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:
The vapour pressure of the solution \(M\) is
- A \(39.3 \mathrm{~mm} \mathrm{Hg}\)
- B \(36.0 \mathrm{~mm} \mathrm{Hg}\)
- C \(29.5 \mathrm{~mm} \mathrm{Hg}\)
- D \(28.8 \mathrm{~mm} \mathrm{Hg}\)
Answer & Solution
Correct Answer
(B) \(36.0 \mathrm{~mm} \mathrm{Hg}\)
Step-by-step Solution
Detailed explanation
Total vapour pressure, \(p=p_A^{\circ} \chi_A\)
\(
p=40 \times 0.9=36 \mathrm{~mm} \text { of } \mathrm{Hg}
\)
In the paragraph, it has been directed to take solute as non-volatile, thus \(\mathrm{H}_2 \mathrm{O}\) do not contribute in the total vapour pressure.
\(
p=40 \times 0.9=36 \mathrm{~mm} \text { of } \mathrm{Hg}
\)
In the paragraph, it has been directed to take solute as non-volatile, thus \(\mathrm{H}_2 \mathrm{O}\) do not contribute in the total vapour pressure.
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