Question

Verify that 0.200 mol of a nonvolatile solute in 125 g of benzene (C6H6) produces a solution

whose boiling point is 84.2 °C.

Answer 1

Number of moles of solute = 0.200 mol

Weight of solvent (Benzene) = 125 g = 0.125 kg

Molality of solution $= \frac{0.20}{0.125} = 1.6 \;m$

Elevation of boiling point $= ΔT_{bp} = K_{bp} m_{solute}$

$K_{bp}$ = Molal elevation boiling point constant

$K_{bp} = 2.53 \; °C/m$ for benzene

$ΔT_{bp} = 2.53 \times 1.6 = 4.048 \;°C$

Boiling point of solution = Boiling point of pure solvent + Elevation in boiling point

= 80.1 °C + 4.048 °C = 84.15 °C

Boiling point of solution = 84.15 °C