Question

An aqueous solution of MgCl₂ freezes at -3.2 ^o C. Calculate the boiling point for this solution. Note that for water, K_b= 0.51^oC/m; K_f= 1.86^oC/m

Answer 1

Here the solute is MgCl₂ and solvent is water.

We have:

Freezing point of this solution, T_f = -3.2 ^oC

Freezing point of water = 0^oC

Therefore, Depression of freezing point, ΔT_f = (0 - (-3.2)^oC = 3.2^oC

Freezing point depression constant of water, K_f = 1.86^oC/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 = 100^oC

Therefore. Boiling point of this solution is given by:

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