$K_{s p}=\left[A g^{+}\right]\left[B r^{-}\right]$

For precipitation to occur

lonic product $>$ Solubility product

$\left[B r^{-}\right]=\frac{K_{m}}{| A g^{+1}}=\frac{5 \times 10^{-13}}{0.05}=10^{-11}$

i.e., precipitation just starts when $10^{-11}$

moles of $K B r$ is added to $1\, \ell \,A g N O_{3}$ solution

$\therefore$ Number of moles of $B r^{-}$ needed

from $K B r=10^{-11}$

$\therefore$ Mass of $K B r=10^{-11} \times 120$

$=1.2 \times 10^{-9} g$