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Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
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A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K

$\begin{array}{1 1}269.06K\\169.06K\\369.06K\\469.06K\end{array} $

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Molar mass of cane sugar $C_{12}H_{22}O_{11}=342gmol^{-1}$
Molality of sugar =$\large\frac{5\times 1000}{342\times 100}$
$\Rightarrow 0.146$
$\Delta T_f$ for sugar solution =273.15-271=$2.15^{\large\circ}$
$\Delta T_f=K_f \times m$
$K_f=\large\frac{2.15}{0.146}$
Molality of glucose solution $=\large\frac{5}{180}\times \frac{1000}{100}$$=0.278$
$\Delta T_f$(Glucose)=$\large\frac{2.15}{0.146}$$\times 0.278$
$\Rightarrow 4.09^{\large\circ}$
$\therefore$ Freezing point of glucose solution =$273.15-4.09^{\large\circ}=269.06K$
answered Aug 5, 2014 by sreemathi.v
 
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