The reaction between gaseous ${\mathrm{NH}}_3$ and $\mathrm{HBr}$ produces a white solid ${\mathrm{NH}}_4\mathrm{Br}$. Suppose that ${\mathrm{NH}}_3$ and $\mathrm{HBr}$ are introduced simultaneously into the opposite ends of an open tube of 1 metre long. Where would you expect the white solid to form?

Assume the distance of white solid from ${\mathrm{NH}}_3$ end to be $\mathrm{x} \mathrm{cm}$.

And the distance of white solid from $\mathrm{HBr}$ end to be $(100-\mathrm{x}) \mathrm{cm}$.

So, the  rates of diffusion would be proportional to these assumed distances.

Mathematically,

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

We know that,

Mol. mass of $\mathrm{HBr}=1+80=81$

Mol. mass of ${\mathrm{NH}}_3=14+3=17$

So, $\frac{x}{(100-x)}=\sqrt{\frac{81}{17}}$

or, $\frac{x}{(100-x)}=2.18$

So, $x=100\times 2.18-2.18x$

or, $3.18\mathrm{x}=100\times 2.18$

So, $x=\frac{100\times 2.18}{3.18}=68.55 \mathrm{cm}$.