A gaseous substance A undergoes first order dissociation to give B, C and D as shown.
$A (g) ⟶ 2 B (g) + C (g) + D (g)$
If molar masses of A, B, C and D are 450, 100, 50 and 200 respectively and rate constant of disappearance of A is $0.693 \times 10^{- 3} \sec^{- 1}$ , then calculate ratio of rate of effusion initially and after 2000 seconds.
[Multiply the answer by $26 \sqrt{52}$ and fill the value in the OMR]

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answer is 32.

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Detailed Solution

$\begin{array}{ccc} A (g) & \to 2 B (g) + C (g) + D (g) \\ 1 & - & - & - \\ 1 - x & 2 x & x & x \end{array}$

$t_{1 / 2} = \frac{\ln 2}{0.693 \times 10^{- 8}} = 1000 \sec$

After 2000 see., $n_{A} = \frac{1}{4}; x = \frac{3}{4}$

$\begin{array}{l} M_{mix} = \frac{\frac{1}{4} \times 450 + \frac{3}{2} \times 100 + \frac{3}{4} \times (50 + 200)}{1 + 3 \times \frac{3}{4}} \end{array}$

$M_{mix} = 138.46$

$\begin{array}{l} \frac{r_{A}}{r_{mix}} = [\sqrt{\frac{M_{mix}}{M_{A}}}] \times \frac{1}{(1 + 3 \times \frac{3}{4})} \\ = [\sqrt{\frac{450 \times 4}{(13) (450)}}] \times \frac{4}{13} = (\frac{2}{\sqrt{13}}) \times \frac{4}{13} \\ = \frac{2}{\sqrt{13}} \times 26 \sqrt{52} \times \frac{4}{13} = 32 \end{array}$