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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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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?

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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?

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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?