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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. (Kb for water = 0.52 K kgmol-1)
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Solution
Given:
Kb = 0.52 K kg mol-1
Mass of MgSO4 (solute) = 4 g
Mass of Water (Solvent) = 100 g
Molarity of solution =
`(4/12)/(100/1000)= 0.33 " mol/L"`
MgSO4 undergoes complete ionisation, So, i = 2
Elevation in boiling point is given as,
ΔTb = i x Kb x m
= 2 x 0.52 x 0.33 = 0.34 K
Tf = 373.15 + 034
= 373.49 K
Boiling point of the solution is 373.49 K.
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