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An antifreeze solution is prepared from 222.6 g of ethylene glycol $(C_2H_6O_2$) 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?

$\begin{array}{1 1}17.95molkg^{-1},9.11molL^{-1}\\27.95molkg^{-1},8.11molL^{-1}\\47.95molkg^{-1},7.11molL^{-1}\\77.95molkg^{-1},4.11molL^{-1}\end{array} $

1 Answer

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Mass of the solute $C_2H_4(OH)_2$=222.6g
Molar mass of $C_2H_4(OH)_2=62g mol^{-1}$
Moles of the solute =$\large\frac{222.6}{62gmol^{-1}}$
$\Rightarrow 3.59$
Mass of the solvent =200g=0.200kg
Molality =$\large\frac{3.59 moles}{0.200kg}$
$\Rightarrow 17.95molkg ^{-1}$
Total mass of the solution =422.6g
Volume of the solution =$\large\frac{422.6g}{1.072gmL^{-1}}$
$\Rightarrow 394.2ml$
$\Rightarrow 0.3942L$
Molarity =$\large\frac{3.59moles}{0.3942L}$
$\Rightarrow 9.11mol L^{-1}$
answered Aug 4, 2014 by sreemathi.v

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