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
$\text{R}\text{a}\text{t}\text{e}\text{o}\text{f}\text{d}\text{i}\text{f}\text{f}\text{u}\text{s}\text{i}\text{o}\text{n}\left(\text{r}\right)\propto \frac{1}{\sqrt{\text{d}}}$
$\text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\left(\text{M}\right)=2\times \text{v}\text{a}\text{p}\text{o}\text{u}\text{r}\text{d}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}$
$\frac{{\text{r}}_1}{{\text{r}}_2}=\sqrt{\frac{{\text{V.D}}_2}{{\text{V.D}}_1}}$
${\left(\text{M}\right)}_{\text{A}}=\left(\frac{100}{2}\right)$
${\left(\text{M}\right)}_{\text{B}}=\left(\frac{64}{2}\right)$
${\text{r}}_{\text{A}}=12\times 10^{-3}\text{a}\text{n}\text{d}{\text{r}}_{\text{B}}=?$
$\frac{{\text{r}}_{\text{A}}}{{\text{r}}_{\text{B}}}=\sqrt{\frac{{\text{d}}_{\text{B}}}{{\text{d}}_{\text{A}}}}=\sqrt{\frac{{\text{V.D}}_{\text{B}}}{{\text{V.D}}_{\text{A}}}}$
$\frac{12\times 10^{-3}}{{\text{r}}_{\text{B}}}=\sqrt{\frac{64/2}{100/2}}=\sqrt{\frac{64}{100}}=\frac{8}{10}$
${\text{r}}_{\text{B}}=\frac{12\times 10^{-3}\times 10}{8}$
$=15\times 10^{-3}$