The density of mercury is `13.6g//mc^(3)`.Estimate the b value.

Answer

To estimate the b value for mercury, we will follow these steps:

### Step 1: Understand the relationship between density, mass, and volume
The density (d) of a substance is defined as its mass (m) per unit volume (V). The formula is:
\[ d = \frac{m}{V} \]

### Step 2: Identify the volume of a single molecule
For a single molecule, the volume can be expressed as:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the molecule.

### Step 3: Relate the density to the volume of one molecule
For \( N_A \) (Avogadro's number) molecules, the total volume can be expressed as:
\[ V_{total} = N_A \cdot V \]
Thus, we can write:
\[ V_{total} = N_A \cdot \frac{4}{3} \pi r^3 \]

### Step 4: Express the radius in terms of density and molecular weight
We can rearrange the density formula to find the radius:
\[ r^3 = \frac{3m}{4 \pi d N_A} \]
Here, \( m \) is the molar mass of mercury, which is approximately 200 g/mol, \( d \) is the density (13.6 g/cm³), and \( N_A \) is Avogadro's number (approximately \( 6.022 \times 10^{23} \) mol⁻¹).

### Step 5: Substitute the known values
Substituting the values into the equation:
- \( m = 200 \, \text{g/mol} \)
- \( d = 13.6 \, \text{g/cm}^3 \)
- \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)

We calculate:
\[ r^3 = \frac{3 \times 200}{4 \pi \times 13.6 \times 6.022 \times 10^{23}} \]

### Step 6: Calculate \( r^3 \)
Calculating \( r^3 \):
1. Calculate the numerator: \( 3 \times 200 = 600 \)
2. Calculate the denominator: \( 4 \times \pi \times 13.6 \times 6.022 \times 10^{23} \approx 1.702 \times 10^{25} \)
3. Thus, \( r^3 \approx \frac{600}{1.702 \times 10^{25}} \approx 3.53 \times 10^{-24} \, \text{cm}^3 \)

### Step 7: Calculate \( b \)
The value of \( b \) is given by:
\[ b = 4 \times V_{total} = 4 \times \frac{4}{3} \pi r^3 N_A \]
Substituting \( r^3 \) and \( N_A \):
\[ b = 4 \times \frac{4}{3} \pi (3.53 \times 10^{-24}) \times (6.022 \times 10^{23}) \]

### Step 8: Final calculation
Calculating \( b \):
1. Calculate \( \frac{4}{3} \pi \approx 4.19 \)
2. Then, \( b \approx 4 \times 4.19 \times (3.53 \times 10^{-24}) \times (6.022 \times 10^{23}) \)
3. After performing the calculations, we find \( b \approx 58.8 \, \text{cm}^3/\text{mol} \)

### Conclusion
The estimated value of \( b \) for mercury is approximately \( 58.8 \, \text{cm}^3/\text{mol} \).

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