Question

1. which of the following will result in 1.00 mole 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 mole...
(clear all button)

Explanation:

Step1: Recall Avogadro's number and mole concept

Avogadro's number is \(N_A = 6.02\times10^{23}\) particles per mole. For a compound \(CaCl_2\), the formula unit has 1 \(Ca^{2+}\) ion and 2 \(Cl^-\) ions. So, 1 mole of \(CaCl_2\) contains 1 mole of \(Ca^{2+}\) ions and 2 moles of \(Cl^-\) ions.

Step2: Analyze each option

• Option 1: 1 atom of Ca and 2 atoms of Cl. This is just 1 formula unit, not 1 mole (which needs \(6.02\times10^{23}\) formula units). So incorrect.
• Option 2: \(6.02\times10^{23}\) ions of \(Ca^{2+}\) (which is 1 mole of \(Ca^{2+}\)) and \(1.20\times10^{24}\) ions of \(Cl^-\) (since \(\frac{1.20\times10^{24}}{6.02\times10^{23}} \approx 2\) moles of \(Cl^-\)). This matches the 1:2 ratio of \(Ca^{2+}\) to \(Cl^-\) in \(CaCl_2\), so this would give 1 mole of \(CaCl_2\).
• Option 3: 1 gram of Ca and 2 grams of Cl. Moles are calculated by mass/molar mass. Molar mass of Ca is ~40 g/mol, Cl is ~35.5 g/mol. 1g Ca is \(\frac{1}{40}\) mol, 2g Cl is \(\frac{2}{35.5}\) mol, not 1 and 2 moles. So incorrect.
• Option 4: The option is incomplete, but also Ca is an element, not a molecule, so this is incorrect from the start.

Answer:

The correct option is the one with \(6.02\times10^{23}\) ions of \(Ca^{2+}\) and \(1.20\times10^{24}\) ions of \(Cl^-\) (the second option in the list).