1. How many grams of ethylene glycol, C2H4(OH)2, are needed per kilogram of wate

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.

Explanation / Answer

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

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