Colligative Properties 1. How many grams of ethylene glycol, C2H4(OH)2, are need
ID: 1005421 • Letter: C
Question
Colligative Properties
1. How many grams of ethylene glycol, C2H4(OH)2, are needed per kilogram of water to protect radiator fluid against freezing down to -10°C? __________g
2. For benzene, C6H6, the freezing point constant, Kf, is 5.12 K oC/m and its normal freezing point is 5.5°C. What is the freezing point of a solution containing 100.0 g of benzene and 20.0 g of naphthalene (C10H8)? ________________°C
3. When 5.0 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? g mol-1 4. If 2.00 moles of a substance is dissolved in 1.00 kg of water, the freezing point of the solution decreases by 7.44°C. Does this substance dissociate to form ions in solution? Explain why. 15 5. Calculate the freezing point of a 0.2 m CaCl2 aqueous solution, assuming it dissociates completely to form ions in solution. ____________________°C
4. If 2.00 moles of a substance is dissolved in 1.00 kg of water, the freezing point of the solution 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 CaCl2 aqueous solution, assuming it dissociates completely to form ions in solution.
_________________ °C
Explanation / Answer
1. DTf = i*Kf*m
10 = 1*1.86*m
m = 5.37 molal
mass of ethylene glycol required = 5.37*62.07 = 333.316 grams
2.
5.5-x = 1*5.12*(20/128.1705)*(1000/100)
x = -2.5 C = freezing point of a solution
3.
1.5 = 1*1.86*(5/x)*(1000/100)
x = molarmass of unknown = 62 g/mol
4.
7.44 = i*1.86*2/1
i = 2 = vanthoff factor, indicates substance dissociates into ions.
5.
0-x = 3*1.86*0.2
x = freezing point of a 0.2 m CaCl2 aqueous solution = -1.116 C