At 25 , ^circ C,a 3.7 , g gas occupies the sa | vedclass

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(B) Since the pressure and volume of the $H_2$ gas and the given gas are the same,we use the ideal gas equation $PV = nRT = \frac{m}{M}RT$.For the giv

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$41.3$

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(B) Since the pressure and volume of the $H_2$ gas and the given gas are the same,we use the ideal gas equation $PV = nRT = \frac{m}{M}RT$.For the given gas: $P V = \frac{3.7}{M} R (25 + 273) = \frac{3.7}{M} R (298)$.For $H_2$ gas: $P V = \frac{0.184}{2} R (17 + 273) = \frac{0.184}{2} R (290)$.Equating both expressions: $\frac{3.7}{M} 	imes 298 = \frac{0.184}{2} 	imes 290$.Solving for $M$: $M = \frac{3.7 	imes 298 	imes 2}{0.184 	imes 290} \approx 41.3 \, g \, mol^{-1}$.

At $25 \, ^\circ C$,a $3.7 \, g$ gas occupies the same volume as $0.184 \, g$ of $H_2$ gas at $17 \, ^\circ C$ under the same pressure. What is the molar mass of the gas?

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