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  4. For NaCl , the Ksp=36 mol2L-2 , the molar concentration of it will be

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Q. For $ NaCl $ , the $ {{K}_{sp}}=36\,\,mo{{l}^{2}}{{L}^{-2}} $ , the molar concentration of it will be

Rajasthan PMTRajasthan PMT 2009

Solution:

For $ NaCl $ $ {{K}_{sp}}={{S}^{2}} $ where, $ S= $ solubility $ \Rightarrow $ $ 36={{S}^{2}} $ $ \therefore $ $ S=6\,\,mol\,\,{{L}^{-1}} $ Hence, molarity $ =6\,\,[M] $