Under identical condition of temperature and pressure, 1 litre of `CH_(4)` weighted 1.2 gram while 2 litre of another gaseous hydrocarbon, `C_(n) H_(2n-2)` weighted 8.1 gram. What is the value of n ?

Answer

To solve the problem, we need to find the value of \( n \) in the hydrocarbon \( C_nH_{2n-2} \) given the weights of \( CH_4 \) and the other hydrocarbon under identical conditions of temperature and pressure.

### Step-by-Step Solution:

1. **Calculate the Molar Mass of \( CH_4 \)**:
- The molar mass of methane (\( CH_4 \)) is calculated as follows:
\[ \text{Molar mass of } CH_4 = 12 \, (\text{C}) + 4 \times 1 \, (\text{H}) = 16 \, \text{g/mol} \]

2. **Determine the Volume of 1 mole of \( CH_4 \)**:
- Under standard conditions, 1 mole of any gas occupies 22.4 liters. However, we are given that 1 liter of \( CH_4 \) weighs 1.2 grams.
- Using the ratio:
\[ 1 \, \text{L} \, CH_4 \text{ weighs } 1.2 \, \text{g} \implies \text{1 mole (16 g)} \text{ corresponds to } \frac{1 \, \text{L} \times 16 \, \text{g}}{1.2 \, \text{g}} = \frac{16}{1.2} \, \text{L} \]
- Thus, the volume of 1 mole of \( CH_4 \) is:
\[ V = \frac{16}{1.2} = \frac{40}{3} \, \text{L} \]

3. **Calculate the Molar Mass of \( C_nH_{2n-2} \)**:
- We know that 2 liters of \( C_nH_{2n-2} \) weighs 8.1 grams.
- Using the ratio for the hydrocarbon:
\[ 2 \, \text{L} \text{ weighs } 8.1 \, \text{g} \implies \text{Molar mass } = \frac{8.1 \, \text{g}}{2 \, \text{L}} \times \frac{40}{3} \, \text{L} \]
- Therefore:
\[ M = \frac{8.1 \times 40/3}{2} = \frac{8.1 \times 20}{3} = \frac{162}{3} = 54 \, \text{g/mol} \]

4. **Set Up the Equation for Molar Mass**:
- The molar mass of \( C_nH_{2n-2} \) can be expressed as:
\[ \text{Molar mass} = 12n + (2n - 2) = 14n - 2 \]
- We set this equal to the calculated molar mass:
\[ 14n - 2 = 54 \]

5. **Solve for \( n \)**:
- Rearranging the equation:
\[ 14n = 54 + 2 = 56 \implies n = \frac{56}{14} = 4 \]

### Final Answer:
The value of \( n \) is \( 4 \).