Question

If the boiling point increases by `"0.126 K"` for a solution of a nonelectrolyte in benzene, what molality was required? `K_b = "2.52 K" cdot "kg/mol"`.

Answer 1

We assume the process is at `"1 atm"`, where the normal boiling point would have been `80.1^@ "C"`. But since we're given `DeltaT_b`, we don't need to use that.

The boiling point elevation is given by

`DeltaT_b = T_b - T_b^"*" = iK_bm`,

where:

• `T_b` is the boiling point.
• `"*"` indicates the pure solvent.
• `i` is the van't Hoff factor, i.e. the effective number of particles per dissociated formula unit.
• `K_b = "2.52 K/m"` is the boiling point elevation constant.
• `m` is the molality in `"mol solute/kg solvent"`.

Benzene is the solvent to this nonelectrolyte. So, the molality is given by

`color(blue)(m) = (DeltaT_b)/(iK_b)`

`~~ (0.126 cancel"K")/(1cdot2.52 cancel"K"cdot"kg/mol")`

`~~` `color(blue)("0.05 mol/kg")`