How many sodium ions are present in 30.0 mL of 0.600 M Na2CO3 solution?

• A. 6.02 × 10^{21} sodium ions
• B. 3.61 × 10^{22} sodium ions
• C. 2.17 × 10^{22} sodium ions
• D. 1.08 × 10^{22} sodium ions

Answer: (C)

Verified step by step guidance

1. Identify the given information: volume of solution = 30.0 mL, molarity of Na2CO3 = 0.600 M. Remember to convert the volume from milliliters to liters because molarity is in moles per liter.
2. Calculate the number of moles of Na2CO3 in the solution using the formula: \(\text{moles} = \text{molarity} \times \text{volume (in liters)}\).
3. Determine the number of moles of sodium ions (Na\textsuperscript{+}) produced by Na2CO3. Since each formula unit of Na2CO3 produces 2 sodium ions, multiply the moles of Na2CO3 by 2.
4. Use Avogadro's number, \$6.022 \times 10^{23}\( particles/mole, to convert moles of sodium ions to the actual number of sodium ions: \)\text{number of ions} = \text{moles of Na}^+ \times 6.022 \times 10^{23}$.
5. Express the final answer in scientific notation and compare it with the given options to identify the correct number of sodium ions.