Chapter 31, Problem 46P

To determine

To find:The mass of $\mathrm{C}$

Answer to Problem 46P

  $0.224\mathrm{\mu }\text{g}$

Explanation of Solution

Given data:

Half-life of $\mathrm{C}$ = 5730 yr

Activity of carbon is $\left(1.0\times {10}^{-6}\text{Ci}\right)\left(\frac{3.70\times {10}^{10}\text{ decays}/\text{s}}{1\text{Ci}}\right)=3.70\times {10}^4\text{ decays}/\text{s}$

  $\begin{array}{l}\Rightarrow \mathrm{N}=\frac{\mathrm{\Delta }\mathrm{N}}{\mathrm{\Delta }\mathrm{t}}\frac{{\mathrm{T}}_{1/2}}{0.693}\\ =\left(3.70\times {10}^4\text{ decays}/\text{s}\right)\frac{5730\text{yr}}{0.693}\left(3.156\times {10}^7\text{s}/\text{yr}\right)\\ =9.655\times {10}^{15}\text{ nuclei}\end{array}$

Since, 1 mole of carbon is 0.014 kg

  $\begin{array}{c}9.655\times {10}^{15}\text{ nuclei }\left(\frac{0.0140\text{kg}}{6.02\times {10}^{23}\text{ nuclei }}\right)\\ =2.24\times {10}^{-10}\text{kg}\\ =0.224\mathrm{\mu }\text{g}\end{array}$

Conclusion:

The mass of $\mathrm{C}$

  $0.224\mathrm{\mu }\text{g}$