Question

The freezing point (in °C) of solution containing 0.18 g of glucose in 100 g water is (Kf = 1.86 K kg mol–1) is

Answer 1

Given

$m_g$ = Mass of glucose = 0.18 g

$m_s$ = Mass of solvent = 100 g

$K_f$ = Molal freezing point of water = 1.86 K kg/mol

To find

Freezing point of solution

Solution

M = Molar mass of glucose = 180.156 g/mol

Number of moles of glucose

$n_g=\dfrac{m_g}{M}\\\Rightarrow n_g=\dfrac{0.18}{180.156}\\\Rightarrow n_g=0.99\times 10^{-3}\ \text{mol}$

Molality is given by

$m=\dfrac{n_g}{m_s}=\dfrac{0.99\times 10^{-3}}{0.1}\\\Rightarrow m=0.0099\ \text{mol/kg}$

Depression in the freezing point of the solution

$\Delta T_f=K_fm\\\Rightarrow \Delta T_f=1.86\times 0.0099\\\Rightarrow \Delta T_f=0.018414\ \text{K}$

Freezing point of water = $0^{\circ}\text{C}$

Freezing point of the solution is $0-0.018414=-0.018414^{\circ}\text{C}$.