Question

The normal freezing point of water, H₂O is 0.00 °C and its K_fp value is 1.86 °C/m.

Assuming complete dissociation of the electrolyte, if 10.87 grams of cobalt(II) nitrate (Co(NO₃)₂, 183.0 g/mol) are dissolved in 289.5 grams of water what is the freezing point of the solution?

Answer 1

Lets calculate molality first

mass of solute = 10.87 g

we have below equation to be used:
number of mol of solute,
n = mass of solute/molar mass of solute
=(10.87 g)/(183 g/mol)
= 5.94*10^-2 mol

mass of solvent = 289.5 g
= 0.2895 kg     [using conversion 1 Kg = 1000 g]

we have below equation to be used:
Molality,
m = number of mol / mass of solvent in Kg
=(5.94*10^-2 mol)/(0.2895 Kg)
= 0.2052 molal

Co(NO3)2 breaks into 1 Co2+ and 2 NO3-
So, number of ions = 3
i=3

lets now calculate deltaTf
deltaTf = i*Kf*m
= 3.0*1.86*0.2052
= 1.1449 oC

This is decrease in freezing point
freezing point of pure liquid = 0.0 oC
So, new freezing point = 0 - 1.1449
= -1.14 oC
Answer: -1.14 oC