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The molar solubility of $\mathrm{CaF}_2$ $\left(\mathrm{K}_{\mathrm{sp}}=5.3 \times 10^{-11}\right)$ in $0.1 \mathrm{M}$ solution of $\mathrm{NaF}$ will be

ChemistryIonic EquilibriumNEETNEET 2019 (Odisha)

Options:

A $5.3 \times 10^{11} \mathrm{~mol} \mathrm{~L}^{-1}$
B $5.3 \times 10^{-8} \mathrm{~mol} \mathrm{~L}^{-1}$
C $5.3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}$
D $5.3 \times 10^{-10} \mathrm{~mol} \mathrm{~L}^{-1}$

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