Question

Calculate the freezing point and boiling point of 543 ppm (1ppm=1mg/L) MgCl₂ solution in water, assuming complete dissociation. Mass of solution is 1000g/L.   (The K_b of water is 0.51 °C kg/mol and the K_f of water is 1.86 °C kg/mol

Answer 1

Apply Colligative properties

This is a typical example of colligative properties.

Recall that a solute ( non volatile ) can make a depression/increase in the freezing/boiling point via:

dTf = -Kf*molality * i

dTb = Kb*molality * i

where:

Kf = freezing point constant for the SOLVENT; Kb = boiling point constant for the SOLVENT;

molality = moles of SOLUTE / kg of SOLVENT

i = vant hoff coefficient, typically the total ion/molecular concentration.

At the end:

Tf mix = Tf solvent - dTf

Tb mix = Tb solvent - dTb

change mg to mass ; 543 mg of MgCl2 --> 0.543 g of MgCl2

MW of MgCl2 = 95.211 g/mol; i = 3 ions, Mg+2 and 2Cl-

kg of solvent = 1000 g/L --> 1 kg/L, assume 1 Liter balance

dTf = -1.86 * (0.543/95.211 ) / 1 * 3

dTf = -0.031823 °C

For BP

dTb = 0.512* (0.543/95.211 ) / 1 * 3

dTb =0.00875 °C

Tb = 100.00875 °C