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Standard XII

Chemistry

Depression in Freezing Point

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Question

Calculate the amount of $K C l$ which must be added to 1 kg of water so that the freezing point is depressed by 2K. $(k_{f}$ for water $= 1.86 K k g m o l^{- 1})$

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Solution

Change in freezing point, $Δ T_{f} = k_{f} \times m$
m= $\frac{Δ T_{f}}{k_{f}}$
$= \frac{2}{1.86} = 1.075 m o l e / k g$
$m = \frac{n u m b e r o f m o l e s o f s o l u t e}{m a s s o f s o l v e n t}$
$⟹ 1.075 = \frac{n}{1}$
$⟹ n = 1.075$

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Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2K. (kf for water =1.86Kkgmol−1)

Change in freezing point, ΔTf=kf×mm=ΔTfkf=21.86=1.075 mole/kgm=number of moles of solutemass of solvent⟹1.075=n1⟹n=1.075

Calculate the amount of $K C l$ which must be added to 1 kg of water so that the freezing point is depressed by 2K. $(k_{f}$ for water $= 1.86 K k g m o l^{- 1})$

Change in freezing point, $Δ T_{f} = k_{f} \times m$
m= $\frac{Δ T_{f}}{k_{f}}$
$= \frac{2}{1.86} = 1.075 m o l e / k g$
$m = \frac{n u m b e r o f m o l e s o f s o l u t e}{m a s s o f s o l v e n t}$
$⟹ 1.075 = \frac{n}{1}$
$⟹ n = 1.075$