Question

The vapour pressure of pure benzene (molar mass 78 g/mol) at a certain temperature is 640 mm Hg. A nonvolatile solute of mass 2.315 g is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. The molar mass of solute is ____________.

Options

• 64 g mol⁻¹

• 72 g mol⁻¹

• 92 g mol⁻¹

• 142 g mol⁻¹

Answer 1

The vapour pressure of pure benzene (molar mass 78 g/mol) at a certain temperature is 640 mm Hg. A nonvolatile solute of mass 2.315 g is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. The molar mass of solute is 72 g mol⁻¹.

Explanation:

`("P"_1^0 - "P"^1)/("P"_1^0) = ("W"_2"M"_1)/("M"_2"W"_1)`

∴ `(640  "mm Hg" - 600  "mm Hg")/(640  "mm Hg") = (2.315  "g" xx 78  "g mol"^-1)/("M"_2 xx 40  "g")`

∴ `(40  "mm Hg")/(640  "mm Hg") = (2.315  "g" xx 78  "g mol"^-1)/("M"_2 xx 40  "g")`

∴ M₂ = `(2.315  "g" xx 78  "g mol"^-1 xx 640  "mm Hg")/(40  "mm Hg" xx 40  "g")`

= 72 g mol⁻¹