Question

What is the new boiling point of a solution made with 35.7 g MgCl₂ and 150 mL water? (density of water is 1 g/mL)

Answer 1

ΔTb = i*Kb*m
i for MgCl2 is 3 as it dissociates into 1 Mg2+ and 2 Cl-

step 1: calculate molality

Molar mass of MgCl2,
MM = 1*MM(Mg) + 2*MM(Cl)
= 1*24.31 + 2*35.45
= 95.21 g/mol


mass(MgCl2)= 35.7 g

use:
number of mol of MgCl2,
n = mass of MgCl2/molar mass of MgCl2
=(35.7 g)/(95.21 g/mol)
= 0.375 mol

m(solvent)= density * volume
= 150 mL * 1 g/ml
= 150 g
= 0.150 kg

use:
Molality,
m = number of mol / mass of solvent in Kg
=(0.375 mol)/(0.15 Kg)
= 2.5 molal

lets now calculate ΔTb
ΔTb = i*Kb*m
= 3.0*0.512*2.4997
= 3.84 oC

So, boiling point = 100 + 3.84
= 103.84 oC