Question

The rate of diffusion of two gases 'A' and 'B' are in the ratio 16 : 3. If the ratio of their masses present in the mixture is 2 : 3, then the ratio of their:

• A. Molar masses is 16 : 1
• B. Molar masses is 1 : 4
• C. Moles present inside the container is 1 : 24
• D. Moles present inside the container is 8 : 3

Answer 1

Answer: (B), (D)

The correct options are
B Molar masses is 1 : 4
D Moles present inside the container is 8 : 3

Given : $r_A:r_B=16:3$, and, $w_A:w_B=2:3$
$\frac{r_A}{r_B}=\frac{n_A}{n_B}\sqrt{\frac{M_B}{M_A}}$ : Graham's law.
Say, $w_A=2w$ and $w_B=3w$ (w : some const.)
then, $n_A=\frac{2w}{M_A}$ and $n_B=\frac{3w}{M_A}$
$\frac{r_A}{r_B}=\frac{\left(\frac{2w}{M_A}\right)}{\left(\frac{3w}{M_B}\right)}\times \sqrt{\frac{M_B}{M_A}}=\frac{2}{3}{\left(\frac{M_B}{M_A}\right)}^{3/2}$
$\therefore \frac{16}{3}=\frac{2}{3}{\left(\frac{M_B}{M_A}\right)}^{3/2}$
$\Rightarrow \frac{M_B}{M_A}=8^{2/3}=4\Rightarrow \frac{M_A}{M_B}=\frac{1}{4}$

Then, $\frac{n_A}{n_B}=\frac{2w}{M_A}\times \frac{M_B}{3w}=\frac{2}{3}\times 4=\frac{8}{3}$

$\therefore$ [B], [D]