Please, I don\'t want the answer I want to know how to do the work 1. You have a

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.

Explanation / Answer

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

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