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? - Chemistry

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Sum

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?

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Solution

Given: Molality of K2SO4 solution = m = 0.1 m

Freezing point of solution = Tf = – 4.3°C

Kf of water = 1.86 K kg mol–1

To find: van’t Hoff factor

Formula: ΔTf = i Kf m

Calculation: 

ΔTf = `"T"_"f"^0` – Tf

= 0 °C – (–4.3 °C) = 4.3 °C = 4.3 K

Now, using formula,

ΔTf = i Kf 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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Chapter 2: Solutions - Exercises [Page 45]

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Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 2 Solutions
Exercises | Q 2.04 | Page 45
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