If two molecules of $A$ and $B$ having mass $100 \mathrm{~kg}$ and $64 \mathrm{~kg}$ and rate of diffusion…
If two molecules of $A$ and $B$ having mass $100 \mathrm{~kg}$ and $64 \mathrm{~kg}$ and rate of diffusion of $A$ is $12 \times 10^{-3}$, then rate of diffusion of $B$ will be
$15 \times 10^{-3}$
$64 \times 10^{-3}$
$36 \times 10^{-3}$
$10 \times 10^{-3}$
Solution
$\begin{aligned} & m_A=\left(\frac{100}{2}\right) \mathrm{kg} / \text { molecule } \\ & m_B=\left(\frac{64}{2}\right) \mathrm{kg} / \text { molecule }\end{aligned}$
Rate of diffusion, $r_B=$ ?
$r_A=12 \times 10^{-3}$
According to Graham's law of diffusion,
$\frac{r_A}{r_B}=\sqrt{\frac{d_B}{d_A}}=\sqrt{\frac{M_B}{M_A}}$
$\frac{r_A}{r_B}=\sqrt{\frac{m_B \times N_A}{m_A \times N_A}}\left(N_A=\right.$ Avogadro's number $)$
$\frac{r_A}{r_B}=\sqrt{\frac{64}{100}}=\frac{8}{10}=0.8$ or $\frac{12 \times 10^{-3}}{r_B}=0.8$
$r_B=\frac{12 \times 10^{-3}}{0.8}=15 \times 10^{-3}$