1) A solution is prepared by dissolving 19.2 g of ordinary sugar (sucrose, C12H2

Question

1) A solution is prepared by dissolving 19.2 g of ordinary sugar (sucrose, C12H22O11, 342 g/mol) in 50.0 g of water. Calculate the boiling point of the solution. Sucrose is a nonvolatile nonelectrolyte.

2) Calculate the freezing point of a 4.20m aqueous solution of ethylene glycol, a nonvolatile nonelectrolyte.

3) An aqueous solution contains 0.180 g of an unknown, nonionic solute in 50.0 g of water. The molal freezing point depression constant for water, Kf, is -1.86°C/m. The solution freezes at -0.0400 °C. What is the molar mass of the solute? Molar mass =  g/mol

4) What is the boiling point of a solution composed of 17.1 g of urea, (NH2)2CO, in 0.100 kg of water? Kb(water) = 0.5121 °C/m, Boiling point = °C

5) Estimate the osmotic pressure associated with 15.8 g of an enzyme of molecular weight 1.47*10^6 g/mol dissolved in 1500. mL of ethyl acetate solution at 30.0°C. Osmotic pressure =  atm

6) Some ethylene glycol, HOCH2CH2OH, is added to your car's cooling system along with 5.0 kg of water. If the freezing point of the water-glycol solution is -10.6 °C, what mass of HOCH2CH2OH must have been added? Kf = 1.86 °C/m Mass =  g

Explanation / Answer

(1)

molality of sucrose solution = (mass of sucrose/molarmass of sucrose) * (1000/mass of solvent)

m = (19.2/342)*(1000/50.0)

m = 0.561 m

Kb of water = 0.512 0C.kg./mol

T0 of water = 1000C

Then, Tb of solution = ?

Elevation in boiling point = Kb*m

Tb - 100 = 0.512 * 0.561

Tb = Boiling point of solution = 100.287 0C

(2)

Depression in freezing point = Kf * m

T0 - Tf = 1.86 * 4.20

0.000 - Tf = 7.56

Tf = Freezing point of solution = - 7.560C

(3)

Depression in freezing point = 1000 * Kf * w / W m

- 0.0400 = 1000 * (-1.86) * 0.180 / (50.0 * m)

m = molar mass of solute = 167. g/mol

(4)

Tb - T0 = 1000 Kb * w / (W * m)

Tb - 100 = 1000 * 0.5121 * 17.1 / (0.100 *1000 * 60)

Tb = 101.460C

(5)

Molarity = (15.8/1.47*106)*(1000/1500.) = 7.16 * 10-6 M

Osmatic pressure = C R T = 7.16 * 10-6 * 0.0821 * 303. = 1.76 * 10-4 atm

(6)

Depression in freezing point = 1000 * Kf * w / (W m)

10.6 = 1000 * 1.86 * w / (5.0 * 1000 * 62)

w = 33.0 kg.

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