States of Matter - Result Question 94
94. 3.7 g of a gas at $25^{\circ} \mathrm{C}$ occupied the same volume as 0.184 g of hydrogen at $17^{\circ} \mathrm{C}$ and at the same pressure. What is the molecular weight of the gas ?
$(1979,2 M)$
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Answer:
Correct Answer: 94. (41.32g)
Solution:
For same $p$ and $V, n \propto \dfrac{1}{T}$
$$ \begin{aligned} & \Rightarrow \quad \dfrac{n(\mathrm{gas})}{n\left(\mathrm{H}_2\right)}=\dfrac{T\left(\mathrm{H}_2\right)}{T(\mathrm{gas})} \\ & n\left(\mathrm{H}_2\right)=\dfrac{0.184}{2}=0.092 \\ & \Rightarrow \quad n \text { (gas) }=\dfrac{290}{298} \times 0.092=0.0895 \\ & \end{aligned} $$
$\because \quad 0.0895$ mole of gas weigh 3.7 g
$\therefore 1$ mole of gas will weigh $\dfrac{3.7}{0.0895}=41.32 \mathrm{~g}$
BETA
