Question

Ratio of times taken for two gases to diffuse equal volumes is $10/20$. If molecular weight of one gas is $49$, then the molecular weight of another gas is

• A. $49$
• B. $98$
• C. $147$
• D. $196$

Answer 1

Answer: (D)

Let the times of diffusion for same volumes of two gases be $t_1$ and $t_2$ respectively. Then
$\frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}$ or $\frac{t_2^2}{t_1^2}=\frac{M_2}{M_1}$
$M_2=M_1\times \frac{t_2^2}{t_1^2}=49\times {\left(\frac{20}{10}\right)}^2$
$=196$