An aqueous solution of MgCl2 freezes at -3.2 o C. Calculate the boiling point for this solution. Note that for water, Kb= 0.51oC/m; Kf= 1.86oC/m
Here the solute is MgCl2 and solvent is water.
We have:
Freezing point of this solution, Tf = -3.2 oC
Freezing point of water = 0oC
Therefore, Depression of freezing point, ΔTf = (0 - (-3.2)oC = 3.2oC
Freezing point depression constant of water, Kf = 1.86oC/m
Hence, Molality (m) of the solution is given by:
Elevation of boiling point of this solution is given by:
We have, Boiling point of water = 100oC
Therefore. Boiling point of this solution is given by:
---- (answer)
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