At (298 ~K ), the vapour pressure of a solution of (7.5 ~g ) of non-volatile solute in (90 ~g ) of water is (2.8 kPa ). If (18 ~g ) of water is added to this solution the vapour pressure becomes (2.81 kPa ) at same temperature, the molar mass of solute in ( g mol ⁻¹ ) is

Step-by-step solution

Key Idea Relation between relative lowering of vapour pressure and molecular mass of solute is given by ( p^ -p p^ = w₂ / M₂ w₁ M₁ ) Given, Weight of non-volatile solute, (w₂=7.5 ~g ) Weight of water, (w₁=90 ~g ) Vapour pressure of solution (=2.8 kPa ) In first case, ( aligned p^ -2.8 p^ & = 7.5 / M₂ 90 / 18 p^ -2.8 p^ & = 1.5 M₂ (i) aligned ) In second case, ( p^ -2.81 p^ = 7.5 / M₂ 108 / 18 = 1.25 M₂ (ii) ) On solving (i) and (ii) we get (M₂=71.5 ~g ~mol ⁻¹ )