Question

A chemist wishing to do an experiment requiring${}^{\mathbf{47}}\mathbf{Ca}^{\mathbf{2}\mathbf{+}}$(half-life =4.5 days) needs $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu g}$of the nuclide. What mass of ${}^{\mathbf{47}}\mathbf{CaCO}_{\mathbf{3}}$must be ordered if it takes 48 h for delivery from the supplier? Assume that the atomic mass of⁴⁷Ca is 47.0 u.

Short answer

The mass of calcium carbonate that contains $6.80\mu g$of calcium is $15.5\mu g$ .

Step-by-step solution

Calculating the rate constant.

The formula for evaluating the rate constant is:

$\mathbf{k}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{693}}{{\mathbf{t}}_{{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}}}$

The known value is substituted.

$\begin{array}{c}\mathrm{k}=\frac{0.693}{4.5days}\\ =0.154{\text{days}}^{-1}\end{array}$

Calculating the mass of calcium

Using the integrated rate law for first-order reaction:

$\mathbf{ln}\frac{{\left[\mathbf{N}\right]}_{\mathbf{t}}}{{\left[\mathbf{N}\right]}_{\mathbf{0}}}\mathbf{=}\mathbf{-}\mathbf{kt}$

The known values are substituted.

$\begin{array}{c}\ln \left(\frac{5.0mg}{x}\right)=-0.154day{s}^{-1}\times 2days\\ ln\left(\frac{5.0mg}{x}\right)=-0.308\\ \ln 5-\ln x=-0.308\\ \ln x=1.917\\ x=6.80\mathrm{\mu g}\end{array}$

Determining the mass of calcium carbonate .

Themass of${}^{47}\mathrm{CaC}{\mathrm{O}}_3$ is evaluated as:

role="math" localid="1663798010262" $\begin{array}{c}Mass=6.80\times \left({10}^{-6}\right)g^{-1}\times \left(\frac{1molCa}{47.0g}\right)\times \left(\frac{1molCaC{O}_3}{1molCa}\right)\times \left(\frac{107.01g}{1molCaC{O}_3}\right)\times \frac{1\mu g}{{10}^{-6}g}\\ =15.5\mathrm{\mu g}\end{array}$

Hence, the mass of calcium carbonate that contains $6.80\mathrm{\mu g}$of calcium is $15.5\mathrm{\mu g}$.