The solubility product $(K_{sp})$ of $AgCl$ at $100 \, ^\circ C$ is $1.44 \times 10^{-4}$. The solubility of $AgCl$ in boiling water is $.......$
- A$0.72 \times 10^{-4} \, M$
- B$1.20 \times 10^{-2} \, M$
- C$0.72 \times 10^{-2} \, M$
- D$1.20 \times 10^{-4} \, M$
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