Question

Ammonia gas diffuse twice as fast as a gas x. Find the molecular weight of gas x.

Answer 1

Let molecular weight of a gas be x

$r_{N{H}_3}=2r_x;M_{N{H}_3}=17$

$\therefore \frac{r_{N{H}_3}}{r_x}=2=\sqrt{\frac{M_x}{M_{N{H}_3}}}$

$2=\sqrt{\frac{M_x}{17}}$

Squaring on both sides

$4=\frac{M_x}{17}$

$M_x=17\times 4$

$M_x=68$

$\therefore$ Molecular weight of gas x is 68.