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Question: Answered & Verified by Expert
$1 \mathrm{~g}$ 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
ChemistrySurface ChemistryKCETKCET 2011
Options:
  • A 2
  • B 5
  • C 3
  • D 1
Solution:
1125 Upvotes Verified Answer
The correct answer is: 3
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 $=\frac{1}{31} \times 100 \simeq 3 \%$

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