Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\), \(23 \mathrm{~g} \mathrm{Na}, 6.0 \times 10^{23} \mathrm{~N}_{2}\) molecules.

To compare the number of atoms, we first need to convert each sample into moles using the atomic weights and the Avogadro constant (6.022 x 10^23). 1. For \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\): Already given in moles, no conversion needed. 2. For \(23 \mathrm{~g} \mathrm{Na}\): Use the atomic weight of Na (\(22.99 \)) to convert the mass (g) to moles: \(\dfrac{23 \mathrm{~g}}{22.99 \mathrm{~g/mol}} \approx 1 \mathrm{~mol}\) Na. 3. For \(6.0 \times 10^{23} \mathrm{~N}_{2}\) molecules: Convert number of molecules to moles using Avogadro's constant (\(6.022 \times 10^{23}\) molecules/mol): \(\dfrac{6.0 \times 10^{23} \mathrm{~molecules}}{6.022 \times 10^{23} \mathrm{~molecules/mol}} \approx 1 \mathrm{~mol} \mathrm{N}_{2}\).

Now, we can find the number of atoms in each sample by multiplying the number of moles by the number of atoms per mole: 1. For \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\): There are three atoms per mole (\(2\) hydrogen atoms and \(1\) oxygen atom), so \(0.50 × 3 = 1.50\) moles of atoms. 2. For \(1 \mathrm{~mol} \mathrm{Na}\): There is one atom of Na per mole, so \(1 × 1 = 1\) mole of atoms. 3. For \(1 \mathrm{~mol} \mathrm{N}_{2}\): There are two atoms of N per mole, so \(1 × 2 = 2\) moles of atoms.

Using the moles of atoms from step 2, we can now rank the samples in order of increasing number of atoms: 1. \(1 \mathrm{~mol} \mathrm{Na}\) (1 mole of atoms) 2. \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) (1.50 moles of atoms) 3. \(1 \mathrm{~mol} \mathrm{N}_{2}\) (2 moles of atoms) So, the final ranking in order of increasing number of atoms is: \(23 \mathrm{~g} \mathrm{Na}\) < \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) < \(6.0 \times 10^{23} \mathrm{~N}_{2}\) molecules.