Question

$50\, mL$ of each gas A and of gas B takes $150$ and $200\, s$ respectively for effusing through a pin hole under the similar conditions. If molecular mass of gas B is $36,$ the molecular mass of gas A will be

• A. 96
• B. 128
• C. 32
• D. 20.2

Answer 1

Answer: (D)

20.2

Answer 2

Given, $V_A=V_B=50ml$
$\, \, \, \, \, \, \, \, \, \, T_A=150s$
$\, \, \, \, \, \, \, \, \, \, T_B=200s$
$\, \, \, \, \, \, \, \, \, \, M_B=36$
$\, \, \, \, \, \, \, \, \, \, M_A=?$
From Graham's law of effusion
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{I_B}{I_A}=\sqrt{\frac{M_A}{M_B}}=\frac{V_B T_A}{T_B V_A}$
$\Rightarrow \, \, \, \, \, \, \, \, \sqrt{\frac{M_A}{36}}=\frac{V_A\times 150}{200\times V_A}$
$or\, \, \, \, \, \, \, \, \, \, \, \frac{\sqrt{M_A}}{36}=\frac{15}{220}=\frac{3}{4}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{M_A}{36}=\frac{9}{16}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, M_A=\frac{9\times 36}{16}=\frac{9\times 9}{4}=\frac{81}{4}=20.2$