The vapour pressure of acetone at 20 o C is 185 torr . When 1 . 2 g of a non-volatile substance was dissolved in 100 g of acetone at 20 o C , its vapour pressure was 183 torr . The molar mass ( g m o l - 1 ) of the substance is:

Step-by-step solution

Δ P = 1 8 5 - 1 8 3 = 2 torr M CH 3 - C O || - CH 3 = 1 5 × 2 + 1 6 + 1 2 = 58 g/mol Δ P P 0 = 185 - 183 185 = 2 1 8 5 = X B = 1.2 M 1.2 M + 1 0 0 5 8 1.2 M << 1 0 0 5 8 ⇒   2 1 8 5 = 1.2 M × 5 8 1 0 0 M = 5 8 × 1.2 1 0 0 × 1 8 5 2 = 64.38 ≈ 64 g/mol