Question

Calculate the freezing point of an aqueous solution containing 10.5 g of Magnesium bromide in 200 g of water, assuming complete dissociation of Magnesium bromide. (Molar mass of Magnesium bromide = 184 g mol–1, Kf for water = 1.86 K kg mol–1).

Answer 1

The freezing point of the solution will be 272.47 K

Explanation :

Given data,

W₂ = 10.5 g

W₁ = 200 g

Molar mass of MgBr, M₂ = 184 g

Kf of water = 1.86

Hence the change in freezing point is given by the relation,

ΔTf = (1000 x Kf x W₂)/(W₁ x M₂)

= (1000 x 1.86 x 10.5)/(200 x 184)

=> ΔTf = 0.53 K

Hence new freezing point ,

Tf = T₀ - ΔTf = 273 - 0.53 = 272.47 K

Hence the freezing point of the solution will be 272.47 K

Answer 2

Given,

Mass of solute(MgBr2),Wb = 10.50 g

Mass of solvent(water),Wa = 200 g

Molar mass of MgBr2,Mb = 184 g/mol

Kf for water = 1.86 K kg / mol

we have to calculare, the freezing point of solution, Tf = ?

we know the freezing point of pure water, Tf° = 0°C = 273 K

because,

∆Tf = Kf * m

∆Tf = Kf * Wb * 1000

Mb * Wa

= 1.86 * 10.50 * 1000

184 * 200

=> ∆Tf = 0.53 K

we also know that,

∆Tf = Tf° - Tf

0.53 = 273 - Tf

=> Tf = 273 - 0.53

= 272.47 K

Hence, freezing point of solution is 272.47 K