Question

Calculate the boiling point of a solution containing 3.5 g sugar (molar mass = 342 g)

dissolved in 150 g of water. Kb for water is 0.52 oC/m.

Answer 1

Moles of sugar $=\frac{3.5}{342}=0.01moles$

Molarity $=\frac{0.01}{0.150}=0.067M$

$∆T_b=0.067×0.52=0.03°c$

Boiling point $=100+0.03=100.03°c$