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JEE Advanced 2012Paper 2
Exam
JEE Advanced
Year
2012
Paper
Paper 2
Subject
Chemistry
Chapter
Solutions

JEE Advanced 2012 · Paper 2 · Chemistry · Solutions

For a dilute solution containing 2.5 ~g of a non-volatile...

For a dilute solution containing 2.5 g2.5 \mathrm{~g} of a non-volatile non-electrolyte solute in 100 g100 \mathrm{~g} of water, the elevation in boiling point at 1 atm pressure is 2∘C2^{\circ} \mathrm{C}. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure ( mm\mathrm{mm} of Hg\mathrm{Hg} ) of the solution is (take Kb=0.76 K kg mol−1K_{b}=0.76 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} )

Options

  1. A724
  2. B740
  3. C736
  4. D718

Correct answer

A. 724

Step-by-step solution

From Raoult's law, p∘−pp∘= No. of moles of solute  No. of moles of solvent + No. of moles of solute \frac{p^{\circ}-p}{p^{\circ}}=\frac{\text{ No. of moles of solute }}{\text{ No. of moles of solvent }+\text{ No. of moles of solute }} When the concentration of solute is much lower than the concentration of solvent, p∘−pp∘= No. of moles of solute  No. of moles of solvent 760−p760=2.5/m100/18...(i)\begin{array}{l} \frac{p^{\circ}-p}{p^{\circ}}=\frac{\text{ No. of moles of solute }}{\text{ No. of moles of solvent }} \\ \frac{760-p}{760}=\frac{2.5 / m}{100 / 18}...(i) \end{array} From elevation in boiling point, ΔTb=Kb×m\Delta T_{b}=K_{b} \times m 2=0.76×mm=20.76...(ii)\begin{array}{l} 2=0.76 \times m \\ m=\frac{2}{0.76}...(ii) \end{array} From(i) and (ii), p=724 mmp=724 \mathrm{~mm}
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