Question

If the boiling point elevation of an aqueous solution containing a non-volatile nonelectrolyte is 0.48◦C, what is the molality of the solution?

Answer in units of molality.

Answer 1

I got `"0.94 molal"` to two sig figs.

---

Boiling point elevation is given by:

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

where:

• `T_b` is the boiling point of the solvent in the context of the solution (in `""^@ "C"`), and `"*"` indicates pure solvent.
• `i` is the van't Hoff factor, i.e. the effective number of solute particles that have dissociated from the original solute particles.
• `K_b = 0.512^@ "C"cdot"kg/mol"` is the boiling point elevation constant of water.
• `m` is the molality of the solution in `"mols solute/kg solvent"`.

The solute is a nonelectrolyte, so `i = 1`. It also is nonvolatile, so it does not alter the vapor pressure other than through means of concentration at the surface.

Therefore,

`DeltaT_b = 0.48^@ "C" = (1)(0.512^@ "C"cdot"kg/mol")cdot m`

`=> color(blue)(m) = (0.48^@ "C")/((1)(0.512^@ "C"cdot"kg/mol"))`

`=` `color(blue)(0.93_8 " mols solute/kg solvent")`