To determine the number of moles of ions formed when 65.25 g of barium nitrate (Ba(NO₃)₂) dissolves in water, follow these steps:
### Step 1: Calculate the molar mass of barium nitrate (Ba(NO₃)₂)
To find the molar mass, sum up the atomic masses of all the atoms in the formula.
1. Atomic mass of Barium (Ba): 137 g/mol
2. Atomic mass of Nitrogen (N): 14 g/mol
3. Atomic mass of Oxygen (O): 16 g/mol
4. Barium Nitrate (Ba(NO₃)₂) comprises:
- 1 Barium atom
- 2 Nitrogen atoms
- 6 Oxygen atoms (since each NO₃ has 3 Oxygens and there are 2 NO₃ groups)
[tex]\[ \text{Molar mass of Ba(NO₃)₂} = 137 + 2(14) + 2(3 \times 16) = 137 + 2(14) + 2(48) = 137 + 28 + 96 = 261 \text{ g/mol} \][/tex]
### Step 2: Calculate the number of moles of barium nitrate (Ba(NO₃)₂)
Use the formula:
[tex]\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} \][/tex]
Given:
- Mass of barium nitrate = 65.25 g
- Molar mass of Ba(NO₃)₂ = 261 g/mol
[tex]\[ \text{Moles of Ba(NO₃)₂} = \frac{65.25 \text{ g}}{261 \text{ g/mol}} = 0.25 \text{ moles} \][/tex]
### Step 3: Determine the number of moles of ions formed
When barium nitrate dissolves in water, it dissociates into ions. The dissociation reaction is:
[tex]\[ \text{Ba(NO₃)₂} \rightarrow \text{Ba}^{2+} + 2\text{NO}_3^- \][/tex]
From this reaction, we see:
- 1 mole of Ba(NO₃)₂ produces 1 mole of Ba²⁺ ions and 2 moles of NO₃⁻ ions.
- Therefore, for each mole of Ba(NO₃)₂, a total of 1 + 2 = 3 moles of ions are produced.
For 0.25 moles of Ba(NO₃)₂:
[tex]\[ \text{Total moles of ions} = 0.25 \text{ moles of Ba(NO₃)₂} \times 3 = 0.75 \text{ moles of ions} \][/tex]
### Conclusion:
When 65.25 g of barium nitrate dissolves in water, it forms 0.75 moles of ions.
