Question

In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (molar mass = 58 g mol⁻¹) at 298 K. The vapour pressure of the solution was found to be 192.5 mm Hg. The molecular weight of the solute is (vapour pressure of acetone = 195 mm Hg) ____________ g mol⁻¹.

पर्याय

• 25.24

• 35.24

• 45.24

• 55.24

Answer 1

In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (molar mass = 58 g mol⁻¹) at 298 K. The vapour pressure of the solution was found to be 192.5 mm Hg. The molecular weight of the solute is (vapour pressure of acetone = 195 mm Hg) 45.24 g mol⁻¹.

Explanation:

`"P"_"A"^0` = vapour pressure of pure solvent (acetone)

= 195 mm Hg,

P = vapour pressure of solution

= 192.5 mm Hg,

W_A = amount of solvent = 100 g,

W_B = amount of solute = 1 g,

M_A = Molar mass of solvent = 58 g mol⁻¹ and

M_B = Molar mass of solute

`("P"_"A"^0 - "P")/("P"_"A"^0) = ("W"_"B""M"_"A")/("M"_"B""W"_"A")`

`(195  "mm Hg" - 192.5  "mm Hg")/(195  "mm Hg") = (1  "g" xx 58  "g mol"^-1)/("M"_"B" xx 100  "g")`

M_B = `195/2.5 xx 58/100` g mol⁻¹

= 45.24 g mol⁻¹