If 1.00 L of an unknown gas at STP has a mass of 5.40 g, what is its molar mass?

May 8, 2016

Approx, $121 \cdot g \cdot m o {l}^{-} 1$

Explanation:

The molar volume at $\text{STP}$ is $22.4 \cdot L \cdot m o {l}^{-} 1$.

Given this, if we divide the mass by the molar quantity of the gas we get an answer in $g \cdot m o {l}^{-} 1$ as required:

$\frac{1.00 \cdot L}{22.4 \cdot L \cdot m o {l}^{-} 1}$ $=$ $0.0446 \cdot m o l$

$\text{Molar mass}$ $=$ $\frac{5.40 \cdot g}{0.0446 \cdot m o l}$ $=$ $121 \cdot g \cdot m o {l}^{-} 1$