The solubility product of PbI₂ is 8.0 × 10⁻⁹. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10⁻⁶. mol/L. The value of x is _ _ _ _ _. (Rounded off to the nearest integer) [Given 2 = 1.41]

Equilibrium: JEE Main 2021 Chemistry question

JEE Main 2021 · 24 Feb, Evening shift · Q12

JEE Main 202124 Feb, Evening shiftChemistryMediumEquilibrium123× since 2002

The solubility product of PbI₂ is 8.0

\times

10^

-

9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x

\times

10^

-

6. mol/L. The value of x is __________. (Rounded off to the nearest integer) [Given

\sqrt 2

= 1.41]

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2021 Equilibrium Question with Answer (24 Feb Evening shift, Q12) | JEE Arc