Calculate the freezing point of a solution containing $60 g$ of glucose (Molar mass $= 180 g m o l^{- 1}$ ) in $250 g$ of water.

( $K f$ of water $= 1.86 K k g m o l^{- 1}$ ).

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Solution

The number of moles of glucose $= \frac{Mass of glucose}{Molecular weight of glucose} = \frac{60 g}{180 g/mol} = 0.333 mol$

The molality of glucose solution $= \frac{Number of moles of glucose}{Mass of solvent (in kg)} = \frac{0.333 mol}{250 g \times \frac{1 kg}{1000 g}} = 1.333 mol/kg$

The depression in the freezing point of glucose solution :

$Δ T_{f} = K_{f} m = 1.86K kg /mol \times 1.333 mol/kg = 2.48 K$

The freezing point of pure water is $0^{o} C$

The freezing point of glucose solution $= 0 - 2.48 = - {2.48}^{o} C$

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(a) A $10 %$ solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of $10 %$ glucose in water if the freezing point of pure water is 273.15 K.
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(Molar mass of glucose = 180 g $m o l^{- 1}$ )
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]

Question 2.32

Calculate the depression in the freezing point of water when 10g of $C H_{3} C H_{2} C H C l C O O H$ is added to 250 g of water, $K_{a} = 1.4 \times 10^{- 3}$ ; $K_{f} = 1.86 K k g m o l^{- 1}$

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Calculate the freezing point of a solution containing 60 g of glucose (Molar mass =180 g mol−1) in 250 g of water. (Kf of water =1.86 K kg mol−1).

Calculate the freezing point of a solution containing $60 g$ of glucose (Molar mass $= 180 g m o l^{- 1}$ ) in $250 g$ of water.

( $K f$ of water $= 1.86 K k g m o l^{- 1}$ ).