Question

Please, I don't want the answer I want to know how to do the work

1. You have added 2.20 kg of ethylene glycol (antifreeze), C2H6O2 MW = 62.068 g/mol, to theradiator of your car that contains 9100 g of water. Calculate the freezing point and the boilingpoint of the solution in the radiator of your car.

2. Calculate the heat of vaporization, Hvap, of diethyl ether if its vapor pressures at 13.0 ºC and 35.0 ºC are 305 mmHg and 760 mmHg, respectively.

3. The electrolyte in a lead storage battery is 3.75 M H2SO4 solution that has a density of 1.230g/mL. The molality of this solution is _________ MW H2SO4 = 98.1 g/mol

a. 3.05 m

b. 4.19 m

c. 4.35 m
d. None of these.

Answer 1

a. Freezing point of solution = = -i.Kf.m

with,

i = 1

Kf = 1.86 oC/m

m = (2.20 x 1000)g/62.07 g/mol x 9.100 kg = 3.895 m

Feed values,

Freezing point = 1 x 3.895 x 1.86 = -7.24 oC

The solution freezes at -7.24 oC

c. Using equation,

ln(P2/P1) = dHv/R[1/T1 - 1/T2]

with,

P1 = 305 mmHg

T1 = 13 + 273 = 286 K

P2 = 760 mmHg

T2 = 35 + 273 = 308 K

R = gas constant

dHv = heat of vaporisation

we get,

ln(760/305) = dHv/8.314[1/286 - 1/308]

dHv = 30.43 kJ/mol

3. Molality of the solution would be,

c. 4.35 m

Details:

molarity of solution = 3.75 M H2SO4

moles of H2SO4 = 3.75 mols in 1000 ml solution

mass of H2SO4 = 3.75 x 98 = 367.5 g

mass of solution = 1000 x 1.230 = 1230 g

mass of solvent = 1230 - 367.5 = 862.5 g

molality of solution = 3.75 mols/0.8625 kg = 4.35 m