Question

If two molecules of $A$ and $B$ having mass $100 \,kg$ and $64\, kg$ and rate of diffusion of $A$ is $12\times 10^{-3},$ then what will be the rate of diffusion of $B$?

• A. $15\times {{10}^{-3}}$
• B. $64\times {{10}^{-3}}$
• C. $~5\times {{10}^{-3}}$
• D. $~46\times {{10}^{-3}}$

Answer 1

Answer: (A)

$15\times {{10}^{-3}}$

Answer 2

According to Grahams law of diffusion At constant pressure and temperature, the rate of diffusion of a gas is inversely proportional to the square root of its vapour density. 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 $r_{B}=?$ $\frac{r_{A}}{r_{B}}=\sqrt{\frac{d_{B}}{d_{A}}}=\sqrt{\frac{M_{B}}{M_{A}}}$ $\frac{12 \times 10^{-3}}{r_{B}}=\sqrt{\frac{64 / 2}{100 / 2}}=\sqrt{\frac{64}{100}}=\frac{8}{10}$ $r_{B}=\frac{12 \times 10^{-3} \times 10}{8}$ $=15 \times 10^{-3}$