:
• Moles of MnO2:

• Molar mass of MnO2 = 54.94 (Mn) + 2 × 16.00 (O) = 86.94 g/mol

• Moles of MnO2 = $$\frac{1 \, g }{86.94 \, g/mol}$$ = 0.0115 mol

• Moles of HCl:

• Molar mass of HCl = 1.01 (H) + 35.45 (Cl) = 36.46 g/mol

• Moles of HCl = ​$$\frac{1 \, g }{36.46 \, g/mol}$$= 0.0274 mol

3. Determine the limiting reactant:

From the balanced equation, we see that 1 mole of MnO2 reacts with 4 moles of HCl. Let’s compare the mole ratio we have to the stoichiometric ratio:

• Available ratio =​$$\frac{0.0274 \ mol \, HCl}{0.0115 \, mol \, MnO2}$$ = 2.38

Since the available ratio (2.38) is less than the stoichiometric ratio (4), HCl is the limiting reactant. This means HCl will be completely consumed before all the MnO2 reacts.

4. Calculate the moles of Cl2 produced:

The balanced equation shows that 4 moles of HCl produce 1 mole of Cl2. Since HCl is the limiting reactant, we use its moles to calculate the moles of Cl2 produced:

• Moles of Cl2 = ​$$0.0274 \text{ mol HCl}× \frac{1 \text{ mol Cl2 }}{4 \text{ mol HCl}} = 0.00685 \text{ mol}$$

5. Calculate the mass of Cl2 produced:

• Molar mass of Cl2 = 2 × 35.45 (Cl) = 70.90 g/mol

• Mass of Cl2 = (0.00685 mol) × (70.90 g/mol ) = 0.4863 g

Therefore, the correct answer is (D) 0.4863 g.

Q50. 5 g of an impure sample of NaCl is treated with excess of $$\mathrm{AgNO}_{3}$$ solution. 9.812 g of AgCl is thus precipitated. So the purity (\%) of the sample of NaCl is-
(A) 60.2
(B) 75.0
(C) 80.0
(D) 82.4

(C) 80.0

1. Write the balanced chemical equation:

NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)

2. Determine the mole ratio between AgCl and NaCl:

From the balanced equation, we see that 1 mole of NaCl reacts to produce 1 mole of AgCl.

3. Calculate the moles of AgCl precipitated:

• Molar mass of AgCl = 107.87 g/mol (Ag) + 35.45 g/mol (Cl) = 143.32 g/mol

• Moles of AgCl = ​$$\frac{ \text{ mass of AgCl}}{ \text{ molar mass of AgCl}}$$

• Moles of AgCl = ​$$\frac{9.812 \text{ g}}{143.32 \text{ g/mol}}$$ = 0.0685 mol

4. Calculate the mass of pure NaCl in the sample:

• Since the mole ratio of NaCl to AgCl is 1:1, there were 0.0685 moles of NaCl in the sample.

• Molar mass of NaCl = 22.99 g/mol (Na) + 35.45 g/mol (Cl) = 58.44 g/mol

• Mass of pure NaCl = moles of NaCl × molar mass of NaCl

• Mass of pure NaCl = 0.0685 mol × 58.44 g/mol = 4.00 g

5. Calculate the percentage purity of NaCl:

• Percentage purity = ​$$\frac{ \text{ mass of pure NaCl}}{ \text{ mass of impure sample}}$$​ × 100%

• Percentage purity = ​$$\frac{4.00 \text{ g}}{5.0 \text{ g}}$$​ × 100% = 80.0%

Therefore, the answer is (C) 80.0%.

Thank You

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