At $298 \ K$ temperature,the solubility product $(K_{sp})$ for $AgCl$ is $1.5 \times 10^{-10}$. Calculate its solubility in $g \ L^{-1}$ in pure water. (Molar mass of $AgCl = 143.5 \ g \ mol^{-1}$)

  • A
    $1.22 \times 10^{-5} \ g \ L^{-1}$
  • B
    $1.75 \times 10^{-3} \ g \ L^{-1}$
  • C
    $1.50 \times 10^{-5} \ g \ L^{-1}$
  • D
    $2.15 \times 10^{-3} \ g \ L^{-1}$

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$A$ $1.0 \ L$ of aqueous solution contains $1 \times 10^{-8} \ M \ NaBr$,$1 \times 10^{-8} \ M \ NaCl$ and $1 \times 10^{-8} \ M \ NaI$. To this solution,$1 \times 10^{-10} \ M$ aqueous $AgNO_3$ solution is added dropwise. The order of precipitation of $AgX$ $(X = Cl, Br, I)$ is:
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