If the solubility product of {\PbS} is 8 \times 10^{-28}, then the solubility of {\PbS} in pure water at 298 {\K} is \times \times 10 -16 {\mol} {\L}^{-1}. The value of x is _______. (Nearest Integer) [Given : \sqrt{2}=1.41 ] [29-Jul-2022-Shift-1]

JEE Mains 29-July-2022 Shift 1 Solved Paper

Section: Chemistry

Question : 86 of 90

Marks: +1, -0

If the solubility product of PbS is 8×10−28, then the solubility of PbS in pure water at 298K is ××10 −16‌mol‌L−1. The value of x is _______. (Nearest Integer)
[Given : √2=1.41 ]

[29-Jul-2022-Shift-1]

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