IIT-JEE 2012 · Paper 2 · Chemistry Q25

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ChemistrySingle correct+3 / -1Question 25 of 60

For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2 °C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take $K_b=0.76$ K kg mol⁻¹)

724

740

736

718

Answer and worked solution

Official answer: A

Molality. $m=\dfrac{\Delta T_b}{K_b}=\dfrac2{0.76}=2.63\ \mathrm{mol\,kg^{-1}}$ .

Mole fraction of solute (dilute, so $n_\text{solute}\ll n_\text{water}$ ):
$x\approx\frac{2.63}{1000/18}=0.0474.$

Raoult's law. At the normal boiling point the pure water has $p^\circ=760$ mm Hg:
$p=760(1-0.0474)\approx724\text{ mm Hg}.$

(A)

Topics:Colligative properties Raoult's law and vapour pressure

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IIT-JEE 2012 · Paper 2, Chemistry question 25