An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/ml; = 62.07 g/mol) and water (d = 1.00 g/ml) at 20°c. the density of the mixture is 1.070 g/ml. express the concentration of ethylene glycol as:

a) Volume percent

Formula: % v/v = [volume solute / volume solution] * 100

Just to make it easy take a base of 50 volume parts of ethylen glycol and 50 volume parts of water to make 100 volumes of mixture (this assumpion will be valid for all the questions):

% v/v =[ 50 ml ethyleneglycol] / [100 ml mixture] * 100 = 50%

50% v/v

b) Mass percent

% m/m = [mass ethylene glycol / mass solution] * 100

mass ethylene glycol = 50 ml * 1.114 g/ ml = 55.7 g

mass of mixture = 100 ml * 1.07 g/ml = 107 g

% m/m = [55.7 / 107 g] * 100 = 52.06 %

52.06%

c) Molarity

M = number of moles of solute / liters of solution

number of moles of solute = mass in grams / molar mass

number of moles of ehtylene glycol = 55.7 g / 62.07 g/mol = 0.8974 mol

liters of solution = 0.1 liter

M = 0.8974 mol / 0.1 liter = 8.974 M

8.974 M

d) Molality

m = number of moles of solute / kg of solvent

number of moles of ethylen glycol = 0.8974 mol

mass of water = 50 ml * 1 g/ml = 50 g = 0.05 kg

m = 0.8974 mol / 0.05 kg = 17.95 m

17.95 m

e) mole fraction

mole fraction = [number of moles of solute] / [number of moles of mixture] * 100

number of moles of ethylen glycol = 0.8974 mol

number of moles of water = 50 g / 18.01 g /mol = 2.776 mol

mole fraction = 0.8974 mol / [0.8974 mol + 2.776 mol] = 0.244

0.244

While unproven, it uses facts to support part of a conclusion. it is still factual and has predictive power, however.