Vapour pressure of pure water at the boiling point (P0)=1 atm =1.013 bar
Vapour pressure of solution (Ps)=1.004bar
Let mass of solution = 100g , then, mass of solute = ( w2 ) = 2g
Mass of solvent (w1)=100 – 2=98g
Applying Raoult’s law for dilute solution,

P0−Ps

P0

=

n2

n1+n2

=

n2

n1

=

w2 M2

w1 M1

=

w2

M2

×

M1

w1

1.013−1.004

1.013

=

2 g

M2

×

18 g mol−1

98 g

or, M2=

2×18

98

×

1.013

0.009

g mol−1 =41.35 g mol−1