A saturated hydrocarbon having molecular formula `C_(n)H_(2n+2)` diffuses through a porous membrane twice as fast as Sulphur dioxide.
Determine the molecular mass of hydrocarbon and name it.

Explanation

To solve the problem, we will use Graham's law of effusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

### Step-by-Step Solution:

1. **Understand the relationship between the rates of diffusion:**
- Let the rate of diffusion of sulfur dioxide (SO₂) be R.
- According to the problem, the rate of diffusion of the hydrocarbon is twice that of sulfur dioxide. Therefore, the rate of diffusion of the hydrocarbon is 2R.

2. **Write down Graham's law:**
\[ \frac{R_{SO_2}}{R_{hydrocarbon}} = \sqrt{\frac{M_{hydrocarbon}}{M_{SO_2}}} \]
where \( R_{SO_2} \) is the rate of diffusion of sulfur dioxide, \( R_{hydrocarbon} \) is the rate of diffusion of the hydrocarbon, \( M_{hydrocarbon} \) is the molar mass of the hydrocarbon, and \( M_{SO_2} \) is the molar mass of sulfur dioxide.

3. **Substitute the known values:**
- The molar mass of sulfur dioxide (SO₂) is 64 g/mol.
- Substitute \( R_{SO_2} = R \) and \( R_{hydrocarbon} = 2R \) into the equation:
\[ \frac{R}{2R} = \sqrt{\frac{M_{hydrocarbon}}{64}} \]
This simplifies to:
\[ \frac{1}{2} = \sqrt{\frac{M_{hydrocarbon}}{64}} \]

4. **Square both sides to eliminate the square root:**
\[ \left(\frac{1}{2}\right)^2 = \frac{M_{hydrocarbon}}{64} \]
\[ \frac{1}{4} = \frac{M_{hydrocarbon}}{64} \]

5. **Cross-multiply to find the molar mass of the hydrocarbon:**
\[ M_{hydrocarbon} = 64 \times \frac{1}{4} = 16 \text{ g/mol} \]

6. **Use the molecular formula of the hydrocarbon:**
- The molecular formula of the saturated hydrocarbon is given as \( C_nH_{2n+2} \).
- The molar mass can be expressed as:
\[ M_{hydrocarbon} = 12n + (2n + 2) = 14n + 2 \]
Set this equal to the calculated molar mass:
\[ 14n + 2 = 16 \]

7. **Solve for n:**
\[ 14n = 16 - 2 \]
\[ 14n = 14 \]
\[ n = 1 \]

8. **Determine the hydrocarbon:**
- If \( n = 1 \), then the hydrocarbon is \( C_1H_{2(1)+2} = CH_4 \), which is methane.

### Final Answer:
- The molecular mass of the hydrocarbon is **16 g/mol** and the hydrocarbon is **methane (CH₄)**.