Question:
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Answer:
Vapour pressure of pure water (p10) =1 atm = 1.013 Bar
Vapour pressure of pure solution (p1) = 1.004 Bar
Mass of non-volatile solute (W2) = 2 gm
Mass of solvent (W1) = 98 gm molar mass of solvent (water) (M1) = 18
molar mass of solute = (M1) =?
Formula of Raoult’s law
(p10 - p1)/ p10 =(W2 x M1)/ (W1 x M2)
(1.013-1.004)/1.013=2x18/98x M2
M2= 2x18x1.013/98 x (1.013-1.004) =2x18x1.013/98x.009 = 41.35 gm/mol
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