##### Exercise | Q 2.04 | Page 45

# Answer the following in one or two sentences.

##### A 0.1 m solution of K2SO4 in water has a freezing point of – 4.3 °C. What is the value of van’t Hoff factor if Kf for water is 1.86 K kg mol–1?

##### Solution**Given:** Molality of K_{2}SO_{4} solution = m = 0.1 m

Freezing point of solution = T_{f} = – 4.3 °C

K_{f} of water = 1.86 K kg mol^{–1}

**To find:** van’t Hoff factor

**Formula:** ΔT_{f} = i K_{f} m

**Calculation: **

ΔT_{f} = $\text{T}}_{\text{f}}^{0$ - T_{f}

= 0 °C - (- 4.3 °C) = 4.3 °C = 4.3 K

Now, using formula,

ΔT_{f} = i K_{f} m

∴ i = $\frac{\u25b3{\text{T}}_{\text{f}}}{{\text{K}}_{\text{f}}\cdot \text{m}}$

$=\frac{4.3\text{K}}{1.86{\text{k kg mol}}^{-1}\times 0.1{\text{mol kg}}^{-1}}$

= 23.1

The value of van’t Hoff factor is 23.1.

**Given:** Molality of K_{2}SO_{4} solution = m = 0.1 m

Freezing point of solution = T_{f} = – 4.3 °C

K_{f} of water = 1.86 K kg mol^{–1}

**To find:** van’t Hoff factor

**Formula:** ΔT_{f} = i K_{f} m

**Calculation: **

ΔT_{f} = _{f}

= 0 °C - (- 4.3 °C) = 4.3 °C = 4.3 K

Now, using formula,

ΔT_{f} = i K_{f} m

∴ i =

= 23.1

The value of van’t Hoff factor is 23.1.