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InterviewSolution
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A 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 "1.072 g mL"^(-1), then what shall be the molarity of the solution? |
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Answer» Solution :`"MASS of the solute, "C_(2)H_(4)(OH)_(4)="22.6 g,Molar mass of "C_(2)H_(4)(OH)_(2)="62 g mol"^(-1)` `therefore"Moles of the solute"=("222.6 g")/("62 g mol"^(-1))="3.59 ,Mass of the solvent =200 g = 0.200 KG"` `"Molality "=("3.59 moles")/("0.200 kg")="17.95 mol kg"^(-1)` `"Total mass of the solution = 422.6 g"` `"Volume of the soltuion "=("422.6 g")/("1.072 g ml"^(-1))="394.2 ml = 0.3942 L , Molarity "=("3.59 moles")/("0.3942 L")="9.11 mol L"^(-1)`. |
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