Consider a solution in which W2

g of solute of molar mass M2 is dissolved in W1

g of solvent of molar mass M1

.

Hence, number of moles of solvent, n1=M1 W1

and,

Number of moles of solute, n1=M2W2

∴ Total number of moles n=n1+n2

Mole fraction of the solvent, xx1 =nn1

Mole fraction of the solute,x x2 =nn 2

In case of a dilute solution, the concentration of number of moles of the solute is very low, i.e., n1>>n2

∴n1 +nn ≃n1

Now, the mole fraction of the solute x2

is given by,

x2 =n1 +n2

n2≃ n1 n2

∴x2 = W1 /M1

W2 /M2

∴x2 =W1 ×M2

W2×M1

lf P1

0

and P are the vapour pressure of pure solvent and a solution respectively, then relative lowering of vapour pressure is given by,

P= P1 0

P1 0−P

By Raoult's law

P1 0

ΔP =P1 0

P1 0 −P =x2

P 1 0

P1 0−P= W1 M2 W2M11

or ∴

P

1

0

ΔP

=

n

1

n

2

=

W

1

/M

1

W

2

/M

2

=

W

1

M

2

W

2

M

1

Hence, by determining the vapour pressure of a pure solvent and a solution, the molar mass of a non-volatile solute can be measured.

Hope It helps...