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How many grams of CaC₂O₄ (molecular weight = 128) on dissolving in distilled water will give a saturated solution
$[K_{sp} ({CaC}_{2} O_{4}) = 2 .5 \times 10^{- 9} {mol}^{2} I^{- 2}]$

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0.0064 g

0.1280 g

0.0128 g

1.2800 g

answer is A.

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Detailed Solution

Solubility of ${CaC}_{2} O_{4} = \sqrt{K_{sp}} = \sqrt{2 .5 \times 10^{- 4}}$
$\begin{array}{l} = 5 \times 10^{- 5} {molL}^{- 1} \\ = 5 \times 10^{- 5} \times 128 = 640 \times 10^{- 5} = 0 .0064 g \end{array}$

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