Question

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

Answer 1

Volume of $N{H}_3\left(V_1\right)$ effused $=100c{m}^3$

Time taken to effuse $\left(t_1\right)=32.5s$

Volume of $N_2$ effused $\left(V_1\right)=60c{m}^3$

Time taken to effuse by $N_2$ be $t_2$

$\therefore \frac{r_1}{r_2}=\frac{\left(V_1/t_1\right)}{\left(V_2/t_2\right)}=\sqrt{\frac{M_{N_2}}{N_{N{H}_3}}}$

$\frac{100\times t_2}{32.5\times 60}=\sqrt{\frac{28}{17}}$

$t_2=\frac{3}{5}\times 32.5\sqrt{\frac{28}{17}}s=25s$