Question

Calcium Chloride is frequently used to melt ice on sidewalks. Calculate the new freezing point and boiling point found when 25.6g of CaCl2 is dissolved in 260.0 grams of water.

Answer 1

a) for freezing point:

Molar mass of CaCl2,

MM = 1*MM(Ca) + 2*MM(Cl)

= 1*40.08 + 2*35.45

= 110.98 g/mol

mass(CaCl2)= 25.6 g

number of mol of CaCl2,

n = mass of CaCl2/molar mass of CaCl2

=(25.6 g)/(110.98 g/mol)

= 0.2307 mol

m(solvent)= 260.0 g

= 0.26 Kg

Molality,

m = number of mol / mass of solvent in Kg

=(0.2307 mol)/(0.26 Kg)

= 0.8872 molal

i for CaCl2 is 3

use:

Δ Tf = i*Kf*mb

Δ Tf = 3.0*1.86 oC/m * 0.8872 m

Δ Tf = 4.95 oC

This is decrease in freezing point

Freezing point of pure water is 0 oC

So,

new freezing point is -4.95 oC

Answer: -4.95 oC

b) for boiling point:

use:

Δ Tb = i*Kb*mb

0.0 = 3.0*0.512*0.8872

Δ Tf = 1.36 oC

This is increase in boiling point

boiling point of pure water is 1000 oC

So,

new freezing point is 101.36 oC

Answer: 101.36 oC