Question

Calculate the boiling point and freezing point of a solution containing 478g of ethylene glycol in 3202g of water. Molar mass of ethylene glycol is 62.01g/mol.

Answer 1

Boiling point of pure water = 100 ºC

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

n(ethylene glycol) $= \frac{478}{62.01} = 7.71 \;mol$

Molality(ethylene glycol) $= 7.71 \;mol \times 3.202 \;kg = 2.41 \; mol/kg$

$ΔT_b = 0.51 \times 2.41 = 1.23 \;ºC$

Boiling Point of Solution = 100 + 1.23 = 101.23 ºC

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

$ΔT_f = T_{f0} - T_f \\ ΔT_f = k_f \times M_s \\ k_f = 1.86 \;C \;kg/mol \\ ΔT_f = 1.86 \times 2.41 = 4.48 \;ºC$

$T_f = 0.0 – 4.48 = -4.48\;ºC$