Question

Calculate the freezing point of a solution containing 13.2 gg FeCl3FeCl3 in 159 gg water.

Calculate the boiling point of the solution above.

Calculate the freezing point of a solution containing 5.9 %% KClKCl by mass (in water).

Calculate the boiling point of the solution above.

Calculate the freezing point of a solution containing 0.151m MgF2

Calculate the boiling point of the solution above.

Answer 1

Freezing point depression of electrolyte

= i K_f × m

K_f of water = 1.86 ^oc/m

Molality = moles of solute /mass of solvent (Kg)

Boiling point elevation

= i × Kb × m

Kb of water = 0.512⁰ c /m

i is van't hoff factor

For strong electrolyte ; i = total number of ions in solution

1.

Moles of FeCl₃ = ( mass /molar mass)

= (13.2/162)

= 0.08

Mass of water = 159 g = (159/1000) = 0.159 Kg

Molality of of FeCl₃ = ( 0.08/0.159) = 0.5

In solution FeCl₃ dissociates as

FeCl₃ Fe³⁺ + 3 Cl^-

Van't hoff factor (i) for FeCl₃ = 4

hence,

= 4 × 1.86 × 0.5

Or, = 3.74^o c

Then , freezing point of solution

= freezing point of pure water -

= 0 - 3.74 = - 3.74⁰c.

Now, Boiling point elevation

= i × Kb × m

= 4 × 0.512 × 0.5

= 1.024 ^oc

Boiling point of solution = (100 + 1.024)⁰c = 101.024⁰ c .

2.

In 5.9% KCl by mass (in water)

Then 5.9 g of KCl present per 100 g solution

KCl dissociates to K⁺ and Cl^- ions, so i = 2

moles of KCl

= ( Mass/molar mass)

= (5.9/74.5)

= 0.079

Mass of water = 100 g - 5.9 g = 94.1 g

= (94.1/1000) = 0.0941 Kg.

Molality (m) = (0.079/0.0941) = 0.84

Freezing point depression

= i × Kf × m

= 2 × 1.86 × 0.84

= 3.13⁰c.

Freezing point of solution

= Freezing point of pure water -

= ( 0 - 3.13)⁰ c

= - 3.13⁰c .

Kb of water = 0.512⁰c

Then,

Boiling point elevation

= i × Kb × m

Or,

= 2 × 0.512 × 0.84

= 0.86 ^oc.

Hence, boiling of solution = boiling point of pure water +

= 100 + 0.86 = 100.86 ⁰c.

3.

m = 0.151

MgF₂ dissociates to Mg²⁺ and 2 F^- ions

So, i = 3

Freezing point depression

= i × Kf × m

= 3 × 1.86 × 0.151

= 0.842^o c.

Then, freezing point of solution

= 0 - 0.842 = - 0.842⁰ c

Now, boiling point elevation is

= i × Kb × m

= 3 × 0.512 × 0.151

= 0.232^o c.

So, boiling point of water

= (100 + 0.232 )⁰ c = 100.232⁰ c .