An antifreeze solution is prepared from | Class 12 Chemistry Chapter Solutions, Solutions NCERT Solutions

Question:

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?

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


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Comments

  • Rakshita
  • 2019-03-14 21:01:50

Y we shouldn't consider the moles of water as we had considered of solute in calculating of total solvent?


  • Aynal Hoque
  • 2018-10-22 10:36:17

Yup it's correct bcs we should always consider the no.of moles....which is more for water here....hence water is solvent...and in las molality can be directly calculated from molarity as molality= 1000 M/1000d-MM' (M = molarity and M'= molar mass of solute)


  • Kane
  • 2018-07-01 19:02:37

Ethylene glycol is solute because it has less no. Of moles in the solution in comparison to water.


  • Abhi
  • 2018-06-23 16:46:45

Why ethylene glycol is solute


  • Pankaj
  • 2017-05-04 16:58:56

Why ethylene glycol is solute here even it is more quantity


Comment(s) on this Question

Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 8: An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calc....