An antifreeze solution is prepared from 222.6 g of ethylene glycol C2H4(OH)2 and 200g of water . Calculate the molality of the solution . If the density of the solution is 1.072 g per / ml -1 , then what shall be the molarity of the solution ?

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Solution:-

Mass of ethylene glycol (solute) = 222.6 g

Molar mass of ethylene glycol ( C2H4(OH)2 = 62g

Number of moles of the solute = 222.6 / 62  = 3.59 moles

Mass of the solvent (water) in solution = 200g = 0.200Kg

Volume of the solution = 422.6 / 1.072 = 394.2 ml

                                                = 0.394L

Molality of the solution = number of moles of the solute/ Mass of solvent in Kg

                                                = 3.59 / 0.2

                                                = 17.95 m

Molarity of the solution = number of moles / volume of the solution in litres

                                                = 3.59 / 0.3942

                                                = 9.11 M

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