1. Tardigrade
  2. Question
  3. Chemistry
  4. An aqueous solution containing 1.248 g of barium chloride (molar mass =208.34g mol-1 ) in 100 g of water boils at 100.0832oC . The degree of dissociation of BaCl2 will be ( Kb for water =0.52K text kg text mol-1 )

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Q. An aqueous solution containing 1.248 g of barium chloride (molar mass $ =208.34g\,mo{{l}^{-1}} $ ) in 100 g of water boils at $ {{100.0832}^{o}}C $ . The degree of dissociation of $ BaC{{l}_{2}} $ will be ( $ {{K}_{b}} $ for water $ =0.52K\text{ }kg\text{ }mo{{l}^{-1}} $ )

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Solution:

$ {{M}_{2}} $ (observe $ =\frac{1000\times 0.52\times 1.248}{100\times 0.0832}=78g\,\,mo{{l}^{-1}} $ $ i=\frac{208.34}{78}=2.67 $ $ \underset{1-\alpha }{\mathop{\underset{1\,mol}{\mathop{BaC{{l}_{2}}}}\,}}\,\xrightarrow{{}}\underset{\alpha }{\mathop{\underset{0}{\mathop{B{{a}^{2+}}}}\,}}\,+\underset{\begin{smallmatrix} 0 \\ 2\alpha \end{smallmatrix}}{\mathop{2C{{l}^{-}}}}\, $ Number of total moles $ =1+2\alpha $ or $ i=1+2\alpha $ or $ \alpha =\frac{i-1}{2}=\frac{2.67-1}{2}=0.835 $