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Ethylene glycol is used as an antifreeze in a cold climate.How much mass of ethylene glycol should be added to $5\;kg$ of water to prevent it form freezing at -$6^{\large\circ}C$ will be : $(K_t$ for water =$1.86Kkg mol^{-1}$ and molar mass of ethylene glycol=$62g mol^{-1})$

$\begin{array}{1 1} 1000\;g \\ 2040\;g \\ 1860\;g \\ 3042 \;g \end{array}$

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Answer: 1000 g.
Depression of Freezing point, $\Delta\; T_f = i\;K_f\; m$, where $i$ is the van't Hoff factor, $K_f$ is the Freezing Point Depression Constant and $m$ is the molality.
Ethylene glycol does not dissociat; hence its van't Hoff factor $i= 1 $
Molality $m$ is the number of moles of solute per kg of solvent, $m = \large\frac{W}{M}$$ \times \large \frac{1000}{V \;mL}$
$\Rightarrow m = \large\frac{W}{62\times5}$$ = \large\frac{W}{310}$
Subsitituting in $\Delta\; T_f = i\;K_f\; m$, we get $6^{\circ} = 1.86 \times \large\frac{W}{310}$
$\Rightarrow W = \large\frac{6 \times 310}{1.86}$$ = 1000\;g$
answered Feb 13, 2014 by sreemathi.v
edited Jul 16, 2014 by balaji.thirumalai

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