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NEET CHEMISTRYSolutionsMedium

Question

0.50.5 molal aqueous solution of a weak acid (HX\text{HX}) is 20%20\% ionised. If KfK_f for water is 1.86 K kg mol11.86 \text{ K kg mol}^{-1}, the lowering in freezing point of the solution is:

A

1.12 K-1.12 \text{ K}

B

0.56 K0.56 \text{ K}

C

1.12 K1.12 \text{ K}

D

0.56 K-0.56 \text{ K}

Step-by-Step Solution

For the weak acid HX\text{HX}, the ionization reaction is: HXH++X\text{HX} \rightleftharpoons \text{H}^+ + \text{X}^- The number of ions produced per molecule upon dissociation, n=2n = 2. Given, the degree of ionization α=20%=0.2\alpha = 20\% = 0.2.

The van't Hoff factor (ii) is calculated as: i=1+α(n1)=1+0.2(21)=1.2i = 1 + \alpha(n - 1) = 1 + 0.2(2 - 1) = 1.2

The depression (lowering) in freezing point is given by the formula: ΔTf=i×Kf×m\Delta T_f = i \times K_f \times m

Given values: Molality, m=0.5 mol kg1m = 0.5 \text{ mol kg}^{-1} Molal depression constant, Kf=1.86 K kg mol1K_f = 1.86 \text{ K kg mol}^{-1}

Substituting the values: ΔTf=1.2×1.86×0.5=1.116 K1.12 K\Delta T_f = 1.2 \times 1.86 \times 0.5 = 1.116 \text{ K} \approx 1.12 \text{ K}.

Therefore, the lowering in freezing point of the solution is 1.12 K1.12 \text{ K}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Solutions. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYSolutionsaqueoussolutiontexthxionisedlowering

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