Please help!

Question

The vapor pressure of water at 35 oC is 42.175 mm Hg. The vapor pressure of ethyl alcohol (C2H5OH) at 35 C is 100.5 mm Hg. What is the vapor pressure of a solution prepared by dissolving 250 g of C2H5OH in 375 g of H2O?

The answer is supposed to be 48.3 mm Hg but I don't know how to get there!

Answer 1

mols EtOH = g/molar mass = ?
mols H2O = g/molar mass = ?
Total mols = mols EtOH + mols H2O = ?
XEtOH = mols EtOH/total mols = ?
XH2O = mols H2O/total mols = ?
pEtOH = X_EtOH*P^o_EtOH = ?
pH2O = X_H2O*P^o_H2O = ?
Then Ptotal above the solution is pEtOH + pH2O
Post your work if you get stuck.

Answer 2

Here's my resulting answer but it isn't 48.3 mm Hg

mols EtOH = g/molar mass = 5.43 mols
mols H2O = g/molar mass = 20.81 mols
Total mols = mols EtOH + mols H2O = 26.24 mols
XEtOH = mols EtOH/total mols = 0.207 mol
XH2O = mols H2O/total mols = 0.793 mol
pEtOH = XEtOH*PoEtOH = 20.80 mmHg
pH2O = XH2O*PoH2O = 33.45 mmHg

Total Pressure = 54.25 mm Hg