Question

At a particular fixed temperature the gas $A_n$ is 52% dissociated according to the following reaction $A_n(g) \rightleftharpoons nA(g)$ The equilibrium mixture effuses 1.25 times slower than pure oxygen gas under identical condition. If atomic wt. of A is 32, then find "n" ?

• A. 4
• B. 3
• C. 5
• D. 2

Answer 1

Answer: (A)

4

Answer 2

1. Relate Effusion Rate and Molar Mass: According to Graham's Law of Diffusion, the rate of effusion ($R$) of a gas is inversely proportional to the square root of its molar mass ($M$). $\frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}}$ Let $R_{mixture}$ and $M_{mixture}$ be the rate and molar mass of the equilibrium mixture, and $R_{O_2}$ and $M_{O_2}$ be for pure oxygen. We are given that the mixture effuses 1.25 times slower than pure oxygen, so $R_{mixture} = \frac{R_{O_2}}{1.25}$. $\frac{R_{mixture}}{R_{O_2}} = \frac{1}{1.25} = \frac{4}{5}$

2. Calculate Molar Mass of the Equilibrium Mixture: $\frac{4}{5} = \sqrt{\frac{M_{O_2}}{M_{mixture}}}$ Squaring both sides: $\left(\frac{4}{5}\right)^2 = \frac{16}{25} = \frac{M_{O_2}}{M_{mixture}}$ The molar mass of oxygen ($O_2$) is $M_{O_2} = 32 \text{ g/mol}$. $M_{mixture} = M_{O_2} \times \frac{25}{16} = 32 \text{ g/mol} \times \frac{25}{16} = 50 \text{ g/mol}$

3. Relate Molar Mass to Degree of Dissociation: Let the initial number of moles of $A_n$ be 1. The reaction is: $A_n(g) \rightleftharpoons nA(g)$ Given that $A_n$ is 52% dissociated, the degree of dissociation ($\alpha$) is 0.52. Initial moles: 1 | 0 Change: -$\alpha$ | +$n\alpha$ Equilibrium moles: $(1-\alpha)$ | $n\alpha$ At equilibrium, moles of $A_n = 1 - 0.52 = 0.48$ mol. Moles of $A = 0.52n$ mol. Total moles at equilibrium = $(1-\alpha) + n\alpha = 0.48 + 0.52n$.

   The molar mass of $A_n$ is $M_{A_n} = n \times (\text{atomic wt. of A}) = n \times 32$. The total mass of the gas mixture remains constant (conservation of mass). Assuming 1 mole of $A_n$ initially, the total mass is $1 \times M_{A_n} = 32n$. The observed molar mass of the mixture is given by: $M_{mixture} = \frac{\text{Total mass}}{\text{Total moles at equilibrium}}$ $50 = \frac{32n}{0.48 + 0.52n}$

4. Solve for 'n': $50(0.48 + 0.52n) = 32n$ $24 + 26n = 32n$ $24 = 32n - 26n$ $24 = 6n$ $n = \frac{24}{6} = 4$