Question

45 g of ethylene glycol (c2h6o2) is mixed with 600 g water. calculate: 1) freezing point depression 2) freezing point of solution given that kf for water = 1.86 k kg/mol.

Answer 1

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Answer 2

Answer: The freezing point of the solution will be 275.24 k.

Explanation:

Mass of the ethylene glycol = 45 g

Weight of of solvent  = 600 g = 0.6 kg

$K_f=1.86Kkg/mol$

$T_o$ = freezing point of pure water = 273 K

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

$=1.86 K kg/mol\times \frac{45}{62.07 g/mol\times 0.6 kg}=2.24 K$

$Delta T_f=2.24 K=T_o-T=273 K-T$

T = 275.24 K

The freezing point of the solution will be 275.24 k.