A sample of sodium phosphate, , weighing 86.4 mg contains radioactive . If 0.164% of the phosphorus — Nuclear Chemistry and Radioactivity Chemistry Question
Question
A sample of sodium phosphate, $\text{Na}_3\text{PO}_4$, weighing 86.4 mg contains radioactive $\text{P}^{32}$. If 0.164% of the phosphorus atoms in the compound is $\text{P}^{32}$, activity of the sample (in order of $10^{11}$ dps) is ($\text{P}^{32}$ has a half-life of 14.0 days, $N_A = 6 \times 10^{23}$, $\ln 2 = 0.7$)
💡 Solution & Explanation
Molar mass of $\text{Na}_3\text{PO}_4 = 3(23) + 31 + 4(16) = 164 \text{ g/mol}$. Moles of $\text{Na}_3\text{PO}_4 = \frac{86.4 \times 10^{-3}}{164}$. Total number of P atoms $= \frac{86.4 \times 10^{-3}}{164} \times 6 \times 10^{23} \approx 3.16 \times 10^{20}$. Number of $\text{P}^{32}$ atoms $N = 0.164\% \times 3.16 \times 10^{20} \approx 5.184 \times 10^{17}$. Decay constant $\lambda = \frac{0.7}{14 \times 24 \times 3600} \text{ s}^{-1}$. Activity $A = \lambda N = \frac{0.7}{1.2096 \times 10^6} \times 5.184 \times 10^{17} = 3 \times 10^{11} \text{ dps}$. The coefficient for $10^{11}$ is 3. Therefore, correct answer is 3.