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JEE Advanced · Chemistry · 16. Solutions

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Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator automobiles.
A solution \(M\) is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixtrue is \(0.9\)
Given Freezing point depression constant of water \(\left(k^{\text {water }}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
Freezing point depression constant of ethanol \(\left(k_f^{\text {ethanol }}\right)=2.0 \mathrm{Kkg} \mathrm{mol}^{-1}\)
Boiling point elevation constant of water \(\left(k_b^{\text {water }}\right)=0.52 \mathrm{Kkg} \mathrm{mol}^{-1}\)
Boiling point elevation constant of ethanol \(\left(k_b^{\text {ethanol }}\right)=1.2 \mathrm{Kkg} \mathrm{mol}^{-1}\)
Standard freezing point of water \(=273 \mathrm{~K}\)
Standard freezing point of ethanol \(=155.7 \mathrm{~K}\)
Standard boiling point of water \(=373 \mathrm{~K}\)
Standard boiling point of ethanol \(=351.5 \mathrm{~K}\)
Vapour pressure of pure water \(=328 \mathrm{~mm}\) of \(\mathrm{Hg}\)
Vapour pressure of pure ethanol \(=40 \mathrm{~mm}\) of \(\mathrm{Hg}\)
Molecular weight of water \(=18 \mathrm{~g} \mathrm{~mol}^{-1}\)
Molecular weight of ethanol \(=46 \mathrm{~g} \mathrm{~mol}^{-1}\)
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.

Question:
The freezing point of the solution \(M\) is

  1. A \(268.7 \mathrm{~K}\)
  2. B \(268.5 \mathrm{~K}\)
  3. C \(234.2 \mathrm{~K}\)
  4. D \(150.9 \mathrm{~K}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(150.9 \mathrm{~K}\)

Step-by-step Solution

Detailed explanation

Solution \(M\) is mixture of ethanol and water.
Mole fraction of ethanol is \(0.9 \Rightarrow\) Solvent is \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)
Mole fraction of water is \(0.1 \Rightarrow \mathrm{H}_2 \mathrm{O}\) is solute
Molality of \(\mathrm{H}_2 \mathrm{O}=\frac{n_2}{n_1 M_1}=\frac{0.1}{0.9 \times 46} \times 1000=2.415\)
\(
\Delta T_f=k_f \cdot m=2 \times 2.415=4.83
\)
or Freezing point of solution \(=155.7-4.83=150.87 \mathrm{~K}\)
From JEE Advanced
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