# What is the molar mass of a gas that at 823*K exerts a pressure of 31.89*kPa in a volume of 56.0*mL?

Sep 18, 2017

Approx.....$120 \cdot g \cdot m o {l}^{-} 1$

#### Explanation:

We use the Ideal Gas equation, and solve for molar mass.....

$n = \frac{P V}{R T} = \frac{\frac{31.89 \cdot k P a}{101.3 \cdot k P a \cdot a t {m}^{-} 1} \times 56.0 \times {10}^{-} 3 L}{0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 823 \cdot K} = 2.61 \times {10}^{-} 4 \cdot m o l$

And thus $\text{molar mass"="Mass of gas"/"Number of moles}$

$= \frac{3.243 \times {10}^{-} 2 \cdot g}{2.61 \times {10}^{-} 4 \cdot m o l} = 124.3 \cdot g \cdot m o {l}^{-} 1$.

And note that this molar mass is consistent with a ${P}_{4}$ molecule.....