Exercise 2
Question 8
An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL-1, then what shall be the molarity of the solution?
Calculation of Molality :
Mass of ethylene glycol = 222.6 (Given)
Molar mass of ethylene glycol [C2H4(OH)2]
= 2 X 12 + 6 x 1 + 2 x 16
= 62
Therefore moles of ethylene glycol
= 222.6g / 62 gmol-1
= 3.59 mol
Mass of water = 200g (Given)
Therefore molality of the solution is = (moles of ethylene glycol / mass of water) x 1000
= (3.59 / 200) x 1000
= 17.95 m
Calculation of Molarity:
Moles of ethylene glycol = 3.59 mol (already calculated)
Total Mass of solution = 200 + 222.6
= 422.6g
Volume of solution = mass / density volume
= 422.6 / 1.072
= 394.22 ml
now molarity of the solution is = (moles of ethylene glycol / volume of solution) x 1000
= (3.59 / 394.22) x 1000
= 9.11 M
