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  3. Chemistry
  4. On chemical analysis, it is found that 200 mL of CaCl2 solution contains 3.01× 1022 chloride ions. Calculate the molarity of this solution.

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Q. On chemical analysis, it is found that 200 mL of $ CaC{{l}_{2}} $ solution contains $ 3.01\times {{10}^{22}} $ chloride ions. Calculate the molarity of this solution.

VMMC MedicalVMMC Medical 2013

Solution:

$ \underset{1\,\text{mol}}{\mathop{CaC{{l}_{2}}}}\,\xrightarrow{{}}C{{a}^{2+}}+\underset{2\,\times \,6.02\,\times \,{{10}^{23}}\text{ions}}{\mathop{2C{{l}^{-}}}}\, $ $ \therefore $ $ 301\times {{10}^{22}}C{{l}^{-}} $ ions will be present in $ CaC{{l}_{2}}=\frac{1\times 3.01\times {{10}^{22}}}{2\times 6.02\times {{10}^{23}}} $ $ =0.025\,mol $ Molarity of solution $ =\frac{0.025\,mol}{0.200\,L}=0.125\,M $