The reaction between gaseous $N H_{3}$ and $H B r$ produces a white solid $N H_{4} B r$ . Suppose a small quantity of gaseous $N H_{3}$ and gaseous $H B r$ are introduced simultaneously into opposite ends of an open tube which is one metre long. Calculate the distance of white solid formed from the end which was used to introduce $N H_{3}$ .

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Solution

Let the distance of white solid from $N H_{3}$ end $= x c m$ .
The distance of white solid from $H B r$ end $= (100 - x) c m$ .
Rates of diffusion shall be proportional to these distances.
$\frac{r_{1}}{r_{2}} = \frac{x}{(100 - x)} = \sqrt{\frac{M_{H B r}}{M_{N H_{3}}}}$
Mol. mass of $H B r = 1 + 80 = 81$
Mol. mass of $N H_{3} = 14 + 3 = 17$
So, $\frac{x}{(100 - x)} = \sqrt{\frac{81}{17}}$
or, $\frac{x}{(100 - x)} = 2.18$
So, $x = 100 \times 2.18 - 2.18 x$
or, $3.18 x = 100 \times 2.18$
So, $x = \frac{100 \times 2.18}{3.18} = 68.55 c m$

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