Calculate the boiling point of solution when 4 g of `MgSO_4 (M = 120 g "mol"^(-1))` was dissolved in 100 g of water, assuming `MgSO_4` undergoes complete ionisation.

Explanation

Apply the relation
` Delta T_b = i xx K_b xx m`
For `MgSO_4 , i = 2`
Molarity of solution = ` (4//120)/(100) xx 1000 = 4/120 xx 10 = 1/3`
Substituting the values in equation (i), we have
` Delta T_b = 2 xx 0.52 xx 1/3 = 0.347`
Boiling point of the solution = `100 + 0.347 = 100.347^@C = 373.347 K `