Question

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 M

(d) molality m

(e) mole fraction

Answer 1

Solution-

Let us assume 10 mL of ethylene glycol and 10 mL of water

mass ethylene glycol = 10 mL x 1.114 g/mL = 11.14 g
mass water = 10 mL x 1.00 g/mL = 10.0 g
mass solution = 11.14 + 10.0= 21.14 g

mass percent = 11.14 x 100/ 21.14 = 52.7

volume solution = 21.14 g / 1.070 g/mL=19.8 mL

volume percent = 10 x 100/ 19.8=50.5

moles ethylene glycol = 11.14 / 62.07 =0.179
M = 0.179/ 0.0198 L=9.06

m = 0.179/ 0.010 Kg = 17.9

moles water = 10.0 g/ 18.02 g/mol=0.555
mole fraction = 0.179/0.179 + 0.555 =0.244