At 100^ the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm . If K_b = 0.52 , the boiling point of this solution will be: - Chemistry MCQ Chemistry | ExamTest.live | ExamTest.live

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At 10 0^{\circ} C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm . If K_{b} = 0.52, the boiling point of this solution will be:

A $10 0^{\circ} C$
B $10 2^{\circ} C$
C $10 3^{\circ} C$
D $10 1^{\circ} C$

Solution & Step-by-step Explanation

Using Raoult's law for relative lowering of vapour pressure: \frac{P ^{\circ} - P _{s}}{P ^{\circ}} = \frac{n _{solute}}{n _{solvent}} At 10 0^{\circ} C, P^{\circ} (water) = 760 mm Hg . \frac{760 - 732}{760} = \frac{6.5/ M}{100/18} \Rightarrow \frac{28}{760} = \frac{6.5 \times 18}{M \times 100} M = \frac{6.5 \times 18 \times 760}{28 \times 100} \approx 31.75 g/mol \approx 32 g/mol Now, elevation in boiling point Δ T_{b} = K_{b} \times m : Δ T_{b} = 0.52 \times \frac{6.5/32}{100/1000} = 0.52 \times \frac{0.203}{0.1} \approx 1.0 5^{\circ} C \approx 1^{\circ} C Boiling point of solution = 100 + 1 = 10 1^{\circ} C .

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At 10 0^{\circ} C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm . If K_{b} = 0.52, the boiling point of this solution will be:

A

10 0^{\circ} C

B

10 2^{\circ} C

C

10 3^{\circ} C

D

10 1^{\circ} C

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