A gas has a density of 3.18*g*L^-13.18gL1 at a temperature of 347*K347K, and a pressure of 1.2*atm1.2atm. What is its molecular mass?

1 Answer
Apr 3, 2017

We use the Ideal Gas equation to get "molecular mass"=75.5*g*mol^-1molecular mass=75.5gmol1

Explanation:

We assume ideality, and so PV=nRTPV=nRT, and thus,

PV=("mass"/"molar mass")RTPV=(massmolar mass)RT because n="mass"/"molar mass"n=massmolar mass

On rearrangement, "molar mass"="mass"/Vxx(RT)/Pmolar mass=massV×RTP

But "mass"/V=rho, "density"massV=ρ,density, and thus...........

"molar mass"=(rhoRT)/P=molar mass=ρRTP=

(3.18*g*L^-1xx0.0821*(L*atm)/(K*mol)xx347*K)/(1.2*atm)3.18gL1×0.0821LatmKmol×347K1.2atm

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*g*mol^-1. Because we have got consistent units, I think our order of operations is correct.