A gas has a density of 3.18*g*L^-1 at a temperature of 347*K, and a pressure of 1.2*atm. What is its molecular mass?

Apr 3, 2017

We use the Ideal Gas equation to get $\text{molecular mass} = 75.5 \cdot g \cdot m o {l}^{-} 1$

Explanation:

We assume ideality, and so $P V = n R T$, and thus,

$P V = \left(\text{mass"/"molar mass}\right) R T$ because $n = \text{mass"/"molar mass}$

On rearrangement, $\frac{\text{molar mass"="mass}}{V} \times \frac{R T}{P}$

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

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

$\frac{3.18 \cdot g \cdot {L}^{-} 1 \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 347 \cdot K}{1.2 \cdot a t m}$

Let's just cancel out the units to see if we have got this right. I am not immune to mistakes.............

"Molar mass"=(3.18*g*cancel(L^-1)xx0.0821*cancel(L*atm)/(cancelK*mol)xx347*cancelK)/(1.2*cancel"atm")

And this gives, I think, an answer of........

$75.5 \cdot g \cdot m o {l}^{-} 1$. Because we have got consistent units, I think our order of operations is correct.