Paragraph: Properties such as boiling point, freezing point and vapour pressure of a pure solvent change…

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. $380.4 \mathrm{~K}$
  2. $376.2 \mathrm{~K}$
  3. $375.5 \mathrm{~K}$
  4. $354.7 \mathrm{~K}$

Solution

$\chi_{\mathrm{H}_2 \mathrm{O}}=0.9$ (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, $T_b=373+3.2=376.2 \mathrm{~K}$

Asked in: JEE Advanced 2008 (Paper 1)

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