At 21.0 °C, a solution of 23.71 g of a nonvolatile, nonpolar compound in 30.64 g of ethylbromide, C2H5Br, had a vapor pressure of 319.7 torr. The vapor pressure of pure ethyl bromide at this temperature is 400.0 torr. Assuming an ideal solution, what is the molecular mass of the compound?

Answer 1

pEtBr = XEtBr*P^o_EtBr
319.7 = X*400
XEtBr = about 0.8 (estimated) see last post.
Xunk = 1-about 0.8 = about 0.2

mols unk = 23.71/molar mass or 23.71/Y.
mols EtBr = 30.64/109 = about 0.281

Then Xunk = mols unk/total mols.
0.2 = [(23.71g/Y)/(23.71Y)+(0.281)]
and solve for Y = molar mass unknown.

Answer 2

How do I solve for the y? I get
0.2=23.71g / 24.71+.281
Y. Y
But I don't know how to solve

Answer 3

Then Xunk = mols unk/total mols.
0.2 = [(23.71g/Y)/(23.71Y)+(0.281)]
and solve for Y = molar mass unknown.
I made a typo in the above. It should be
0.2 = [23.71/Y/(23.71/Y)+ (0.281)].

First the numerator = 0.2*denominator.
23.71/Y = [0.2*(23.71/Y) + (0.281)]

23.71/Y = (4.74/Y)+ 0.0562
Multiply through by Y
23.71 = 4.74 + 0.0562Y
23.71-4.74 = 0.0562Y = 18.97
Y = 18.97/0.0562 = 337.5 = molar mass of unknown.

You can check that, and should.

molar mass ethyl bromide = about 109
molar mass unknownh = about 337.5

mols ethyl bromide 30.64/109 = 0.281
mols unknown = 23.71/337.5 = 0.0702
total mols = sum = 0.351

Then Xethylbromide = 0.281/0.351 = 0.800
and
Xunknown = 0.0702/0.351 = 0.200
so it checks.