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Question

A very small mass of radioactive isotope of 213Pb 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 10cm3 of the supernatent liquid gave a residue having a radioactivity 124000 of that of the original quantity of 213Pb. Calculate the solubility of lead chromate in mol dm3 :

A
3.0×107moldm3
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B
2.0×107moldm3
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C
4.0×107moldm3
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D
None of these
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Solution

The correct option is D 2.0×107moldm3
Since, the solution shows a radioactivity only 124000 of the original mixture, therefore, the fraction of the radioactive lead obtained after evaporation of the supernatent liquid =124000. Since, almost lead has been precipitated in form of lead chromate an insoluble salt and therefore,
Fraction of nonradioactive lead may also be taken =124000
The lead salt originally contained 0.01 g of Pb
The mass of non-radioactive lead obtained from
10 c.c.=0.01×124000 g
The molar mass of Pb is 207 g/mol.
Moles of non-radioactive lead obtained from 10 cc is the product of the mass and the molar mass of Pb. It is
0.0124000×207 mol
From 1 litre, moles of non-radioactive lead obtained is
0.0124000×207×100010=2.0×107mol. This is also equal to the number of moles of lead chromate present in the solution.
The solubility of lead chromate is the number of moles of lead chromate per litre of solution. It is
2.0×107moldm3

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