On gram of silver gets distributed between `10cm^(3)` of molten zinc and 100 `cm^(3)` of molten lead at `800^(@)C`. The percentage of silver still left in the lead layer is approximately

Options

A. 2
B. 5
C. 3
D. 1

Answer

C

Explanation

According to Nernst distribution law,
`("Conc. (%) of Ag in molten Zn")/("Conc. (%) of Ag in molten Pb")=K" (distribution coeff.)"`
= 300
If at equilibrium, x g of Ag is left in `100cm^(3)` of molten lead, Ag that goes into `10cm^(3)` zinc `=(1-x)g`
`therefore" "(10(1-x))/(x)=300" or "10-10x=300x`
`"or "310x=10" or "x=(1)/(31)g`
Thus, out of 1 g of Ag, Ag present in Pb layer
`=(1)/(31)g`
`therefore" % of Ag left in the Pb layer "=(1)/(31)xx100~~3%`