If 1.5 grams of a non-volatile solute (MW = 100) is added to 200 ml pure CS2 (ρ = 1.3 g/cc) whose vapor pressure is 400 mm of Hg at 28.0°C, what is the resulting vapor pressure of the dilute solution?(a) 401.246 mm Hg(b) 398.754 mm Hg(c) 401.754 mm Hg(d) 398.246 mm HgThe question was posed to me during an online exam.The question is from Colligative Properties and Determination of Molar Mass in division Solutions of Chemistry – Class 12

If 1.5 grams of a non-volatile solute (MW = 100) is added to 200 ml pu

1.	If 1.5 grams of a non-volatile solute (MW = 100) is added to 200 ml pure CS2 (ρ = 1.3 g/cc) whose vapor pressure is 400 mm of Hg at 28.0°C, what is the resulting vapor pressure of the dilute solution?(a) 401.246 mm Hg(b) 398.754 mm Hg(c) 401.754 mm Hg(d) 398.246 mm HgThe question was posed to me during an online exam.The question is from Colligative Properties and Determination of Molar Mass in division Solutions of Chemistry – Class 12
Answer» The correct option is (d) 398.246 mm Hg For EXPLANATION: Given, Mass of solute, m2 = 1.5 grams Molar mass of solute, M2 = 100 Vapor pressure of pure CS2, p^01 = 400 mm Hg Volume of solvent, ρ = 200 ml Density of solvent, ρ = 1.3 g/cc Number of moles of solute, N2 = m2/MW = 1.5/100 = 0.015 mole From the law of relative LOWERING of vapor pressure, Δp1 = p^01X2, where X2 is the mole fraction of solute and Δp1 is the difference in pressure. Mass of solvent, m1 = ρ x V = 1.3 x 200 = 260 grams Number of moles of solvent, n2 = 260g/[(12 + 32 + 32)g/mole]= 3.421 Since the solution is dilute we can APPROXIMATE X2 = n2/(n1 + n2) ≈ n2/n1 (since n2<< n1) Using ∆p1 = p^01X2, ∆p1 = (400 mg of Hg) (0.015/3.421) = 1.754 mm of Hg Using Δp1 = p^01 – p1, we get p1 = p^01–Δp1 HENCE, resulting lowered vapor pressure, p1 = 400 mm Hg – 1.754 mm Hg = 398.246 mm Hg.

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