Question

If `"NaCl"` is dissolved in water to raise its boiling point by `2.10^@ "C"`, what is the molarity? `K_b = 0.512^@ "C/m"`.

Answer 1

I assume you mean molality.

The boiling point elevation `DeltaT_b` is defined in terms of the molality `m`, boiling point elevation constant `K_b`, and van't Hoff factor `i`:

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

The van't Hoff factor for `"NaCl"` is ideally `2` (the actual value is `1.9`), as 100% dissociation leads to two ions per formula unit, and thus the molality is:

`DeltaT_b = 102.10^@ "C" - 100.00^@ "C" = (2)(0.512^@ "C"cdot"kg/mol")m`

`=>m = (2.10cancel(""^@ "C"))/(2 cdot 0.512cancel(""^@ "C")cdot"kg/mol")`

`=>color(blue)(m = "2.051 mol solute/kg solvent")`

If you want the molarity, you'll have to provide the new volume of the solution.