Question

A very small mass 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 $10c{m}^3$ of the supernatent liquid gave a residue having a radioactivity $\frac{1}{24000}$ of that of the original quantity of ${}^{213}Pb$. Calculate the solubility of lead chromate in mol $d{m}^{-3}$ :

• A. $3.0\times {10}^{-7}mold{m}^{-3}$
• B. $2.0\times {10}^{-7}mold{m}^{-3}$
• C. $4.0\times {10}^{-7}mold{m}^{-3}$
• D. None of these

Answer 1

Answer: (B)

$2.0\times {10}^{-7}mold{m}^{-3}$

Since, the solution shows a radioactivity only $\frac{1}{24000}$ of the original mixture, therefore, the fraction of the radioactive lead obtained after evaporation of the supernatent liquid $=\frac{1}{24000}$. Since, almost lead has been precipitated in form of lead chromate an insoluble salt and therefore,
Fraction of non$-$radioactive lead may also be taken $=\frac{1}{24000}$
The lead salt originally contained 0.01 g of Pb
$\therefore$ The mass of non-radioactive lead obtained from
$10$ c.c.$=0.01\times \frac{1}{24000}$ g
The molar mass of Pb is 207 g/mol.
$\therefore$ Moles of non-radioactive lead obtained from 10 cc is the product of the mass and the molar mass of Pb. It is
$\frac{0.01}{24000\times 207}$ mol
From 1 litre, moles of non-radioactive lead obtained is
$\frac{0.01}{24000\times 207}\times \frac{1000}{10}=2.0\times {10}^{-7}mol$. 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
$2.0\times {10}^{-7}mold{m}^{-3}$