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

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Solution

$i = 2$
$W_{A} = 1 k g = 1000 g$
$Δ T_{f} = 2 k$
$K_{f} = 1.86 k k g m o l^{- 1}$
$H_{B} = 74.5 g / m o l^{- 1}$
substituting those values in expansion
$W_{B} = \frac{(Δ T_{f}) (H_{B}) (W_{A})}{i K_{f} (1000)}$
$= \frac{(2) (745) (1000)}{(2) (186) (1000)}$
$∴$ amount of $K_{d}$ to added $= 40.05 m o l e s$

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