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Questions 4-15 · Q13

Q.A mixture of benzene and toluene contains 30% by mass of toluene. At 30⁰C, vapour pressure of pure toluene is 36.7 mm Hg and that of pure benzene is 118.2 mm Hg. Assuming that the two liquids form ideal solutions, calculate the total pressure and partial pressure of each constituent above the solution at 30⁰C. (86.7 mm, P = 96.5 mm)

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Step 1. Take 100 g of mixture: 30 g toluene (M=92 g/mol), 70 g benzene (M=78 g/mol).

Step 2. Moles: ntol=30/92=0.3261n_{tol}=30/92=0.3261 mol; nben=70/78=0.8974n_{ben}=70/78=0.8974 mol; total = 1.2235 mol.

Step 3. Mole fractions: xtol=0.3261/1.2235=0.2665x_{tol}=0.3261/1.2235=0.2665; xben=0.8974/1.2235=0.7335x_{ben}=0.8974/1.2235=0.7335.

Step 4. Since the solution is ideal (obeys Raoult's law), each component's partial pressure is Pi=xiPi0P_i=x_iP_i^0: Ptol=0.2665×36.7=9.78P_{tol}=0.2665\times36.7=9.78 mm Hg; Pben=0.7335×118.2=86.7P_{ben}=0.7335\times118.2=86.7 mm Hg. …

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