Question

IN NEED OF HELP!
A solution is prepared from 27.4 grams of a nonvolatile, non dissociating solute and 85.0 grams of water. The vapor pressure of the solution at 60 degrees Celsius is 142 torr. The vapor pressure of water at 60 degrees Celsius is 150. torr. What is the molar mass of the solute?
A. 28.9
B. 5.81
C. 7.28
D. 169
E. 103

Answer 1

I think you have to use the
Raoult's Law

http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions_and_Mixtures/Ideal_Solutions/Changes_In_Vapor_Pressure,_Raoult's_Law

Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present:

Psolution=Xsolvent Posolvent

Answer 2

I'm still kind of lost.

Answer 3

I need calculations to figure it out.

Answer 4

PH2O = 150 torr

Psoluc = 142 torr

Xsolvent = Psoluc / PH20 = 142 torr / 150 torr = 0.946

nH2O= mass/MM => 85g/18g = 4.72 moles of H20

Xsolvent=0.946 = 4.72 mol H20/ (4.72 mol H20 + n mol solute)


Solving for n:

n = (4.72 mol H20 - (0.946 * 4.72 mol H20) )/ 0.946 =0.266 mol

Therefore:

27.4g solute = 0.266 mol solute

27.4g * (1/0.266mol) = 103.0 g/mol