An antifreeze solution is prepared from 2226 g of ethylene glycol,C2H4

1.	An antifreeze solution is prepared from 2226 g of ethylene glycol,C2H4(OH2 and 200 g of water. Calculate the molality of the solution. If thedensity of the solution is 1-072 gmL-1, then what shall be the molarity ofthe solution ?
Answer» 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

An antifreeze solution is prepared from 2226 g of ethylene glycol,C2H4(OH2 and 200 g of water. Calculate the molality of the solution. If thedensity of the solution is 1-072 gmL-1, then what shall be the molarity ofthe solution ?

Answer»

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