Question

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?

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)?

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?

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.

Answer 1

1) we know that

depression in freezing point is given by

dTf = kf x m

so

10 = 1.86 x m

m = 5.376

now

molality = moles of solute / mass of solvent (kg)

so

5.376 = moles of EG / 1

moles of EG = 5.376

now

mass = moles x molar mass

so

mass of EG = 5.376 x 62

mass of EG = 333.33

so

333.33 g of enthylene glycol is needed

2)

moles of napthalene = 20/128

moles of napthalene = 0.15625

now

molality = 0.15625 / 0.1

molality = 1.5625

now

dTF = kf x m

so

dTf = 5.12 x 1.5625

dTf = 8

so

5.5 - T = 8

T = -2.5

so

the freezing point of the solution is -2.5 C

3)

given

dTf = 1.5

so

1.5 = 1.86 x m

m = 0.806

now

molality = moles of solute / mass of water (kg)

0.806 = moles of solute / 0.1

moles of solute = 0.0806

now

moles = mass / molar mass

so

0.0806 = 5 / molar mass

molar mass = 62 g

4)

molality = 2 / 1

molality =2

now

dTf = 7.44

so

dTf= i x kf x m

so

7.44 = i x 1.86 x 2

i = 2

so

as the i value is 2

the substance dssocites to form ions

5)

given

molality = 0.2

also

CaCl2 ---->> Ca+2 + 2Cl-

three particles are formed

so

i = 3

now

dTf = i x kf x m

= 3 x 1.86 x 0.2

= 1.116

0 - T = 1.116

so

T = -1.116 C

so

the freezing point is -1.116 C