Question

\(5 \mathrm{ml}\) 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 Pbl \(_{2}\) precipitated will be about
(A) \(10^{-2} \mathrm{~mol}\)
(B) \(10^{-4} \mathrm{~mol}\)
(C) \(2 \times 10^{-4} \mathrm{~mol}\)
(D) \(10^{-3} \mathrm{~mol}\)

Answer 1

Ans: (B)
\(\begin{array}{rlccc} & \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\text { aq. }) & +2 \mathrm{Kl}(\text { (aq) }) & \longrightarrow & \mathrm{Pbl}_{2} \downarrow & \downarrow 2 \mathrm{KNO}_{3}\end{array}\) (aq)
Final milimole \(0.1\) of \(\mathrm{Pbl}_{2} \mathrm{PPt}=\frac{0.1}{1000}=1 \times 10^{-4}\)

Answer 2

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