Question

What is the freezing point of a solution prepared from 75.0 g ethylene glycol (C₂H₆O₂) and 100.0 g H₂O?  K_f of water is 1.86°C/m.

	A.	22.5°C
	B.	225°C
	C.	-1.50°C
	D.	-22.5°C

Answer 1

Lets calculate molality first

Molar mass of C2H6O2,

MM = 2*MM(C) + 6*MM(H) + 2*MM(O)

= 2*12.01 + 6*1.008 + 2*16.0

= 62.068 g/mol

mass(C2H6O2)= 75 g

use:

number of mol of C2H6O2,

n = mass of C2H6O2/molar mass of C2H6O2

=(75 g)/(62.07 g/mol)

= 1.208 mol

m(solvent)= 100 g

= 0.1 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(1.208 mol)/(0.1 Kg)

= 12.08 molal

lets now calculate ΔTf

ΔTf = Kf*m

= 1.86*12.0835

= 22.5 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 22.5

= -22.5 oC

Answer: D