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QUESTION 2.8

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 m L1, then what shall be the molarity of the solution?

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Solution

Step I: Calculation of molality of the solution
Mass of ethylene glycol = 222.6 g
Molar mass of ethylene glycol (C2H6O2)
=(12×2)+(1×6)+(16×2)
=24+6+32=62 g mol1
Mass of solution = 200g +222.6 g = 422.6 g = 0.4226kg
Molality (m) =Mass of ethylene glycolMolar massMass of solvent in kg
=(222.6g)(62 g mol1)0.4226 kg =8.49molkg1

Step II: Calculation of molarity of the solution
Density of the solution =1.072 g mL1
Mass of solution = Mass of solute + Mass of solution
=222.6g + 200g = 422.6g
Volume =MassDensity=422.6g1.072 g mL1=394.2 mL
=0.3942 L
Molarity (M) =Mass of ethylene glycolMolar massVolume in litres
=(222.6g)(62 g mol1)0.3942L
=9.10 mol L1=9.10 M


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