Question

Different Ways of Expressing Concentration An aqueous antifreeze solution is 52.5% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.067 g/cm3. Calculate the molality of the ethylene glycol. m 1pts Tries 0/5 Calculate the molarity of the ethylene glycol. M 1pts Tries 0/5 Calculate the mole fraction of ethylene glycol.

Answer 1

a)molality = moles of solute / kg solvent

Assume you have exactly 1 kg of solution. 52.5% of that, means 525 grams of the solution is C2H6O2, and 475 g of the solution is water.

Moles of Ethylene glycol = 525 g / 62.0 g/mol = 8.47 mol C2H6O2

molality = 8.47 mol / 0.475 kg H2O = 17.83 molal

b) Molarity = moles of solute/L of solution
That 1 kg of solution has a volume of:

1000 g / 1.067 g/mL = 937.2 mL = 0.937 L

Molarity = 8.47 mol / 0.937 L = 9.04 M

c) Mole fraction ethylene glycol = moles ethylene glycol / (moles ethylene glycol + moles H2O)

moles ethylene glycol = 8.47
moles H2O = 475g / 18.0 g/mol = 26.38 mol H2O

mole fraction = 8.47/ ( 8.47 + 26.38) = 0.243