Chemistry · 2019
NEET 2019 · 20 May · Code G1 · Q123
The molar solubility of CaF_2( K_sp =5.3 × 10^-11) in 0.1 M solution of NaF will be -
The molar solubility of $\displaystyle \mathrm{CaF}_2\left(\mathrm{~K}_{\mathrm{sp}}=5.3 \times 10^{-11}\right)$ in $\displaystyle 0.1$ M solution of NaF will be -
Official answer
(3)
$\displaystyle 5.3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}$
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NEET (UG) 2019 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.