An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d=1.114 g/mL; M=62.07 g/mL) and water (d=1.00 g/mL) at 20 degrees C. The density of the mixture is 1.070 g/mL. Expreess the concentration of ethylene glycol as volume percent. I've tried this a couple times but keep getting it wrong if u could tell me step by step i'd really appreciate it

Answer 1

I tried this.
Assume we mix 50.00 mL EG with 50.00 mL H2O.
mass EG = VD = 50.00 mL x 1.114 g/mL = 55.70 grams
mass H2O = VD = 50.00 mL x 1.00 g/mL = 50.00 grams.
Total mass = 55.70 + 50.00 = 105.70 grams.
V of total = m/d = 105.70/1.070 = 98.785 mL.
Then % v/v = 50.00 mL/98.785 mL = 50.61% EG

Answer 2

All the Quetions can be :

a-Volume percent ?
b-Mass percent ?
c-molarity ?
d-molality ?
e-mole fraction ?

answers:

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