Potassium iodide has cubic unit cell with cell edge of 705 pm. The density of KI is `3.12 g cm^(-3)`. How many `K^(+) and I^(-)` ions are contained in the unit cell ?

Explanation

To determine the number of \( K^+ \) and \( I^- \) ions contained in the unit cell of potassium iodide (KI), we can follow these steps:

### Step 1: Convert the edge length from picometers to centimeters.
The edge length \( a \) is given as 705 pm. To convert this to centimeters:
\[ a = 705 \, \text{pm} = 705 \times 10^{-12} \, \text{m} = 705 \times 10^{-10} \, \text{cm} = 7.05 \times 10^{-8} \, \text{cm} \]

### Step 2: Calculate the volume of the unit cell.
The volume \( V \) of the cubic unit cell can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of \( a \):
\[ V = (7.05 \times 10^{-8} \, \text{cm})^3 = 3.51 \times 10^{-23} \, \text{cm}^3 \]

### Step 3: Use the density to find the mass of the unit cell.
The density \( \rho \) of KI is given as 3.12 g/cm³. The mass \( m \) of the unit cell can be calculated using the formula:
\[ m = \rho \times V \]
Substituting the values:
\[ m = 3.12 \, \text{g/cm}^3 \times 3.51 \times 10^{-23} \, \text{cm}^3 = 1.095 \times 10^{-22} \, \text{g} \]

### Step 4: Calculate the molar mass of KI.
The molar mass \( M \) of potassium iodide (KI) can be calculated as follows:
- Molar mass of \( K \) = 39 g/mol
- Molar mass of \( I \) = 127 g/mol
\[ M = 39 + 127 = 166 \, \text{g/mol} \]

### Step 5: Calculate the number of formula units in the unit cell.
Using the formula relating mass, molar mass, and the number of formula units \( Z \):
\[ Z = \frac{m \times N_A}{M} \]
where \( N_A \) is Avogadro's number \( (6.022 \times 10^{23} \, \text{mol}^{-1}) \).
Substituting the values:
\[ Z = \frac{1.095 \times 10^{-22} \, \text{g} \times 6.022 \times 10^{23} \, \text{mol}^{-1}}{166 \, \text{g/mol}} \approx 4 \]

### Step 6: Determine the number of \( K^+ \) and \( I^- \) ions in the unit cell.
Since each formula unit of KI contains one \( K^+ \) ion and one \( I^- \) ion, and we have determined that there are 4 formula units in the unit cell:
- Number of \( K^+ \) ions = 4
- Number of \( I^- \) ions = 4

### Final Answer:
The unit cell of potassium iodide contains 4 \( K^+ \) ions and 4 \( I^- \) ions.

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