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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

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Solution :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%`


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