The required amount of $KBr$ (molar mass $= 119 \, g/mol$) in grams 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
    $2.97 \times 10^{-10} \, g$

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