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JEE Advanced · Chemistry · 16. Solutions

Paragraph:
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:
Water is added to the solution \(M\) such that the mole fraction of water in the solution becomes \(0.9\) The boiling point of this solutions is

  1. A \(380.4 \mathrm{~K}\)
  2. B \(376.2 \mathrm{~K}\)
  3. C \(375.5 \mathrm{~K}\)
  4. D \(354.7 \mathrm{~K}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(376.2 \mathrm{~K}\)

Step-by-step Solution

Detailed explanation

\(\chi_{\mathrm{H}_2 \mathrm{O}}=0.9\) (solvent) \(\chi_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}=0.1\) (solute)
\(
\Delta T_b=k_b \times m=0.52 \times \frac{1.0 \times 1000}{0.9 \times 18}=3.2 \mathrm{~K}
\)
Boiling point, \(T_b=373+3.2=376.2 \mathrm{~K}\)
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