# Question 7b124

Sep 23, 2014

The solution should contain 21 % sucrose by mass.

This is really two problems: (a) What molality of solution will give the observed boiling point? (b) What is the percent composition of this solution?

Step 1. Calculate the molality of the solution.

ΔT_"b" = iK_"b"m

 ΔT_"b" = (100 – 99.60) °C = 0.40 °C (Technically, the answer should be 0 °C, because your target boiling point has no decimal places).

${K}_{\text{b}}$ = 0.512 °C·kg·mol⁻¹

m = (ΔT_"b")/(iK_"b") = "0.40 °C"/("1 × 0.512 °C·kg·mol"^-1) = 0.78 mol·kg⁻¹

Step 2. Calculate the percent composition.

Moles of sucrose = 0.78 mol sucrose × $\text{1 mol sucrose"/"342.30 g sucrose}$ = 270 g sucrose

So you have 270 g sucrose in 1 kg water.

% by mass ="mass of sucrose"/"mass of sucrose + mass of water" × 100 % ="270 g"/"270 g + 1000 g" × 100 % =

"270 g"/"1270 g" × 100 %# = 21 %