How would i do these chem math problems step by step?

1) A 200 g sample of of a large biomolecule was dissolved in 15 g of carbon tetrachloride. The boiling point of this solution was determined to be 77.85 degree c. Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling point of pure carbon tetrachloride is 76.50 degree c.

2) What volume of ethylene glycol (C2H6O2) , a nonelectrolyte, must be added to 15 L of water produce an antifreeze solution with a freezing point of -25 degree c. what is the boiling point of this solution? the density of ethylene glycol is 1.11 g/cm^3 and the density of water is 1 g/cm^3)

Answer:
1) this is called "boiling point elevation"

the equation to use is

dT = Kb x m
where dT is the change in boiling point
Kb is a molal boiling point elevation constant = 5.03 °C/molal
for carbon tetrachloride ( CCl4 )

m = dT/ Kb = (77.85 °C - 76.50 °C) / 5.03 °C/molal
m = 0.2684 molal

but molality = moles solute / kg solvent
and kg solvent = 15 g CCl4 x (1 kg / 1000 g ) = 0.015 kg CCl4

so moles solute = molality x kg solvent = 0.2684 moles / kg x 0.015 kg = .00403 moles

moles = weight / molecular weight
molecular weight = weight / moles = 200 g / .00403 moles
= 49,700 g/mole

2) this is called freezing point depression.

applicable equation is

dT = Kf x m
dT is change in freezing point
Kf = molal freezing point depression constant = 1.86 °C / molal for water
m = molality of solution = moles solute / kg solvent

so

m = dT / Kf = (25°C) / 1.86 °C/molal = 13.44 molal

15 L of water x (1000 ml / L ) x (1 cc/ml) x (1 g / cc) x (1kg / 1000 g) = 15 kg water

since molality = moles solute / kg solvent
moles solute = molality x kg solvent = 13.44 moles / kg x 15 kg = 201.6 moles ethylene glycol

moles = weight / mw --> weight = moles x mw = 201.6 moles x 62.07 g/mole = 12514 g

density = mass / volume
volume = mass / density = 12514 g / 1.11 g/cm^3 = 11274 cc

finally

11274 cc x (1ml / cc) x (1 L / 1000 ml) = 11.27 liters

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