A very small amount of radioactive isotope of .^(213)Pb was mixed with a non-radioactive lead salt containing 0.01 g of Pb (atomic mass 207). The whole lead was brought into solution and lead chromate was precipitated by addition of a soluble chromate. Evaporation of 10 cm^(3) of the supernature liquid gave a residue having a radioactivity 1/24000 of that of the original quantity of .^(213)Pb. If the solubility of lead chromate is x xx 10^(-y) mol dm^(-3), then value of x is

A very small amount of radioactive isotope of .^(213)Pb was mixed with

1.	A very small amount of radioactive isotope of .^(213)Pb was mixed with a non-radioactive lead salt containing 0.01 g of Pb (atomic mass 207). The whole lead was brought into solution and lead chromate was precipitated by addition of a soluble chromate. Evaporation of 10 cm^(3) of the supernature liquid gave a residue having a radioactivity 1/24000 of that of the original quantity of .^(213)Pb. If the solubility of lead chromate is x xx 10^(-y) mol dm^(-3), then value of x is
Answer» SOLUTION :Since the activity of solution is `(1)/(24000)` times to the original mixture, THEREFORE the fraction of the radioactive lead obtained after evaporation of the supernatant liquid will be `(1)/(24000)`. Because almost whole amount lead is precipitated in the form of `Pb CrO_(4)` an insoluble salt, `:.` Fraction of non-radioactive lead MAY also be taken `= (1)/(24000)` The mass of non-radioactive lead obtained from `100c.c. = (0.01)/(24000)` `:.` Moles of non-radioactive lead obtained from 1 L i.e., solution `= (0.01)/(207) xx (1)/(24000) xx (1000)/(10) = 2.0 xx 10^(-7) mol. dm^(-3)` `:. X = 2`