Question

Calculate the freezing point and the boiling point of a solution that contains 35 g of C₁₂H₂₂O₁₁ in 370 g of water.

Answer 1

Moles of $C_{12}H_{22}O_{11}=\frac{35}{342}=0.102moles$

Mass of water $= 370g = 0.37kg$

Molarity $= \frac{0.102}{0.37}=0.28M/kg$

Kf= 1.86°c

Depression = $0.28×1.86=0.52°c$

Freezing point $=0-0.52=-0.52 °c$

Kb=0.52°c

Elevation = $0.28×0.52=0.15$

Boiling point =$100+0.15=100.15°c$