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Question
Calculate the freezing point of solution when 1.9 g of MgCl2 (M = 95 g mol−1) was dissolved in 50 g of water, assuming MgCl2 undergoes complete ionization.
(Kf for water = 1.86 K kg mol−1)
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Solution
Given:
Kf = 1.86 K kg mol−1
Mass of solute = 1.9 g
Mass of solvent = 50 g
Therefore,
`"Molality of the solution, "m=1.9/95xx1000/50=0.4m`
Also, MgCl2 undergoes complete ionisation and thereby yielding 3 moles of constituent ions for every mole of MgCl2.
∴ i = 3
Now, depression in freezing point is given as
∆Tf = iKfm
= 3×1.86×0.4
= 2.232 K
Tf = 273.15−2.232
=270.918 K
Hence, the new freezing point of the solution is 270.92 K.
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