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 $\frac{t_{2}^{2}}{t_{1}^{2}}=\frac{M_{2}}{M_{1}}$
$M_{2}=M_{1}\times\frac{t_{2}^{2}}{t_{1}^{2}}=49\times\left(\frac{20}{10}\right)^{2}$
$=196$