Question

$100c{m}^3$ of $N{H}_3$ diffuses through a fine hole in $32.5$ seconds. How much time will $60cc$ of $N_2$ take to diffuse under the same conditions?

Answer 1

$\frac{RateofdiffusionofN{H}_3}{Rateofdiffusionof{N}_2}=\sqrt{\frac{M_{N_2}}{M_{N{H}_3}}}$

$\frac{V_{N{H}_3}/t_{N{H}_3}}{V_{N_2}/t_{N_2}}=\sqrt{\frac{28}{17}}\Rightarrow \frac{100/32.5}{60/t_{N_2}}=\sqrt{\frac{28}{17}}$

$t_{N_2}=\frac{32.5\times 60}{100}\sqrt{\frac{28}{17}}=25sec$