In an experiment during the analysis of carbon compound `145 cm^3` of `H_2` was collected at 760 mm Hg pressure and `27^@C` temperature. The weight of `H_2` is nearly

Options

A. 10 mg
B. 12 mg
C. 24 mg
D. 6 mg

Answer

To find the weight of hydrogen gas (H₂) collected in the experiment, we can use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = Pressure (in atm)
- \( V \) = Volume (in liters)
- \( n \) = Number of moles of gas
- \( R \) = Ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = Temperature (in Kelvin)

### Step-by-Step Solution:

1. **Convert Volume from cm³ to L**:
\[ V = 145 \, \text{cm}^3 = \frac{145}{1000} \, \text{L} = 0.145 \, \text{L} \]

2. **Convert Pressure from mm Hg to atm**:
\[ P = 760 \, \text{mm Hg} = 1 \, \text{atm} \quad (\text{since } 760 \, \text{mm Hg} = 1 \, \text{atm}) \]

3. **Convert Temperature from Celsius to Kelvin**:
\[ T = 27^\circ C + 273 = 300 \, \text{K} \]

4. **Substitute Values into the Ideal Gas Law**:
\[ PV = nRT \implies n = \frac{PV}{RT} \]
\[ n = \frac{(1 \, \text{atm})(0.145 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)})(300 \, \text{K})} \]

5. **Calculate the Number of Moles (n)**:
\[ n = \frac{0.145}{0.0821 \times 300} = \frac{0.145}{24.63} \approx 0.00588 \, \text{mol} \]

6. **Calculate the Weight of H₂**:
- The molecular weight of H₂ is approximately \( 2 \, \text{g/mol} \).
\[ \text{Weight} = n \times \text{Molecular Weight} = 0.00588 \, \text{mol} \times 2 \, \text{g/mol} = 0.01176 \, \text{g} \]

7. **Convert Weight to Milligrams**:
\[ 0.01176 \, \text{g} = 11.76 \, \text{mg} \]

### Final Answer:
The weight of \( H_2 \) is approximately \( 11.76 \, \text{mg} \).