Question

The molal boiling point elevation constant for water is 0.512 °C/m. A solution is prepared by dissolving 26.927 g of CaCl2 x H2O in 55.6 g of water. What is the boiling point of this solution (at 1.000 atm pressure)?

Show all work please. Thank you!

Answer 1

Lets calculate molality first

Molar mass of CaCl2H2O = 1*MM(Ca) + 2*MM(Cl) + 2*MM(H) + 1*MM(O)

= 1*40.08 + 2*35.45 + 2*1.008 + 1*16.0

= 128.996 g/mol

mass of CaCl2H2O = 26.927 g

we have below equation to be used:

number of mol of CaCl2H2O,

n = mass of CaCl2H2O/molar mass of CaCl2H2O

=(26.927 g)/(128.996 g/mol)

= 0.2087 mol

mass of solvent = 55.6 g

= 5.56*10^-2 kg [using conversion 1 Kg = 1000 g]

we have below equation to be used:

Molality,

m = number of mol / mass of solvent in Kg

=(0.2087 mol)/(0.0556 Kg)

= 3.754 molal

lets now calculate deltaTb

deltaTb = Kb*m

= 0.512*3.7544

= 1.9222 oC

This is increase in boiling point

boiling point of pure liquid = 100.0 oC

So, new boiling point = 100 + 1.9222

= 101.9222 oC

Answer: 101.9 oC