Question

If two molecules of $\text{A} \, \text{a}\text{n}\text{d} \, \text{B}$ having mass 100 amu and 64 amu respectively and rate of diffusion of $\text{A} \, \text{i}\text{s} \, 12\times 10^{- 3},$ then what will be the rate of diffusion of $\text{B}?$

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

Answer 1

Answer: (A)

According to Graham's law of diffusion
$\text{R}\text{a}\text{t}\text{e} \, \text{o}\text{f} \, \text{d}\text{i}\text{f}\text{f}\text{u}\text{s}\text{i}\text{o}\text{n} \, \left(\text{r}\right) \propto \frac{1}{\sqrt{\text{d}}}$
$\text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r} \, \text{w}\text{e}\text{i}\text{g}\text{h}\text{t} \, \left(\text{M}\right)=2\times \text{v}\text{a}\text{p}\text{o}\text{u}\text{r} \, \text{d}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}$
$\frac{\text{r}_{1}}{\text{r}_{2}}=\sqrt{\frac{\text{V.D}_{2}}{\text{V.D}_{1}}}$
$\left(\text{M}\right)_{\text{A}}=\left(\frac{100}{2}\right) \, $
$\left(\text{M}\right)_{\text{B}}=\left(\frac{64}{2}\right) \, $
$\text{r}_{\text{A}}=12\times 10^{- 3} \, \text{a}\text{n}\text{d} \, \text{r}_{\text{B}}=?$
$\frac{\text{r}_{\text{A}}}{\text{r}_{\text{B}}}=\sqrt{\frac{\text{d}_{\text{B}}}{\text{d}_{\text{A}}}}=\sqrt{\frac{\text{V.D}_{\text{B}}}{\text{V.D}_{\text{A}}}}$
$\frac{12 \times 10^{- 3}}{\text{r}_{\text{B}}}=\sqrt{\frac{64 / 2}{100 / 2}}=\sqrt{\frac{64}{100}}=\frac{8}{10}$
$\text{r}_{\text{B}}=\frac{12 \times 10^{- 3} \times 10}{8}$
$=15\times 10^{- 3}$