Question

Question #7b124

Answer 1

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_"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 × `"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 %

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