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Sum

**Answer the following in one or two sentences.**

A 0.1 m solution of K_{2}SO_{4} in water has a freezing point of –4.3°C. What is the value of van’t Hoff factor if K_{f} for water is 1.86 K kg mol^{–1}?

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#### 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} = `"T"_"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 = `(triangle "T"_"f")/("K"_"f" * "m")`

`= (4.3 "K")/(1.86 "k kg mol"^-1 xx 0.1 "mol kg"^-1)`

= 2.311

The value of van’t Hoff factor is 2.311.

Concept: Colligative Properties of Electrolytes

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