Question

A solution consists of 10.3 g of the nonelectrolyte glucose,C6H12O6 dissolved in 250.0 g of water. What is the freezing point depression of the solution?

Answer 1

The freezing point depression

$ΔT_f =k_fm \\ k_f = 1.86 \;°C/m$

Molality

$m = \frac{Number \;of \;moles\; of\; glucose}{Weight \;of \;solvent\; in\; kg}$

M(glucose) = 180 g/mol

Molality

$n(glucose) =\frac{10.3}{180} = 0.05722 \\ m= \frac{0.05722}{250 \times 10^{-3}} = 0.2288 \;m \\ ΔT_f =1.86 \times 0.2288 = 0.425 \;°C$