Question

Be sure to answer all parts. An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d - 1.114 g/mL; M-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 %v/v (b) mass percent % w/w (c) molarity d) malality

Answer 1

Let us assume that the volume of ethylene glycol and water are 50 mL each (equal volume). The total volume of the mixture = 100 mL

a) volume percent of ethylene glycol =( volume of ethylene glycol/volume of mixture)*100

= (50mL/100mL)*100

Volume percent = 50%

b) mass of antifreeze mixture = density* volume of the mixture

= 1.070 g/mL * 100 mL = 107 g

Mass of 50mL of ethylene glycol = density* 50mL = 1.114 g/mL * 50 mL = 55.7 g

Mass percent =( mass of ethylene glycol/ mass of the mixture)*100

Mass percent = (55.7/107)*100= 52.05%

c) molarity = moles of ethylene glycol/ volume of solution in L

Moles of ethylene glycol = mass/molar mass = 55.7/62.07 = 0.897 mol

Molarity = 0.897 mol/ 0.1 L = 8.97 M

d) molality = moles of ethylene glycol/ mass of solvent in Kg

Mass of water = density* volume = 1g/mL*50 mL = 50 g = 0.05 kg

Molality = 0.897/0.05 = 17.94 m

e) mole fraction = moles of ethylene glycol/ (moles of ethylene glycol + moles of water)

Moles of water = mass / molar mass = 50 g/18 g/mol = 2.78 mol

Mole fraction = 0.897/(0.897+2.78)= 0.244