`180cm^(3)` of a hydrocarbon diffuses in 15 min, while under the same conditions, `120cm^(3)` of sulphur dioxide diffuse in 20 min. If the molecular mass of `SO_(2)` is 64, what is the molecular formula of the hydrocarbon?

Explanation

To solve the problem, we will use Graham's law of diffusion, 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. **Calculate the Rate of Diffusion for Sulfur Dioxide (SO₂)**:
- The volume of SO₂ diffused = 120 cm³
- Time taken = 20 minutes
- Rate of diffusion of SO₂ = Volume / Time = \( \frac{120 \, \text{cm}^3}{20 \, \text{min}} = 6 \, \text{cm}^3/\text{min} \)

2. **Calculate the Rate of Diffusion for the Hydrocarbon**:
- The volume of the hydrocarbon diffused = 180 cm³
- Time taken = 15 minutes
- Rate of diffusion of the hydrocarbon = Volume / Time = \( \frac{180 \, \text{cm}^3}{15 \, \text{min}} = 12 \, \text{cm}^3/\text{min} \)

3. **Apply Graham's Law of Diffusion**:
- According to Graham's law:
\[ \frac{\text{Rate of Hydrocarbon}}{\text{Rate of SO}_2} = \sqrt{\frac{M_{\text{SO}_2}}{M_{\text{Hydrocarbon}}}} \]
- Substituting the values:
\[ \frac{12}{6} = \sqrt{\frac{64}{M_{\text{Hydrocarbon}}}} \]

4. **Simplify the Equation**:
- The left side simplifies to 2:
\[ 2 = \sqrt{\frac{64}{M_{\text{Hydrocarbon}}}} \]

5. **Square Both Sides**:
- Squaring both sides gives:
\[ 4 = \frac{64}{M_{\text{Hydrocarbon}}} \]

6. **Rearranging to Find Molar Mass of Hydrocarbon**:
- Rearranging the equation:
\[ M_{\text{Hydrocarbon}} = \frac{64}{4} = 16 \, \text{g/mol} \]

7. **Determine the Molecular Formula of the Hydrocarbon**:
- The molar mass of the hydrocarbon is 16 g/mol. The simplest hydrocarbon with this molar mass is methane (CH₄).
- Carbon (C) has a molar mass of 12 g/mol and Hydrogen (H) has a molar mass of 1 g/mol. Thus:
\[ \text{Molar mass of CH}_4 = 12 + (1 \times 4) = 12 + 4 = 16 \, \text{g/mol} \]

### Final Answer:
The molecular formula of the hydrocarbon is **CH₄** (methane).

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