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 %
