Question

Calculate the boiling point of solution when 4g of MgSO₄ (M= 120 g mol^-1) was dissolved in 100g of water, assuming MgSO4 undergoes complete ionization. (K_b for water = 0.52 K kgmol^-1)

Answer 1

Given:

K_b =  0.52 K kg mol^-1

Mass of MgSO₄ (solute) = 4 g

Mass of Water (Solvent) = 100 g

Molarity of solution = 

${\displaystyle \frac{\frac{4}{12}}{\frac{100}{1000}}=0.33\text{mol/L}}$

MgSO₄ undergoes complete ionisation, So, i = 2

Elevation in boiling point is given as,

ΔT_b = i x K_b x m

= 2 x 0.52 x 0.33 = 0.34 K

T_f  = 373.15 + 034

= 373.49 K

Boiling point of the solution is 373.49 K.