Question

(gnaH to noibbe ( (5 pts) Normal boiling point of benzene ( C6H6) is 80.1 °C. If you add 0.15 moles of a nonvolatile solute to 250.0 g of benzene, what is the vapor pressure of the benzene at 80.1 °C? SSTICL TID nctanoo is lezas gnossi0 amilov gi285159b (b nimondo noiibbe (o (5 pts) Ethylene glycol, (C2H6O2) used as an antifreeze. How many grams of ethylene glycol must be added to 1.0 L of water to reduce the freezing point of the glycol-water mixture to -10.0 °C? The freezing point depression constant of water is 1.86 deg/molal)

Answer 1

- 3.20 mole. 0 mole of benzene - 250.0 g 78.11 osmole x2 = 0.15 (0.15 +3.20 - 0.0448 POP, X2 pressure at boiling point & 760 mmHg. PO 760-P 0.0448 760 P = 726 mm Hg. of bentene = vapour pressure 726 mmHg. 1.0t water = 100 kg water, xg ethylene glycol added mole of ebleylene glycol = (282) mole. molality = (2) mele/1.0 kg = 662) (). ATS = ix kam 10.0 = 4 x 1.86 x (72) x = 33,2 g. Grams of ethylene glycol = 33.3 g