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Solutions

Question
CBSEENCH12005728

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?

Solution

Answer:

Consider ethylene glycol as solute and water as a solvent.
Weight of solute, WB = 222.6 g
Molar mass, MB = 24 + 6 + 32 = 62
WB = 200 g = 0.200 kg.

Moles of solute,       nB = WBMB = 222.662 = 3.59

Molaity of ethylene glycol in H
2O

mb =Number of moles of soluteKg. of solvent       = 3.59WA = 3.590.20 = 17.95 molal       = 17.95 m


Total mass of solution
= 222.6 + 200 = 422.6 g
Density of solution
= 1.072 g/ ml Volume of solution

= MassDensity = 422.61.072 = 394.2 mL = 0.3942 L

Molarity of solution,


Mb = Moles of soluteVol. of solution in litres       = 3.590.3942 = 9.1M.