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  • 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?

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

Q8.

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

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What is the correct answer to: 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....

How do you solve 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? step by step?

Step-by-step explanation:
• Calculation of Molality :

• Mass of ethylene glycol = 222
• 6 (Given)

What common mistakes should be avoided in this question?

Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.

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Student Discussion

Rakshita
Class · · , · Mar 14, 2019
Y we shouldn't consider the moles of water as we had considered of solute in calculating of total solvent?
👍 👎 Reply
Aynal Hoque
Class · · , · Oct 22, 2018
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)
👍 👎 Reply
Kane
Class · · , · Jul 01, 2018
Ethylene glycol is solute because it has less no. Of moles in the solution in comparison to water.
👍 👎 Reply
Abhi
Class · · , · Jun 23, 2018
Why ethylene glycol is solute
👍 👎 Reply
Pankaj
Class · · , · May 04, 2017
Why ethylene glycol is solute here even it is more quantity
👍 👎 Reply

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