Number of `Ca^(+2)" and "Cl^(–)` ion in 111 g of anhydrous `CaCl_(2)` are -

Options

A. `N_(A), 2N_(A)`
B. `2N_(A), N_(A)`
C. `N_(A), N_(A)`
D. None

Answer

To find the number of \( \text{Ca}^{2+} \) and \( \text{Cl}^{-} \) ions in 111 g of anhydrous \( \text{CaCl}_2 \), we can follow these steps:

### Step 1: Calculate the Molar Mass of \( \text{CaCl}_2 \)
The molar mass of \( \text{CaCl}_2 \) can be calculated using the atomic masses of calcium (Ca) and chlorine (Cl):
- Atomic mass of Ca = 40 g/mol
- Atomic mass of Cl = 35.5 g/mol
- Since there are 2 Cl atoms in \( \text{CaCl}_2 \), the total mass of Cl is \( 2 \times 35.5 = 71 \) g/mol.

Now, adding these together:
\[ \text{Molar mass of } \text{CaCl}_2 = 40 + 71 = 111 \text{ g/mol} \]

### Step 2: Calculate the Number of Moles of \( \text{CaCl}_2 \)
Using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
Substituting the values:
\[ \text{Number of moles of } \text{CaCl}_2 = \frac{111 \text{ g}}{111 \text{ g/mol}} = 1 \text{ mole} \]

### Step 3: Determine the Number of Ions in 1 Mole of \( \text{CaCl}_2 \)
In 1 mole of \( \text{CaCl}_2 \):
- There is 1 \( \text{Ca}^{2+} \) ion.
- There are 2 \( \text{Cl}^{-} \) ions.

### Step 4: Calculate the Total Number of Ions
Using Avogadro's number (\( N_A = 6.022 \times 10^{23} \)):
- Number of \( \text{Ca}^{2+} \) ions:
\[ \text{Number of } \text{Ca}^{2+} \text{ ions} = 1 \text{ mole} \times N_A = 1 \times 6.022 \times 10^{23} = 6.022 \times 10^{23} \text{ ions} \]

- Number of \( \text{Cl}^{-} \) ions:
\[ \text{Number of } \text{Cl}^{-} \text{ ions} = 2 \text{ moles} \times N_A = 2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24} \text{ ions} \]

### Final Answer
- Number of \( \text{Ca}^{2+} \) ions = \( 6.022 \times 10^{23} \)
- Number of \( \text{Cl}^{-} \) ions = \( 1.2044 \times 10^{24} \)