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An antifreeze solution is prepared from 222.6 g of ethylene glycol (C_(2)H_(6)O_(2)) 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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Answer» SOLUTION :Molar mass of ethylene glycol `[C_(2)H_(4)(OH)_(2)]` `= 2xx12+6xx1+2xx16` = 62 g `MOL^(-1)` Number of moles of ethylene glycol `= (222.6g)/(62 g mol^(-1))` = 3.59 mol Therefore, molality of the solution `= (3.59 mol)/(0.20 kg)` = 17.95 m Total mass of the solution `= (222.6+200)g` = 422.6 g Given, Density of the solution `=(422.6 g)/(1.072 g mL^(-1))` = 394.2 mL `= 0.3942xx10^(-3)L` MOLARITY of the solution `= (3.59 mol)/(0.39422xx10^(-3)L)` = 9.11 M. |
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