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5.18 2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas?

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Let the molar mass of the gas be $M_{gas}$ ,

Then, given that $2.9g$ of a gas at $95^{\circ}C$ occupied the same volume as

$0.184g$ of $H_{2}$ at $17^{\circ }C$

So, we have the relation, $P_{1}=P_{2}$ and $V_{1} = V_{2}$

Therefore $P_{1}V_{2} = P_{2} V_{2}$

which gives, $n_{1}RT_{1} =n_{2}RT_{2}$

$\implies n_{1}T_{1} =n_{2}T_{2}$

Or, $\frac{2.9}{M_{gas}}\times (95+273) = \frac{0.184}{2}\times(17+273)$

Or, $M_{gas} = \frac{2.9\times368\times 2 }{0.184\times 290} = 40g\ mol^{-1}$

Posted by

Divya Prakash Singh

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