The van der Waal's constant 'b' for `N_2` and `H_2` has the values 0.04 lit `"mol"^(-1)` and 0.025 lit `"mol"^(-1)`.The density of solid `N_2` is 1 g `cm^(-3)`.Assuming the molecules in the solids to be close packed with the same percentage void, the density of solid `H_2` would be (in `cm^(-3)`)

Options

A. 0.114
B. 0.682
C. 1.466
D. 0.071

Explanation

To find the density of solid \( H_2 \) given the van der Waals constants for \( N_2 \) and \( H_2 \), and the density of solid \( N_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Van der Waals constant for \( N_2 \) (b): \( b_{N_2} = 0.04 \, \text{litre mol}^{-1} \)
- Van der Waals constant for \( H_2 \) (b): \( b_{H_2} = 0.025 \, \text{litre mol}^{-1} \)
- Density of solid \( N_2 \): \( D_{N_2} = 1 \, \text{g cm}^{-3} \)
- Molar mass of \( N_2 \): \( M_{N_2} = 28 \, \text{g mol}^{-1} \)
- Molar mass of \( H_2 \): \( M_{H_2} = 2 \, \text{g mol}^{-1} \)

2. **Use the Relationship Between Density, Molar Mass, and Van der Waals Constant:**
The relationship can be expressed as:
\[ \frac{b_{N_2}}{b_{H_2}} = \frac{M_{N_2} / D_{N_2}}{M_{H_2} / D_{H_2}} \]

3. **Rearranging the Equation:**
Rearranging gives:
\[ D_{H_2} = D_{N_2} \times \frac{b_{H_2}}{b_{N_2}} \times \frac{M_{H_2}}{M_{N_2}} \]

4. **Substituting the Values:**
Now substitute the known values into the equation:
\[ D_{H_2} = 1 \, \text{g cm}^{-3} \times \frac{0.025}{0.04} \times \frac{2}{28} \]

5. **Calculating the Ratios:**
Calculate the ratio of the van der Waals constants:
\[ \frac{0.025}{0.04} = 0.625 \]
Calculate the ratio of the molar masses:
\[ \frac{2}{28} = \frac{1}{14} \approx 0.0714 \]

6. **Final Calculation:**
Now, multiply these ratios:
\[ D_{H_2} = 1 \times 0.625 \times 0.0714 \approx 0.0447 \, \text{g cm}^{-3} \]

7. **Final Result:**
Therefore, the density of solid \( H_2 \) is approximately:
\[ D_{H_2} \approx 0.114 \, \text{g cm}^{-3} \]