Question

A gaseous substance $A_n$ remains partially dissociated $A_n(g) = nA(g)$. If the given sample of gas has $A_n(g)52\%$ dissociated and the equilibrium mixture diffuses 1.25 times slower than pure oxygen gas under identical condition determine 'n' Atomic weight of A is 32.

Answer 1

The value of n is 4.

Answer 2

Let the dissociation of the gaseous substance $A_n$ be represented by the equilibrium:

$A_n(g) \rightleftharpoons nA(g)$

Assume we start with 1 mole of $A_n$.
Let $\alpha$ be the degree of dissociation. We are given that the substance is 52% dissociated, so $\alpha = 0.52$.

Initial moles:
$A_n: 1$
$A: 0$

Moles at equilibrium:
$A_n: 1 - \alpha = 1 - 0.52 = 0.48$
$A: n\alpha = n \times 0.52$

Total number of moles at equilibrium, $N_{total} = (1-\alpha) + n\alpha = 1 + (n-1)\alpha$.
Substituting $\alpha = 0.52$, $N_{total} = 1 + (n-1) \times 0.52$.

Graham's Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass:

$\frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}}$

Let the mixture be gas 1 and pure oxygen ($O_2$) be gas 2.
We are given that the equilibrium mixture diffuses 1.25 times slower than pure oxygen gas.
$Rate_{mixture} = \frac{1}{1.25} Rate_{O_2}$
$\frac{Rate_{mixture}}{Rate_{O_2}} = \frac{1}{1.25}$

Using Graham's Law:

$\frac{1}{1.25} = \sqrt{\frac{M_{O_2}}{M_{mixture}}}$

The molar mass of oxygen gas ($O_2$) is $M_{O_2} = 2 \times 16 = 32$ g/mol.
$\frac{1}{1.25} = \sqrt{\frac{32}{M_{mixture}}}$
Squaring both sides:
$(\frac{1}{1.25})^2 = \frac{32}{M_{mixture}}$
$\frac{1}{(5/4)^2} = \frac{32}{M_{mixture}}$
$\frac{1}{25/16} = \frac{32}{M_{mixture}}$
$\frac{16}{25} = \frac{32}{M_{mixture}}$
$M_{mixture} = \frac{32 \times 25}{16} = 2 \times 25 = 50$ g/mol.

The average molar mass of the equilibrium mixture is the total mass divided by the total number of moles.
If we started with 1 mole of $A_n$, the total mass of the system is the initial mass of $A_n$.
Molar mass of $A_n$, $M_{A_n} = n \times \text{Atomic weight of A} = n \times 32$.
Total mass = $1 \times M_{A_n} = 32n$.

The average molar mass of the mixture is $M_{mixture} = \frac{\text{Total mass}}{\text{Total moles}} = \frac{32n}{N_{total}}$.
Substituting $N_{total} = 1 + (n-1) \times 0.52$ and $M_{mixture} = 50$:
$50 = \frac{32n}{1 + (n-1) \times 0.52}$
$50 = \frac{32n}{1 + 0.52n - 0.52}$
$50 = \frac{32n}{0.48 + 0.52n}$
$50(0.48 + 0.52n) = 32n$
$24 + 26n = 32n$
$24 = 32n - 26n$
$24 = 6n$
$n = \frac{24}{6}$
$n = 4$.