Calculate the freezing point of solution when 1.9 g of $MgC l_{2}$ $(M - 95g mo l^{- 1})$ was dissolved in 50g of water, assuming $MgC l_{2}$ undergoes complete ionisation. $[K_{f} for water= 1 .86 K Kg mo l^{- 1}]$

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Solution

$Δ T_{f} = i K_{f} m$

$Δ T_{f} = \frac{(3) (1.86) \times 2}{5}$

$Δ T_{f} = 2.232$

$T_{i n i t i a l} - T_{f i n a l} = 2.232$

$T_{f i n a l} = T_{i n i t i a l} - 2.232$

$T_{f i n a l} = 0^{o} C - {2.232}^{o} C$

$T_{f i n a l} = - {2.232}^{o} C$

Note:
$K_{f} = 1.86 K K g . m o l^{- 1}$

$i = 3 [M g C l_{2} ⇌ M g^{2 +} + 2 C l^{-} i = \frac{2 + 1}{1} = 3]$

$m = \frac{G i v e n m a s s}{M o l e c u l a r m a s s} \times \frac{1}{w e i g h t o f s o l v e n t (k g)}$

$m = \frac{1.9}{95} \times \frac{1}{50} \times 1000 m o l / k g$

$m = \frac{20}{50} = \frac{2}{5} m o l / k g$

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