Question

A certain gas takes three times as long to effuse out as helium. Its molecular mass will be :

• A. 27u
• B. 36u
• C. 64u
• D. 1u

Answer 1

The correct option is B 36u

$\begin{array}{c}\text{We know, rate of effusion (r)}\alpha \frac{1}{\sqrt{M}}\text{, where}M\text{is the molecular}\\ \text{weight of gas. Let, the molecular mass of required gas be M.}\end{array}$

$\begin{array}{c}\text{Then,}\frac{r_m}{r_{He}}=\frac{1}{3}\\ \Rightarrow \sqrt{\frac{4}{M}}=\frac{1}{3}\Rightarrow \frac{4}{M}=\frac{1}{9}\Rightarrow M=36\\ \text{Thus, molecular mass of required}gas=36u\end{array}$

Option B is correct answer