Question

REVIEW QUESTIONS 1. How many grams of ethylene glycol, C,H,(OH)2, are needed per kilogram of water to protect radiator fluid against freezing down to -15°C? For water, the freezing point constant, Kp, is 1.86°C kg mol-! CH (OH), does not dissociate in aqueous solution, i = 1. 2. For benzene, CH, the freezing point constant, K, is 5.12°C kg mol-1 and its normal freez- ing point is 5.5°C. What is the freezing point of a solution containing 100.0 g of benzene, and 30.0 g of naphthalene (C10Hg)? 3. When 10.0 g of an unknown substance is dissolved in 100.0 g of water, the freezing point of the solution decreases by 1.5°C. What is the molar mass of the unknown substance if the compound does not dissociate to form ions in solution, that is, i = 1? 4. If 2.00 moles of a substance is dissolved in 1.00 kg of water, the freezing point of the solu- tion decreases by 7.44°C. Does this substance dissociate to form ions in solution? Explain why. 5. Calculate the freezing point of a 0.2 m CaCl, aqueous solution, assuming it dissociates completely to form ions in solution.

Answer 1

ethylene gly mest f = -15°C To=0°c of water) Af = Tf - Tf - 15°c off kfx molality 15°c o 1.86 cryx in m= 15 mal/kg 1.86 m = 0.064 mall kg molality = mass of substance molar mass mass of solvent (kg) 8.064 mol = 62107 g/mol 1 Kg W = 8.064 mol x 62.07 g x 1 kg homol (w = 500g (for benzene, 2) Kf = 5.12 °C kg/mal If = 5.5°c If If a Kofam 515°c. If = 5:12°c kg x 30 g 128.27 g/mol mol - 100 1000 128:17 5.5°C -If & 5.12°C X 30 x 10 5.56 -Tf - 11.98 °C If = -6148°C]

3) Kf = 1,86 °C kg. mol unknown substance w= 10g mass of solvent = loog Alf= 1.5°C molar mass = ? ДТ = kx m 1.5 t= 1.86 C kq xm mol m= 1.5 mol/kg. 1.86 ma 0.806 mal / kg molality = = molarmass mass of solvent (kg) 10g 0.806 mal ка M . 100/1000 erg) almal M = 10x10 01806 M = 124.06 g/mal
I have solved questions 1-3