The density of `KCl` is `1.9893g cm^(-3)` and the length of a side unit cell is `6.29082Å` as determined by `X-`ray diffraction. Calculation the value of Avogadro's number.

Options

A. `6.017xx10^(23)`
B. `6.023xx10^(23)`
C. `6.034xx10^(23)`
D. `6.017xx10^(19)`

Answer

To calculate Avogadro's number using the given density of KCl and the length of the unit cell, we can follow these steps:

### Step 1: Understand the relationship between density, molar mass, unit cell volume, and Avogadro's number.
The formula relating these quantities is:
\[ \text{Density} = \frac{Z \times \text{Molar Mass}}{a^3 \times N_A} \]
where:
- \( Z \) = number of formula units per unit cell
- Molar Mass = mass of one mole of the substance (for KCl, it is 74.5 g/mol)
- \( a \) = edge length of the unit cell in cm
- \( N_A \) = Avogadro's number

### Step 2: Convert the edge length from Ångströms to centimeters.
Given:
\[ a = 6.29082 \, \text{Å} = 6.29082 \times 10^{-8} \, \text{cm} \]

### Step 3: Identify the value of \( Z \) for KCl.
KCl crystallizes in a face-centered cubic (FCC) lattice, which has:
\[ Z = 4 \]

### Step 4: Substitute the known values into the density formula.
We know:
- Density = 1.9893 g/cm³
- Molar Mass = 74.5 g/mol
- Edge length \( a = 6.29082 \times 10^{-8} \, \text{cm} \)

Substituting these values into the density formula:
\[ 1.9893 = \frac{4 \times 74.5}{(6.29082 \times 10^{-8})^3 \times N_A} \]

### Step 5: Rearrange the equation to solve for \( N_A \).
Rearranging gives:
\[ N_A = \frac{4 \times 74.5}{1.9893 \times (6.29082 \times 10^{-8})^3} \]

### Step 6: Calculate \( (6.29082 \times 10^{-8})^3 \).
Calculating the cube:
\[ (6.29082 \times 10^{-8})^3 = 2.496 \times 10^{-23} \, \text{cm}^3 \]

### Step 7: Substitute back to find \( N_A \).
Now substituting this value back into the equation:
\[ N_A = \frac{4 \times 74.5}{1.9893 \times 2.496 \times 10^{-23}} \]
Calculating the numerator:
\[ 4 \times 74.5 = 298 \]
Calculating the denominator:
\[ 1.9893 \times 2.496 \times 10^{-23} \approx 4.973 \times 10^{-23} \]
Now substituting these values:
\[ N_A = \frac{298}{4.973 \times 10^{-23}} \approx 5.99 \times 10^{23} \]

### Step 8: Final calculation and rounding.
Rounding gives:
\[ N_A \approx 6.017 \times 10^{23} \, \text{mol}^{-1} \]

### Conclusion:
The value of Avogadro's number is approximately \( 6.017 \times 10^{23} \, \text{mol}^{-1} \).