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
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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....
Comments
Y we shouldn't consider the moles of water as we had considered of solute in calculating of total solvent?
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)
Ethylene glycol is solute because it has less no. Of moles in the solution in comparison to water.
Why ethylene glycol is solute
Why ethylene glycol is solute here even it is more quantity