Question

Vapour pressure of pure water is 40 mm is a non volatile solute is added to it vapour pressure falls
by 4mm hence molality of solution is
1) 6.173m
2) 3.0864
3) 1.543
4) 0.772m

​

Answer 1

40 - 4 = 36 

p = p°X 

36 = 40 X 

X = mole fraction of the water = 0.9 

0.9 / 1 = 0.9 

this means that moles H2O = 0.9 

mass H2O = 0.9 mol x 18 g/mol=16.2 g => 0.0162 Kg 

moles solute = 1 - 0.9 = 0.1 

m = 0.1 / 0.0162 =6.17
Option A