# A 0.20*g mass of gas occupies a 0.200*L volume at one atmosphere, and 300*K. What is the molar mass of the gas?

Mar 26, 2017

$\text{Molar mass}$ $=$ $24.6 \cdot g \cdot m o {l}^{-} 1$

#### Explanation:

We use the Ideal Gas Equation.......

$P V = n R T$

And thus $n = \text{Mass"/"Molar Mass} = \frac{P V}{R T}$

And so, "Molar Mass"=("Mass"xxRT)/(PV)

$= \frac{0.20 \cdot g \times 0.0821 \cdot \cancel{L} \cdot \cancel{a t m} \cdot \cancel{{K}^{-} 1} \cdot m o {l}^{-} 1 \times 300 \cdot \cancel{K}}{1 \cdot \cancel{a t m} \times 0.200 \cdot \cancel{L}}$

$= 24.6 \cdot g \cdot m o {l}^{-} 1$. I think this is dimensionally correct, however, it does not correspond to a gas I can think of right now.........