Question

A solution containing 34.2 gram of cane sugar dissolved in 500 cm3 of water froze at -0.374degree celcius calculate the freezing point depression constant of water?

Answer 1

Answer: $1.87^0Ckgmol^{-1}$

Explanation:

Volume of solvent = 500 ml

Density of solvent = 1 g/ml

Mass of solvent =$Density\times Volume=1g/ml\times 500ml= 500 g=0.5kg$ (1 kg=1000 g)

Molar mass of cane sugar $(C_{12}H_{22}O_{11})$= 342 g/mol

Mass of cane sugar added = 34.2 g

$\Delta T_f=K_f\times \frac{\text {mass of cane sugar}}{\text {molar mass of cane sugar}}\times \text{weight of solvent in kg}}$

$\Delta T_f=T^{o}_f-T_f=0^0C-(-0.374)^0C=0.374^0C$

$0.374^0C=K_f\times \frac{34.2}{342\times 0.5}$

$K_f=1.87^0Ckgmol^{-1}$