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