Solubility product of silver bromide is $5.0 \times 10^{–13}$. The quantity of potassium bromide (molar mass taken as $120 \;g \;of\; mol^{–1}$) to be added to $1 \;litre\; of\; 0.05\; M$ solution of silver nitrate to start the precipitation of AgBr is

Question

( A ) $1.2 \times 10^{-10}\;g$
( B ) $1.2 \times 10^{-9}\;g$
( C ) $5.0 \times 10^{-8}\;g$
( D ) $6.2 \times 10^{-5}\;g$