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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
  $4 \times {10^{ - 11}}\,g$
• ✓
  $5.95 \times {10^{ - 10}}\,g$
• D
  $2.97 \times {10^{ - 10}}\,g$