Question

38.4 g of unknown substance (molar mass 384 g mol^-1) and 116 g of acetone is used to prepare a solution at 313 K. If vapour pressure of pure acetone (molar mass 58 g mol^-1) is 0.842 atmosphere, what is the vapour pressure of solution?

Options

• 0.7999 atm

• 0.880 atm

• 0.650 atm

• 0.958 atm

Answer 1

0.7999 atm

Explanation:

`"n"_2 = 38.4/384` = 0.1 mol solute

`"n"_1 = 116/58` = 2 mol solvent (acetone)

Mole fraction ; `x_1 = "n"_1/("n"_1 + "n"_2) = 2/(2 + 0.1)` = 0.95

By Raoult's law, the vapour pressure of the solution is given by

P = x₁ × P₀ = 0.95 × 0.842 atm

= 0.7999 atm