The heaviest member of the alkaline earth metals is radium (Ra), a naturally radioactive element discovered by Pierre and Marie Curie in \(1898 .\) Radium was initially isolated from the uranium ore pitchblende, in which it is present as approximately \(1.0 \mathrm{g}\) per 7.0 metric tons of pitchblende. How many atoms of radium can be isolated from \(1.75 \times 10^{8} \mathrm{g}\) pitchblende (1 metric ton = 1000 kg)? One of the early uses of radium was as an additive to paint so that watch dials coated with this paint would glow in the dark. The longest-lived isotope of radium has a half-life of \(1.60 \times 10^{3}\) years. If an antique watch, manufactured in \(1925,\) contains \(15.0 \mathrm{mg}\) radium, how many atoms of radium will remain in \(2025 ?\)

First, let's determine the amount of radium in the given mass of pitchblende. To do this, we'll use the given proportion: 1.0 g radium per 7.0 metric tons of pitchblende Given mass of pitchblende: \(1.75 \times 10^8\) g Now, we'll convert the mass of pitchblende to metric tons: \(1.75 \times 10^8 \text{ g} \div 1000 \text{ g/kg} \div 1000\text{ kg/metric ton} = 175 \text{ metric tons}\) Next, we'll use the proportion to find the mass of radium in the given mass of pitchblende: \(\text{Mass of radium} = 175 \text{ metric tons pitchblende} \times \frac{1.0 \text{ g radium}}{7.0 \text{ metric tons pitchblende}} = 25 \text{ g radium}\) Since we know the mass of radium, we can calculate the number of atoms using the molar mass of radium (226 g/mol) and Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol): \(\text{Moles of radium} = \frac{25 \text{ g radium}}{226 \text{ g/mol}} = 0.1106 \text{ mol}\) \(\text{Number of radium atoms} = 0.1106 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 6.66 \times 10^{22} \text{ atoms}\) So, there are \(6.66 \times 10^{22}\) radium atoms in \(1.75 \times 10^8\) g of pitchblende.

Initially, the antique watch contains 15.0 mg of radium, which is equivalent to: \(\text{Mass of radium} = 15.0 \text{ mg} \times \frac{1 \text{ g}}{1000 \text{ mg}} = 0.015 \text{ g}\) Now, we'll convert the mass of radium into moles and then into atoms: \(\text{Moles of radium} = \frac{0.015 \text{ g radium}}{226 \text{ g/mol}} = 6.637 \times 10^{-5} \text{ mol}\) \(\text{Initial number of radium atoms} = 6.637 \times 10^{-5} \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 3.995 \times 10^{19} \text{ atoms}\) We know that the half-life of radium is \(1.60 \times 10^3\) years. The watch was manufactured in 1925, and we want to find the remaining radium atoms in 2025. The elapsed time is: \(\text{Elapsed time} = 2025 - 1925 = 100 \text{ years}\) Using the half-life formula: \(\text{Final number of radium atoms} = \text{Initial number of radium atoms} \times \left(\frac{1}{2}\right)^{\frac{\text{Elapsed time}}{\text{Half-life}}}\) \(\text{Final number of radium atoms} = 3.995 \times 10^{19} \text { atoms} \times \left(\frac{1}{2}\right)^{\frac{100 \text{ years}}{1.60 \times 10^3\text{ years}}} = 3.995 \times 10^{19} \text{ atoms} \times 0.949\) \(\text{Final number of radium atoms} = 3.789 \times 10^{19} \text{ atoms}\) So, there will be approximately \(3.789 \times 10^{19}\) atoms of radium remaining in the antique watch in 2025.