Question

which of the following will result in 1.00 moles of calcium chloride, cacl₂?
1 atom of ca and 2 atoms of cl
6.02 × 10²³ ions of ca²⁺ and 1.20 × 10²⁴ ions of cl⁻
1 gram of ca and 2 grams of cl
6.02 × 10²³ molecules of ca and 6.02 × 10²³ molecules of cl

Explanation:

Step1: Recall Avogadro's number and formula units

1 mole of any substance contains \(6.02\times10^{23}\) formula units (for ionic compounds like \(CaCl_2\)). In \(CaCl_2\), the ratio of \(Ca^{2+}\) to \(Cl^-\) is 1:2.

Step2: Analyze each option

• Option 1: 1 atom and 2 atoms is for 1 formula unit, not 1 mole. Eliminate.
• Option 2: For 1 mole of \(CaCl_2\), moles of \(Ca^{2+}=1\) mole, so number of \(Ca^{2+}\) ions \(= 6.02\times10^{23}\) (since 1 mole = \(6.02\times10^{23}\) particles). Moles of \(Cl^- = 2\) moles, so number of \(Cl^-\) ions \(= 2\times6.02\times10^{23}=1.204\times10^{24}\approx1.20\times10^{24}\). This matches.
• Option 3: Masses depend on molar masses. Molar mass of \(Ca\) is ~40 g/mol, \(Cl\) is ~35.5 g/mol. 1 mole \(CaCl_2\) has 40 g \(Ca\) and \(2\times35.5 = 71\) g \(Cl\), not 1g and 2g. Eliminate.
• Option 4: \(Ca\) and \(Cl\) are ions in \(CaCl_2\), not molecules. Eliminate.

Answer:

\(6.02\times 10^{23}\) ions of \(Ca^{2+}\) and \(1.20\times 10^{24}\) ions of \(Cl^-\) (the second option)