Question

the freezing point of 0.2 molal K2SO4 is -1.1°c calculate percentage degree of dissociation of K2SO4(kf for water is 1.86°)​

Answer 1

Answer:

Depression in Freezing Point =∆Tb x m. Where ∆Tb= Cryoscopic Constant= 1.86° C/m for water.

So Depression in Freezing Point =1.86 x 0.1= 0.186° C.

Freezing Point of Water=0°C.

Freezing Point of 0.1 m Urea

=0°C-(0.186) =-0.186°C= 273 0.186=272.814 K

Answer 2

Given:

Molality, m = 0.2 m

Freezing point, Tf = - 1.1 °C

Kf = 1.86

To Find:

The percentage degree of dissociation of K2SO4.

Calculation:

- Depression in freezing point, ΔTf = 0 -(-1.1)

⇒ ΔTf = 1.1

- For K2SO4, we have:

ΔTf = i × Kf × m

⇒ i = ΔTf / (Kf × m)

⇒ i = 1.1 / (1.86 × 0.2)

⇒ i = 1.1 / 0.372

⇒ i = 2.957

- For K2SO4, n = 3

- Degree of dissociation is given as:

α = (i - 1)/(n -1)

⇒ α = (2.957 - 1)/(3-1)

⇒ α = 1.957/2

⇒ α = 0.9785

⇒ % α = 0.9785 × 100

⇒ % α = 97.85 %

- So, the percentage degree of dissociation of K2SO4 is 97.85%.