4.4 g of a gas at 0^{circ} C and 0.82 atm p | vedclass

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Question

(D) Using the ideal gas equation: $PV = nRT = \frac{w}{M} RT$ Where $M$ is the molar mass of the gas. Given: $w = 4.4 \ g$, $T = 0^{\circ} C = 273 \ K

Vedclass · Accepted Answer

$CO_2$

Vedclass · Answer

(D) Using the ideal gas equation: $PV = nRT = \frac{w}{M} RT$ Where $M$ is the molar mass of the gas. Given: $w = 4.4 \ g$, $T = 0^{\circ} C = 273 \ K$, $R = 0.082 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1}$, $P = 0.82 \ atm$, $V = 2.73 \ L$. Substituting the values: $0.82 	imes 2.73 = \frac{4.4}{M} 	imes 0.082 	imes 273$ $M = \frac{4.4 	imes 0.082 	imes 273}{0.82 	imes 2.73}$ $M = \frac{4.4 	imes 22.386}{2.2386} = 44 \ g/mol$. The molar mass of $CO_2$ is $12 + 2 	imes 16 = 44 \ g/mol$. Therefore, the gas is $CO_2$.

$4.4 \ g$ of a gas at $0^{\circ} C$ and $0.82 \ atm$ pressure occupies a volume of $2.73 \ L$. The gas can be:

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