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$5 \mathrm{~m}$ of $0.1 \mathrm{M} ~ \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ is mixed with $10 \mathrm{ml}$ of $0.02 \mathrm{M} ~ \mathrm{KI}$. The amount of $\mathrm{Pbl}_{2}$ precipitated will be about
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The correct answer is:
$10^{-4} \mathrm{~mol}$
Hint:
mole of $\mathrm{Pbl}_{2} \quad \mathrm{PPt}=\frac{0.1}{1000}=1 \times 10^{-4}$
mole of $\mathrm{Pbl}_{2} \quad \mathrm{PPt}=\frac{0.1}{1000}=1 \times 10^{-4}$
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