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Calculate the boiling point of solution when 4g of MgSO_{4} (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})

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#### Solution

Given:

K_{b} = 0.52 K kg mol^{-1}

Mass of MgSO_{4} (solute) = 4 g

Mass of Water (Solvent) = 100 g

Molarity of solution =

`(4/12)/(100/1000)= 0.33 " mol/L"`

MgSO_{4} 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.

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