# 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. - Chemistry

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)

#### Solution

Given:

Kf = 1.86 K kg mol1

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

T273.152.232

=270.918 K

Hence, the new freezing point of the solution is 270.92 K.

Concept: Colligative Properties and Determination of Molar Mass - Depression of Freezing Point
Is there an error in this question or solution?