# What is the atomic mass of a gas that has rho=2.50*g*L^-1 at a pressure of 0.974*atm, and a temperature of 371*K?

Jan 24, 2017

Given $P V = n R T$..........$\text{molar mass}$ $\cong$ $80 \cdot g \cdot m o {l}^{-} 1$

#### Explanation:

$\frac{P}{R T} = \frac{n}{V} = \frac{\text{mass"/"molar mass}}{V}$

And thus $\frac{\text{molar mass"="mass of gas}}{P V} \times R T$

All I have done is manipulate the equation,

So $\frac{\text{molar mass"="mass}}{V} \times \frac{R T}{P}$

But $\text{mass"/V="density} , \rho$

Finally, $\text{molar mass} = \rho \times \frac{R T}{P}$

$\frac{2.50 \cdot g \cdot \cancel{{L}^{-} 1} \times 0.0821 \cdot \cancel{L} \cdot \cancel{a t m} \cdot \cancel{{K}^{-} 1} \cdot m o {l}^{-} 1 \times 371 \cdot \cancel{K}}{0.974 \cdot \cancel{a t m}}$

=??*g*mol^-1