A) \[0.017g\text{ }litr{{e}^{-1}}~~~\]
B) \[0.071\text{ }g\text{ }litr{{e}^{-1}}\]
C) \[0.17g\text{ }litr{{e}^{-1}}\]
D) \[1.7\text{ }g\text{ }litr{{e}^{-1}}\]
Correct Answer: A
Solution :
Sol. \[{{K}_{sp}}=4\times {{10}^{-11}}=4{{s}^{3}}\] \[s={{[10]}^{1/3}}\times {{10}^{-4}}mole/litre\] \[=2.2\times {{10}^{-4}}mole/litre\] (Molecular weight of\[Ca{{F}_{2}}=78\]) \[=0.017g\text{ }litr{{e}^{-1}}\]
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