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?

पर्याय

• 0.7999 atm

• 0.880 atm

• 0.650 atm

• 0.958 atm

Answer 1

0.7999 atm

Explanation:

${\displaystyle {\text{n}}_2=\frac{38.4}{384}}$ = 0.1 mol solute

${\displaystyle {\text{n}}_1=\frac{116}{58}}$ = 2 mol solvent (acetone)

Mole fraction ; ${\displaystyle x_1=\frac{{\text{n}}_1}{{\text{n}}_1+{\text{n}}_2}=\frac{2}{2+0.1}}$ = 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