A chemist wishing to do an experiment requiring \(^{47} \mathrm{Ca}^{2+}\) (half- life \(=4.5\) days needs 5.0\(\mu \mathrm{g}\) of the nuclide. What mass of \(^{47} \mathrm{CaCO}_{3}\) must be ordered if it takes 48 \(\mathrm{h}\) for delivery from the supplier? Assume that the atomic mass of \(^{47} \mathrm{Ca}\) is 47.0 \(\mathrm{u} .\)

Short Answer

Expert verified
In conclusion, the chemist must order 150.6 µg of \(^{47} \mathrm{CaCO}_{3}\) for the experiment.

Step by step solution

01

Calculate the amount of \(^{47} \mathrm{Ca}^{2+}\) needed at the time of delivery

We are given that the experiment requires 5.0 µg of the nuclide and the delivery takes 48 hours. Since the half-life of \(^{47} \mathrm{Ca}^{2+}\) is 4.5 days, we need to calculate how many half-lives have passed during this delivery period. First, convert the 48 hours delivery period into days: \[48 \;\text{hours} = 48\;\text{hours} \times \frac{1\;\text{day}}{24\;\text{hours}} = 2\;\text{days}\] Next, calculate the number of half-lives that passed during these 2 days: \[\frac{2\;\text{days}}{4.5\;\text{days/half-life}} = 0.4444\;\text{half-lives}\] Now, we calculate how much \(^{47} \mathrm{Ca}^{2+}\) is needed at the time of delivery: \[5.0\;\mu\text{g} \times 2^{0.4444} = 7.079\;\mu\text{g}\]
02

Calculate the moles of \(^{47} \mathrm{Ca}^{2+}\) required

Now that we know how much \(^{47} \mathrm{Ca}^{2+}\) is needed at the time of delivery, we can calculate the required moles by dividing the mass by the atomic mass of \(^{47} \mathrm{Ca}\): \[7.079\;\mu\text{g} \times \frac{1\;\text{mol}}{47.0\;\text{u}} = 1.506\;\text{µmol}\]
03

Calculate the moles of \(^{47} \mathrm{CaCO}_{3}\) needed

Since one mole of \(^{47} \mathrm{CaCO}_{3}\) contains one mole of \(^{47} \mathrm{Ca}^{2+}\), the moles of \(^{47} \mathrm{CaCO}_{3}\) needed will be equal to the moles of \(^{47} \mathrm{Ca}^{2+}\) required: \[1.506\;\text{µmol}\; \mathrm{Ca}^{2+} = 1.506\;\text{µmol}\;\mathrm{CaCO}_{3}\]
04

Determine the mass of \(^{47} \mathrm{CaCO}_{3}\) needed to be ordered

To determine the mass of \(^{47} \mathrm{CaCO}_{3}\) needed, multiply the moles by the molar mass of \(^{47} \mathrm{CaCO}_{3}\). The molar mass of \(^{47} \mathrm{CaCO}_{3}\) is approximately 47.0 u (for \(^{47} \mathrm{Ca}\)) + 12.0 u (for C) + 3 × 16.0 u (for 3O) = 100.0 u: \[1.506\;\text{µmol}\;\mathrm{CaCO}_{3} = 1.506\;\text{µmol} \times \frac{100.0\;\text{u}}{1\;\text{mol}} = 150.6\;\mu\text{g}\; \mathrm{CaCO}_{3}\] In conclusion, the chemist must order 150.6 µg of \(^{47} \mathrm{CaCO}_{3}\) for the experiment.

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