Among $10^{-9} \mathrm{~g}$ (each) of the following elements, which one will have the highest number of atoms?

Element: $\mathrm{Pb}, \mathrm{Po}, \mathrm{Pr}$ and Pt

<p>To determine which element has the highest number of atoms among 10^-9 grams of the elements Pb, Po, Pr, and Pt, we use the formula for calculating the number of atoms:</p> <p>$ \text{Number of atoms} = \frac{\text{Mass}}{\text{Molar Mass (g/mol)}} \times N_A $</p> <p>where $ N_A $ is Avogadro's number.</p> <p>From this formula, it's clear that for a given mass, the element with the smallest molar mass will have the greatest number of atoms, as it appears in the denominator.</p> <p>Below are the molar masses of the elements:</p> <p><p>Molar mass of Pr (Praseodymium), $ M_{\text{Pr}} = 141 $ g/mol</p></p> <p><p>Molar mass of Pt (Platinum), $ M_{\text{Pt}} = 195 $ g/mol</p></p> <p><p>Molar mass of Pb (Lead), $ M_{\text{Pb}} = 207 $ g/mol</p></p> <p><p>Molar mass of Po (Polonium), $ M_{\text{Po}} = 209 $ g/mol</p></p> <p>Since Praseodymium (Pr) has the smallest molar mass ($ M_{\text{Pr}} = 141 $ g/mol), 10^-9 grams of Pr will contain the highest number of atoms compared to the other elements listed.</p>