Question

The quantity 45.20 g of sucrose is dissolved in 316.0 g of water. Calculate (a) boiling point and(b) freezing point of the solution

Answer 1

Boiling point of pure water = 100 ºC

$ΔT_b = k_b\times m_s \\ k_b = 0.51 \;C \;kg/mol$

M(sucrose) = 342.3 g/mol

$n(sucrose) = \frac{45.20}{342.3} = 0.132 \;mol \\ Molality(sucrose) = \frac{0.132 \;mol}{0.316 \;kg} = 0.417 \; mol/kg \\ ΔT_b = 0.51 \times 0.417 = 0.213 \;ºC$

Boiling Point of Solution = 100 + 0.213 = 100.21 ºC

$ΔT_f$ = freezing point of a solvent - freezing point of a solution

$ΔT_f = T_{f_0} – T_f \\ ΔT_f = k_f \times M_s \\ k_f = 1.86 \;C \;kg/mol \\ ΔT_f = 1.86 \times 0.417 = 0.77 \;ºC \\ T_f = 0.0 – 0.77 = -0.77 ºC$