Question

54.5 mg of $N{a}_{3}P{O}_4$ contains ${}^{32}\text{P}$ (15.6%) and ${}^{31}\text{P}$ atoms. Assuming only ${}^{32}\text{P}$ atoms radioactive, the rate of decay for the given sample of $N{a}_{3}P{O}_4$ is :
(Half-life period for ${}^{32}\text{P}=14.3day$; molar mass of $N{a}_{3}P{O}_4=161.2$)

• A. $6.25\times {10}^{13}dps$
• B. $1.78\times {10}^{13}dps$
• C. $6.37\times {10}^{13}dps$
• D. none of these

Answer 1

Answer: (B)

$1.78\times {10}^{13}dps$

Mole of $N{a}_{3}P{o}_4=\frac{54.5\times {10}^{-3}}{161.2}$
Mole of $P$ atoms $=\frac{54.5\times {10}^{-3}}{161.2}$
$\therefore$ Mole of $P^{32}$ atoms $=\frac{54.5\times {10}^{-3}}{161.2}\times \frac{15.6}{100}=5.27\times {10}^{-5}$
$\therefore$ Atoms of $P^{32}=5.27\times {10}^{-5}\times 6.023\times {10}^{23}$
Now, Rate $=\lambda N=\frac{0.693}{14.3\times 24\times 60\times 60}\times 5.27\times {10}^{-5}\times 6.023\times {10}^{23}=1.78\times {10}^{13}dps$