Required amount of KBr (molar mass = 119 ) in gram to start precipitation of AgBr in 500,mL solution

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6-2.Equilibrium-II (Ionic Equilibrium)

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The required amount of $KBr$ (molar mass $= 119$ ) in gram to start the precipitation of $AgBr$ in $500\,mL$ solution of $0.05\,M\,\,AgNO_3$ will be :- ( $K_{SP}$ of $AgBr = 5 \times 10^{-13}$ )

A

$1.19\times 10^{-9}\,g$

B

$4\times 10^{-11}\,g$

C

$5.95\times 10^{-10}\,g$

D

$2.97\times 10^{-10}\,g$

Solution

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Standard 11

Chemistry

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