Vapour pressure of 0.5 molal solution of non-volatile solute in organic solvent at 30°C would be : (Given : vapour pressure of pure organic solvent is 100 torr; the molecular weight of organic solvent is 100 g mol-1).
90.25 torr
Raoult's law for a non-volatile solute gives \(P = x_{solvent}\,P^{\circ}\), so the calculation is really about converting molality to mole fraction.
Step 1 — moles of solute. 0.5 molal means 0.5 mol of solute per 1 kg of solvent.
Step 2 — moles of solvent. With \(M = 100\) g mol-1, one kilogram contains \(\frac{1000}{100} = 10\) mol.
Step 3 — mole fraction of solvent. \(x_{solvent} = \frac{10}{10 + 0.5} = \frac{10}{10.5} = 0.952\).
Step 4 — apply Raoult's law. \(P = 0.952 \times 100 = 95.2\) torr.
Two sanity checks confirm the method. A non-volatile solute can only lower the vapour pressure, so any answer above 100 torr is impossible on physical grounds — which excludes both 125.50 and 119.95 torr immediately. And the solution is dilute, so the lowering should be modest, of the order of 5 torr.
The correctly computed value of 95.2 torr does not appear among the four options; 90.25 torr is the nearest, and 100.25 torr would require the vapour pressure to rise.
Note: the SPPU final answer key marked this question with code 9 ("No option is correct or the question is wrong"), consistent with the calculation above. The answer recorded here is the closest available option and is flagged as key-disputed.
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