The freezing point of a solution of 20.5 g NH 4 Br in 100 g of water is − 6. 2 ∘ C . The freezing point of a solution of 3.42 g cane-sugar in 100 g of water is − 0.18 5 ∘ C . The degree of ionisation of salt is -

Δ T f = im k f ⇒ 0.185 = 342 × 0.1 1 × 3.42 × k f ⇒ k f = 1.85 6.2 = 98 × 0.1 i × 20.5 × 1.85 ⇒ i = 1.6 i = 1 + α ( n = 2 ) 1.6 = 1 + α ⇒ α = 0.6 or 60%

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The freezing point of a solution of $20.5 g NH_{4} Br$ in 100 g of water is $- 6. 2^{\circ} C$ . The freezing point of a solution of 3.42 g cane-sugar in 100 g of water is $- 0.18 5^{\circ} C$ . The degree of ionisation of salt is -

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Answer:

Solution:

$Δ T_{f} = im k_{f}$

$\Rightarrow 0.185 = \frac{1 \times 3.42 \times k _{f}}{342 \times 0.1} \Rightarrow k_{f} = 1.85 6.2 = \frac{i \times 20.5 \times 1.85}{98 \times 0.1} \Rightarrow i = 1.6 i = 1 + α (n = 2) 1.6 = 1 + α \Rightarrow α = 0.6 or 60%$

Stream:JEESubject:ChemistryTopic:Liquid SolutionSubtopic:Colligative Properties

⏱ 2mℹ️ Source: QB

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