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Calculate the temperature at which a solution containing 54g of glucose,$C_6H_{12}O_6$ in 250g of water will freeze ($K_f$ for water =$1.86K kg mol^{-1}$)

$\begin{array}{1 1}-2.23^{\large\circ}C\\-3.23^{\large\circ}C\\-1.23^{\large\circ}C\\-0.23^{\large\circ}C\end{array} $

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Molecular mass of glucose
$\;\;\;\;\;\;=180g\; mol^{-1}$
$\Delta T_f=\large\frac{K_f\times W_B\times 1000}{M_B\times w_B}$
$\Rightarrow \large\frac{1.86\times 54\times 1000}{180\times 250}$
$\Rightarrow 2.23$
Freezing point of solution =0-$2.23=-2.23^{\large\circ}C$
answered Aug 7, 2014 by sreemathi.v
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