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Calculate the freezing point of a solution containing 0.520g glucose $(C_6H_{12}O_6)$ dissolved in 80.20g of water.For water $K_f=1.86Kkg mol^{-1}$

$\begin{array}{1 1}0.067\\0.057\\0.077\\0.087\end{array} $

1 Answer

Molecular mass of glucose
$C_6H_{12}O_6=72+12+96$
$\Rightarrow 180g mol^{-1}$
$\Delta T_f=\large\frac{K_f\times W_B\times 1000}{M_B\times W_A}$
$\Rightarrow \large\frac{1.86\times 0.520\times 1000}{180\times 80.20}$
$\Rightarrow 0.067$
answered Aug 8, 2014 by sreemathi.v
 

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