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#?

1 Answer
Jan 24, 2017

Answer:

Given #PV=nRT#..........#"molar mass"# #~=# #80*g*mol^-1#

Explanation:

#P/(RT)=n/V=("mass"/"molar mass")/V#

And thus #"molar mass"="mass of gas"/(PV)xxRT#

All I have done is manipulate the equation,

So #"molar mass"="mass"/(V)xx(RT)/P#

But #"mass"/V="density", rho#

Finally, #"molar mass"=rhoxx(RT)/P#

#(2.50*g*cancel(L^-1)xx0.0821*cancel(L)*cancel(atm)*cancel(K^-1)*mol^-1xx371*cancel(K))/(0.974*cancel(atm))#

#=??*g*mol^-1#