Explanation

(a)
According to Graham’s law of diffusion
$Rate\ of\ diffusion\ (r) \propto \frac{1}{\sqrt{d}}$
$Molecular\ weight\ (M) = 2 \times vapour\ density$
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$
$M_{A} = \left( \frac{100}{2} \right)\ kg/molecule$
$M_{B} = \left( \frac{64}{2} \right)\ kg/molecule$
$r_{A} = 12 \times 10^{- 3}\ and\ r_{B} = ?$
$\frac{r_{A}}{r_{B}} = \sqrt{\frac{d_{B}}{d_{A}}} = \sqrt{\frac{M_{B}}{M_{A}}}$
$\frac{12 \times 10^{- 3}}{r_{B}} = \sqrt{\frac{64/2}{100/2}} = \sqrt{\frac{64}{100}} = \frac{8}{10}$
$\ \ \ \ \ \ \ \ r_{B} = \frac{12 \times 10^{- 3} \times 10}{8}$
$\ \ \ \ \ \ \ \ = 15 \times 10^{- 3}$