Question

A 100.0-mL aqueous sodium chloride solution is 14.3% NaCl by mass and has a density of 1.12 g/mL.

What mass of solute would you add to make the boiling point of the solution 104.4 ∘C? (Use i = 1.8 for NaCl)

Express your answer to two significant figures and include the appropriate units.

Answer 1

The mass of 100 mL solution will be 1.12 x 100 = 112 g

112 g of this solution will have ( 14.3 x 112) / 100 = 16.016 g of NaCl

mass of water = 112-16.016 = 95.984 g

Boiling point elevation = i.kb.m

where kb is boiling point constant of water = 0.512 K Kg/mol

m is the molality

4.4 = 1.8 x 0.512 x m

m = 4.77 m

m = no. of moles / mass of solvent in kg

4.77 = n / 0.095984

n = 0.458 moles

0.458 moles of NaCl = 0.458 x 58.44 = 26.8 g

So the amount of NaCl to be added into the solution will be

26.8 - 16.016 = 10.784 grams