Question

A solution is made by dissolving 49.3 g Na₂CO₃ IN 550.0g of water. What is the boiling point of this solution?

Answer 1

first find the molality

m = (49.3 g / 106 g/mol ) x 1000 / 550 g

m = 0.465 x 1.82 = 0.85 m

use the formula

Tb = m x Kb x i

where

Tb = boiling point of the solution - boiling point of the pure wate

m = molality = 0.85

Kb for watere = 0.52 ºC/ m

i = vant haf factor = 3 for Na2CO3 (when dissociate in water it will give 2Na+ ions and one CO32-)

plug in values in above equation

boiling point of solution - 100 ºC = 0.85 m x 0.52 ºC / m x 3 = 1.326

boiling point of the solution = 100 ºC + 1.326 ºC = 101.326 ºC