JEE MAIN - Chemistry (2022 - 29th July Morning Shift - No. 20)
If the solubility product of PbS is 8 $$\times$$ 10$$-$$28, then the solubility of PbS in pure water at 298 K is x $$\times$$ 10$$-$$16 mol L$$-$$1. The value of x is __________. (Nearest Integer)
[Given : $$\sqrt2$$ = 1.41]
Answer
282
Explanation
$$\mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{2}$$
$$\mathrm{S}=\sqrt{K_{s p}}=\sqrt{8 \times 10^{-28}}=2 \sqrt{2} \times 10^{-14}$$
$$=2.82 \times 10^{-14}$$
$$=282 \times 10^{-16}$$
$$ \therefore $$ Ans. 282
$$\mathrm{S}=\sqrt{K_{s p}}=\sqrt{8 \times 10^{-28}}=2 \sqrt{2} \times 10^{-14}$$
$$=2.82 \times 10^{-14}$$
$$=282 \times 10^{-16}$$
$$ \therefore $$ Ans. 282
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