A 3.7 g sample of a gas occupies the same vo | vedclass

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(C) Given: For the unknown gas,$w_1 = 3.7 \ g$,$T_1 = 25 + 273 = 298 \ K$. For $H_2$ gas,$w_2 = 0.184 \ g$,$T_2 = 17 + 273 = 290 \ K$,$M_2 = 2 \ g/mol

Vedclass · Accepted Answer

$41.326$

Vedclass · Answer

(C) Given: For the unknown gas,$w_1 = 3.7 \ g$,$T_1 = 25 + 273 = 298 \ K$. For $H_2$ gas,$w_2 = 0.184 \ g$,$T_2 = 17 + 273 = 290 \ K$,$M_2 = 2 \ g/mol$. Since pressure and volume are the same,we use the ideal gas law $PV = nRT = (w/M)RT$. Thus,$(w_1/M_1)T_1 = (w_2/M_2)T_2$. Substituting the values: $(3.7/M_1) 	imes 298 = (0.184/2) 	imes 290$. $M_1 = (3.7 	imes 298 	imes 2) / (0.184 	imes 290) = 2205.2 / 53.36 \approx 41.326 \ g/mol$.

$A$ $3.7 \ g$ sample of a gas occupies the same volume at $25 \ ^\circ C$ as $0.184 \ g$ of $H_2$ gas at $17 \ ^\circ C$ at the same pressure. The molecular weight of the gas is:

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