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Question
Calculate the boiling point elevation for a solution prepared by adding 10 g of MgCl2 to 200 g of water, assuming MgCl2 is completely dissociated.
(Kb for Water = 0.512 K kg mol−1, Molar mass MgCl2 = 95 g mol−1)
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Solution
We know that
ΔTb = iKbm
\[\ce{MgCl2 -> \underset{1 + 2}{Mg^{2+} + 2Cl^-}}\]
i = 3
Here, m = `("W"_"B" xx 1000)/("M"_"B" xx "W"_"A")` = 0.526
ΔTb = iKbm
= 3 × 0.512 × 0.526
= 0.808 K
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