Question

The rate of diffusion of $2$ gases ${}^{\prime }{A}^{\prime }$ and ${}^{\prime }{B}^{\prime }$ are in the ratio $16:3$. If the ratio of their masses present in the mixture is $2:3$. Then:

• A. The ratio of their molar masses is $16:1$
• B. The ratio of their molar masses is $1:4$
• C. The ratio of their moles present inside the container is $1:24$
• D. The ratio of their moles present inside the container is $8:3$

Answer 1

Answer: (B), (D)

The ratio of their moles present inside the container is $8:3$

Let, $R_A$ be the rate of diffusion of gas $A$
and $r_B$ be the rate of diffusion of gas $B$
$w_A$ and $w_B$ be the weight of gas $A$ and $B$ respectively.
Given,
$\frac{r_A}{r_B}=\frac{16}{3};\frac{w_A}{w_B}=\frac{2}{3}$
We kniw that,
$\frac{r_A}{r_B}=\frac{n_A}{n_B}\sqrt{\frac{M_B}{M_A}}$
Where,
$n_A=\text{number of moles of gas A}$
$n_B=\text{number of moles of gas B}$
$M_A=\text{Molar mass of gas A}$
$M_B=\text{Molar mass of gas B}$
Substituting the values, we get,
$\frac{16}{3}=\frac{w_A}{M_A}\frac{M_B}{w_B}\sqrt{\frac{M_B}{M_A}}$
$\frac{16}{3}=\frac{2}{3}{\left(\frac{M_B}{M_A}\right)}^{\frac{3}{2}}\Rightarrow {\left(\frac{M_B}{M_A}\right)}^{\frac{3}{2}}=8$
$\Rightarrow \frac{M_B}{M_A}=4$
$\therefore \text{mole ratio}=\frac{2}{3}\times 4=\frac{8}{3}$