The molar solubility of {\CaF}_2( {\K}_{ { sp}}=5.3 \times 10^{-11}) in 0.1 {\M} solution of {\NaF} will be [NEET OD 2019]

Correct Answer is : 5.3×10−9molL−1

5.3 \times 10^{-11} {\mol} {\L}^{-1}

5.3 \times 10^{-8} {\mol} {\L}^{-1}

5.3 \times 10^{-9} {\mol} {\L}-1

5.3 \times 10^{-10} {\mol} {\L}^{-1}

Class 11 NEET Chemistry Equilibrium Questions Part 2

Question : 27 of 28

Marks: +1, -0

The molar solubility of CaF2(K sp=5.3×10−11) in 0.1M solution of NaF will be

[NEET OD 2019]

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