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NEET CHEMISTRYSolutionsMedium

Question

The correct option for the value of vapour pressure of a solution at 45 °C with benzene to octane in a molar ratio 3:2 is: [At 45 °C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]

A

336 mm of Hg

B

350 mm of Hg

C

160 mm of Hg

D

168 mm of Hg

Step-by-Step Solution

According to Raoult's law, for an ideal solution of volatile liquids, the total vapour pressure (ptotalp_{total}) is the sum of the partial vapour pressures of the components . ptotal=p10x1+p20x2p_{total} = p_1^0 x_1 + p_2^0 x_2

Let benzene be component 1 and octane be component 2. Given molar ratio n1:n2=3:2n_1 : n_2 = 3 : 2. Mole fraction of benzene, x1=n1n1+n2=33+2=35=0.6x_1 = \frac{n_1}{n_1 + n_2} = \frac{3}{3 + 2} = \frac{3}{5} = 0.6 Mole fraction of octane, x2=n2n1+n2=23+2=25=0.4x_2 = \frac{n_2}{n_1 + n_2} = \frac{2}{3 + 2} = \frac{2}{5} = 0.4

Given vapour pressures of pure components: p10=280 mm Hgp_1^0 = 280 \text{ mm Hg} p20=420 mm Hgp_2^0 = 420 \text{ mm Hg}

Total vapour pressure, ptotal=(280×0.6)+(420×0.4)p_{total} = (280 \times 0.6) + (420 \times 0.4) ptotal=168+168=336 mm Hgp_{total} = 168 + 168 = 336 \text{ mm Hg}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Solutions. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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