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$50 \mathrm{~mL}$ of each gas $A$ and of gas $B$ takes 150 and $200 \mathrm{~s}$ 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
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Verified Answer
The correct answer is:
20.25
Given,
$\begin{aligned}
V_A & =V_B=50 \mathrm{~mL} \\
T_A & =150 \mathrm{~s} \\
T_B & =200 \mathrm{~s} \\
M_B & =36 \\
M_A & =?
\end{aligned}$
From Graham's law of effusion
$\begin{aligned}
\frac{r_B}{r_A} & =\sqrt{\frac{M_A}{M_B}}=\frac{V_B T_A}{T_B \cdot V_A} \\
\Rightarrow \sqrt{\frac{M_A}{36}} & =\frac{V_A \times 150}{200 \times V_A} \\
\text {or } \sqrt{\frac{M_A}{36}} & =\frac{15}{20}=\frac{3}{4} \\
\frac{M_A}{36} & =\frac{9}{16} \\
M_A & =\frac{9 \times 36}{16}=\frac{9 \times 9}{4}=\frac{81}{4}=20.25
\end{aligned}$
$\begin{aligned}
V_A & =V_B=50 \mathrm{~mL} \\
T_A & =150 \mathrm{~s} \\
T_B & =200 \mathrm{~s} \\
M_B & =36 \\
M_A & =?
\end{aligned}$
From Graham's law of effusion
$\begin{aligned}
\frac{r_B}{r_A} & =\sqrt{\frac{M_A}{M_B}}=\frac{V_B T_A}{T_B \cdot V_A} \\
\Rightarrow \sqrt{\frac{M_A}{36}} & =\frac{V_A \times 150}{200 \times V_A} \\
\text {or } \sqrt{\frac{M_A}{36}} & =\frac{15}{20}=\frac{3}{4} \\
\frac{M_A}{36} & =\frac{9}{16} \\
M_A & =\frac{9 \times 36}{16}=\frac{9 \times 9}{4}=\frac{81}{4}=20.25
\end{aligned}$
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