The reaction between gaseous NH₃ and HBr produces a white solid NH₄Br. Suppose that NH₃ and HBr are introduced simultaneously into the opposite ends of an open tube of 1 metre long. Where would you expect the white solid to form?

Let the distance of white solid from ${\mathrm{NH}}_3$end = x cm

The distance of white solid from HBr end = (100 - x)cm

Rates of diffusion shall be proportional to these distances.

$\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{\mathrm{x}}{\left(100-\mathrm{x}\right)}=\sqrt{\frac{{\mathrm{M}}_{\mathrm{HBr}}}{{\mathrm{M}}_{{\mathrm{NH}}_3}}}$

Mol. mass of HBr = 1 + 80 = 81

Mol. mass of NH₃ = 14 + 3 = 17

$$\begin{gathered} \mathrm{So}, \frac{\mathrm{x}}{\left(100-\mathrm{x}\right)}=\sqrt{\frac{81}{17}} \\ \mathrm{or}, \frac{\mathrm{x}}{\left(100-\mathrm{x}\right)}=2.18 \\ \mathrm{So}, \mathrm{x}=100\times 2.18-2.18\mathrm{x} \\ \mathrm{or}, 3.18\mathrm{x}=100\times 2.18 \\ \mathrm{So}, \mathrm{x}=\frac{100\times 2.18}{3.18}=68.55\mathrm{cm} \end{gathered}$$