University:
Kentucky State University
Course:
CHE 302 | Organic Chemistry II
Academic year:

2024
Views:

143

Pages:

3
Author:

hekeldigzjbz

**Benzene phenol** has a freezing point of **49.00°C**. A 0.450 molal solution of urea in this solvent has a freezing point of 44.59°C. What is the freezing point depression constant (Kf) of benzene? ΔT = Kf * m (49 - 44.59) = Kf * 0.450 Kf = 9.8 **The freezing point of a solution that contains 1g of a compound (A) dissolved in 10g of benzene is found to be 2.07°C. The freezing point of pure benzene is 5.48°C. The molal freezing point depression constant of benzene is 5.12. What is the molecular weight of (A)?** ΔT = Kf * m 5.48 - 2.07 = 5.12 * m m = 0.666 = molality moles = 0.666 * 0.01 = 6.66 x 10^-3 mass = 1g MW = mass/moles = 1 / 6.66 x 10^-3 = 150 Dalton **Another approach:** MW = Kf * w2 * 1000 / w1 * ΔT MW = 5.12 * 1000 * 1 / 10 * (5.48 - 2.07) MW = 150 Dalton **Calculate the vapor pressure of a solution (in mmHg) made by dissolving 50g glucose in 500g of water. The vapor pressure of pure water is 47.1 mmHg at 37°C, and the molar mass (Mwt) of glucose is 180.2 and of water is 18.** moles of glucose = 50 / 180.2 = 0.278 moles of water = 500 / 18 = 27.778 X° = 0.278 / (27.778 + 0.278) X° = 0.0099 Pa = X° * P°A Pa = 0.0099 * 47.1 = 0.4663 **What are the total vapor pressure (in mmHg) of a mixture of benzene and toluene in a solution in which the mole fraction of benzene is 0.6, the vapor pressure of pure benzene is 95.1 mmHg, and the vapor pressure of pure toluene is 28.4 mmHg at 25°C.** Pbenzene = Xbenzene * P°benzene = 0.6 * 95.1 = 57.06 Ptoluene = Xtoluene * P°toluene = (1-0.6) * 28.4 = 11.36 Ptotal = Pbenzene + Ptoluene = 57.06 + 11.36 = 68.42 A solution is prepared by dissolving 35 g of hemoglobin in enough water to make 1 L in volume. The osmotic pressure of the solution is 10 mmHg at 25°C. Calculate the molecular weight of hemoglobin. **R = 0.08206 L atm / mol K** T = 25°C = 298 K 760 mmHg = 1 atm 10 mmHg = 10/760 atm π = 10/760 = 0.01316 atm Mwt = πV / nRT Mwt = 35 x 10^-3 x 0.08206 x 298 / 0.01316 **Mwt = 65037 Dalton** π = MRT 0.01316 = M x 0.08206 x 298 M = 5.38 x 10^-4 moles/L moles = 5.38 x 10^-4 moles volume = 1 L moles = 5.38 x 10^-4 mass = 35 g Mwt = 35 / 5.38 x 10^-4 = 65037 Dalton