If Ksp for HgSO4 is 6.4 × 10-5, then solubility of this substance in mole per m3 is

Ksp = s2 s = Ksp = 6.4×10-5s = 8 × 10-3 s = 8 mol/m3

Ksp = s2 s = Ksp = 6.4×10-5s = 8 × 10-3 s = 8 mol/m3

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If $K_{sp} for {HgSO}_{4}$ is $6.4 \times 10^{- 5},$ then solubility of this substance in mole per $m^{3}$ is

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$6.4 \times 10^{- 5}$

none of these

$8 \times 10^{- 6}$

$8$

answer is A.

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$K_{sp} = s^{2}$ $s = \sqrt{K_{sp}} = \sqrt{6.4 \times 10^{- 5}}$

$s = 8 \times 10^{- 3}$ $s = 8 mol / m^{3}$

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