Question

A 15.3 g sample of a compound known to contain 3 ion species is dissolved in 110 g of water. The boiling point of the solution is 103.0 degrees Celsius. What is the molar mass of the compound?

Additional Info: Physical Properties of water

Freezing point: 0.00 degrees Celsius

Boiling point: 100 degrees Celsius

Kf: 1.86 Degrees Celsius/m

Kb: 0.512 degrees Celsius/m

Answer 1

Answer:

Boiling point depression is defined as

∆Tb=i x Kb x m

Where ∆Tb=(Tb)solution-(Tb)solvent,

i=Vant Hoff's factors=number of ions

Kb=boiling point constant

m= Molarity=(weight/molecular weight)(1000/weight of solvent).

Given (Tb)solution=103 °C, (Tb)solvent=100 °C,

i=3, Kb=0.512 °C/m, weight of sample=15.3 g, weight of solvent (water)=110 g.

Substitute all these values in above expression

∆Tb= i x Kb x m

(Tb) solution-(Tb) solvent= i x Kb x (weight/molecular weight)(1000/weight of solvent)

(103°C-100°C)=3 x 0.512 °C/m x (15.3g/molecular weight)(1000/110g).

Molecular weight=(213.643)/3=71.215 g/mol

Molar mass of sample =71.215 g/mol.