# An antifreeze solution is prepared from 222.6 g of ethylene glycol [C_(2)H_(4)(OH)_(2)] and 200 g of water. Calculate the molality of the soluti

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An antifreeze solution is prepared from 222.6 g of ethylene glycol [C_(2)H_(4)(OH)_(2)] and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072g mL^(-1) then what shall be the molarity of the solution?

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Mass of solute =222*6g
Molar mass of solute, C_(2)H_(4)(OH)_(2)
=12xx2+4+2(12+1)=62g mol^(-1)
therefore"Moles of solute"=(222*6)/(62)=3*59
Mass of solvent =200g
thereforeMolality=(3*59)/(200)xx1000=17*95" mol "kg^(-1)
Total mass of solution =422*5g
volume of solution =(442*6)/(1*072)=394*21mL.
therefore"Molarity"=(3*59)/(394*2)xx1000=9*1" mol "L^(-1)