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The freezing point (in °C) of a solution containing 0.1 g of K3[Fe(CN)6] in 100 g water in which it is 80% ionised is [Kf (H20) = 1.86° mol-1 kg]
  • a)
    -2.3 x 10-2
  • b)
    -5.7 x 10-2
  • c)
    -5.7 x 10-3
  • d)
    -1.92 x 10-2
Correct answer is option 'D'. Can you explain this answer?
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Solution:

Given,
Mass of K3[Fe(CN)6] = 0.1 g
Mass of water = 100 g
% ionisation = 80%
Kf (H2O) = 1.86 mol-1kg

Molar mass of K3[Fe(CN)6] = 329 g mol-1

Number of moles of K3[Fe(CN)6] = 0.1/329 = 3.040 x 10^-4 mol

Number of moles of ions produced = 3 x 3.040 x 10^-4 = 0.000912 mol

Total number of moles in the solution = 0.000912 + 0.100 = 0.100912 mol

Molality = Number of moles of solute/ Mass of solvent in kg
= 0.000912/0.100 = 0.00912 mol kg^-1

Freezing point depression, ΔTf = Kf × molality

ΔTf = 1.86 × 0.00912 = 0.01699 K

The solution is 80% ionised, so the number of ions produced is 1.6 times the number of moles of solute.

Number of moles of ions produced = 1.6 x 0.000912 = 0.0014592 mol

Total number of moles in the solution = 0.100912 + 0.0014592 = 0.1023712 mol

Molality = 0.0014592 /0.100 = 0.014592 mol kg^-1

ΔTf = Kf × molality = 1.86 × 0.014592 = 0.027202 K

The freezing point depression is 0.027202 K, which is equal to 27.202°C.

The freezing point of the solution is equal to the freezing point of the solvent minus the freezing point depression.

Freezing point of water = 0°C

Freezing point of the solution = 0 - 27.202 = -27.202°C

Hence, the freezing point of the given solution is -27.202°C, which is equal to -1.92 × 10^-2°C (rounded off to 3 significant figures).

Therefore, the correct option is (D).
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The freezing point (in C) of a solution containing 0.1 g of K3[Fe(CN)6] in 100 g water in which it is 80% ionised is [Kf (H20) = 1.86 mol-1kg]a)-2.3 x 10-2b)-5.7 x 10-2c)-5.7 x 10-3d)-1.92 x 10-2Correct answer is option 'D'. Can you explain this answer?
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The freezing point (in C) of a solution containing 0.1 g of K3[Fe(CN)6] in 100 g water in which it is 80% ionised is [Kf (H20) = 1.86 mol-1kg]a)-2.3 x 10-2b)-5.7 x 10-2c)-5.7 x 10-3d)-1.92 x 10-2Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The freezing point (in C) of a solution containing 0.1 g of K3[Fe(CN)6] in 100 g water in which it is 80% ionised is [Kf (H20) = 1.86 mol-1kg]a)-2.3 x 10-2b)-5.7 x 10-2c)-5.7 x 10-3d)-1.92 x 10-2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The freezing point (in C) of a solution containing 0.1 g of K3[Fe(CN)6] in 100 g water in which it is 80% ionised is [Kf (H20) = 1.86 mol-1kg]a)-2.3 x 10-2b)-5.7 x 10-2c)-5.7 x 10-3d)-1.92 x 10-2Correct answer is option 'D'. Can you explain this answer?.
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