20.0 g each of helium and an unknown diatomic gas are combined in a 1500. ml container. lf the temperature is 298 K, and the pressure inside is 86.11 atm, what is the unknown gas?

1 Answer
Dec 15, 2016

Answer:

We assume (possibly unreasonably) #"ideality"#, and get......#Cl_2#, #"molecular chlorine"#, as the unknown gas..........

Explanation:

#P=(nRT)/(V)#........

And solve for #n=(PV)/(RT)#

#=# #(86.11*atmxx1.500*L)/(0.0821*L*atm*K^-1*mol^-1xx298*K)#

#n=5.28*mol#

But #n=n_"He"+n_"unknown gas"#,

i.e. #5.28*mol=(20*g)/(4*g*mol^-1)+n_"unknown gas"#

And thus #n_"unknown gas"=(5.28-5)*mol=0.280*mol#.

And thus #"molecular mass of unknown gas"# #=# #(20*g)/(0.280*mol)=71.5*g*mol^-1#.

And clearly, since the gas is diatomic, i.e. #X_2#, #X# has an atomic mass of #~=35.5*g*mol^-1#, and thus (finally!) #X=Cl#, which certainly forms a #Cl_2# molecule as required.