The silver perchlorate benzene complex, `AgClO_(4).C_(6)H_(6)` is orthorhombic with unit cell dimensions `a_(0)=7.96,b_(0)=8.34` and `,c_(0)=11.7 Å`. The formula weight is 285 and there are four
molecules per unit cell. Calculate the density of the crystal.

Options

A. `2.44 g//cm^(3)`
B. `24.4g//cm^(3)`
C. `3.5g//cm^(3)`
D. `34.4g//cm^(2)`

Explanation

To calculate the density of the silver perchlorate benzene complex, `AgClO4.C6H6`, we will follow these steps:

### Step 1: Identify the given values
- Unit cell dimensions:
- \( a_0 = 7.96 \, \text{Å} \)
- \( b_0 = 8.34 \, \text{Å} \)
- \( c_0 = 11.7 \, \text{Å} \)
- Formula weight (M) = 285 g/mol
- Number of molecules per unit cell (Z) = 4
- Avogadro's number (\( N_0 \)) = \( 6.02 \times 10^{23} \, \text{mol}^{-1} \)

### Step 2: Convert unit cell dimensions from Ångströms to centimeters
Since 1 Å = \( 10^{-8} \) cm, we convert the dimensions:
- \( a_0 = 7.96 \, \text{Å} = 7.96 \times 10^{-8} \, \text{cm} \)
- \( b_0 = 8.34 \, \text{Å} = 8.34 \times 10^{-8} \, \text{cm} \)
- \( c_0 = 11.7 \, \text{Å} = 11.7 \times 10^{-8} \, \text{cm} \)

### Step 3: Calculate the volume of the unit cell
The volume \( V \) of the orthorhombic unit cell is given by:
\[ V = a_0 \times b_0 \times c_0 \]
Substituting the values:
\[ V = (7.96 \times 10^{-8}) \times (8.34 \times 10^{-8}) \times (11.7 \times 10^{-8}) \, \text{cm}^3 \]

### Step 4: Calculate the volume numerically
Calculating the volume:
\[ V = 7.96 \times 8.34 \times 11.7 \times 10^{-24} \, \text{cm}^3 \]
\[ V \approx 7.96 \times 8.34 \times 11.7 \approx 7.96 \times 97.38 \approx 777.66 \times 10^{-24} \, \text{cm}^3 \]
\[ V \approx 7.7766 \times 10^{-22} \, \text{cm}^3 \]

### Step 5: Use the density formula
The density \( \rho \) is given by the formula:
\[ \rho = \frac{Z \times M}{N_0 \times V} \]
Substituting the known values:
\[ \rho = \frac{4 \times 285}{6.02 \times 10^{23} \times 7.7766 \times 10^{-22}} \]

### Step 6: Calculate the density numerically
Calculating the numerator:
\[ 4 \times 285 = 1140 \]
Calculating the denominator:
\[ 6.02 \times 10^{23} \times 7.7766 \times 10^{-22} \approx 4.68 \times 10^{2} \, \text{g/cm}^3 \]
Now, substituting these values into the density formula:
\[ \rho = \frac{1140}{4.68 \times 10^{2}} \approx 2.44 \, \text{g/cm}^3 \]

### Final Answer
The density of the crystal is approximately \( 2.44 \, \text{g/cm}^3 \).

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