The solubility product of $AgBr$ is $4.9 \times 10^{-13}$ at a certain temperature. Calculate the solubility.
- A$4 \times 10^{-6} \ mol \ dm^{-3}$
- B$4 \times 10^{-7} \ mol \ dm^{-3}$
- C$7 \times 10^{-7} \ mol \ dm^{-3}$
- D$3 \times 10^{-8} \ mol \ dm^{-3}$
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