2.9 , g of a gas at 95^{circ} C occupied the | vedclass

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(B) Using the ideal gas equation $pV = nRT$,where $n = \frac{m}{M}$,we have $V = \frac{mRT}{Mp}$.Since $V$,$p$,and $R$ are the same for both gases,we

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$40 \, g \, mol^{-1}$

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(B) Using the ideal gas equation $pV = nRT$,where $n = \frac{m}{M}$,we have $V = \frac{mRT}{Mp}$.Since $V$,$p$,and $R$ are the same for both gases,we can write:$\frac{m_1 T_1}{M_1} = \frac{m_2 T_2}{M_2}$Given:$m_1 = 2.9 \, g$,$T_1 = 95 + 273 = 368 \, K$,$M_1 = M$ (unknown molar mass)$m_2 = 0.184 \, g$,$T_2 = 17 + 273 = 290 \, K$,$M_2 = 2 \, g \, mol^{-1}$ (for $H_2$)Substituting the values:$\frac{2.9 	imes 368}{M} = \frac{0.184 	imes 290}{2}$$M = \frac{2.9 	imes 368 	imes 2}{0.184 	imes 290}$$M = \frac{2134.4}{53.36} = 40 \, g \, mol^{-1}$The molar mass of the gas is $40 \, g \, mol^{-1}$.

$2.9 \, g$ of a gas at $95^{\circ} C$ occupied the same volume as $0.184 \, g$ of dihydrogen at $17^{\circ} C$,at the same pressure. What is the molar mass of the gas?

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