KCET · Chemistry · Surface Chemistry
\(1 \mathrm{~g}\) of silver gets distributed between \(10 \mathrm{~cm}^{3}\) of molten zinc and \(100 \mathrm{~cm}^{3}\) of molten lead at \(800^{\circ} \mathrm{C}\). The percentage of silver still left in the lead layer is approximately
- A 2
- B 5
- C 3
- D 1
Answer & Solution
Correct Answer
(C) 3
Step-by-step Solution
Detailed explanation
Partition coefficient
\[
=\frac{\text { Conc. of } \mathrm{Ag} \text { in molten } \mathrm{Zn}}{\text { Conc. of } \mathrm{Ag} \text { in molten } \mathrm{Pb}}
\]
Mass of \(\mathrm{Ag}\) in lead at equilibrium \(=1-x\)
Mass of Ag in \(\mathrm{Zn}\) at equilibrium \(=x\)
\[
\begin{aligned}
300 &=\frac{x / 10}{1-x / 100}=\frac{10 x}{1-x} \\
10 x &=300(1-x) \\
10 x &=300-300 x \\
310 x &=300 \\
x &=\frac{300}{310} \\
x &=\frac{30}{31}
\end{aligned}
\]
Amount of silver in molten lead
\[
=1-x=1-\frac{30}{31}=\frac{31-30}{31}=\frac{1}{31}
\]
\(\therefore \quad \%\) of silver \(=\frac{1}{31} \times 100 \simeq 3 \%\)
\[
=\frac{\text { Conc. of } \mathrm{Ag} \text { in molten } \mathrm{Zn}}{\text { Conc. of } \mathrm{Ag} \text { in molten } \mathrm{Pb}}
\]
Mass of \(\mathrm{Ag}\) in lead at equilibrium \(=1-x\)
Mass of Ag in \(\mathrm{Zn}\) at equilibrium \(=x\)
\[
\begin{aligned}
300 &=\frac{x / 10}{1-x / 100}=\frac{10 x}{1-x} \\
10 x &=300(1-x) \\
10 x &=300-300 x \\
310 x &=300 \\
x &=\frac{300}{310} \\
x &=\frac{30}{31}
\end{aligned}
\]
Amount of silver in molten lead
\[
=1-x=1-\frac{30}{31}=\frac{31-30}{31}=\frac{1}{31}
\]
\(\therefore \quad \%\) of silver \(=\frac{1}{31} \times 100 \simeq 3 \%\)
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