The solubility of $AgCl$ is $1.435 \times 10^{-5} \ g \ L^{-1}$ at $30^{\circ} C$. Calculate its solubility product $(K_{sp})$. (Molar mass of $AgCl = 143.5 \ g \ mol^{-1}$)
- A$1.0 \times 10^{-10}$
- B$1.0 \times 10^{-14}$
- C$1.0 \times 10^{-5}$
- D$2.0 \times 10^{-10}$
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