20 dm^3 of an unknown gas diffuse through a porous partition in 60 s, whereas `14.1 dm^3` of `O_2` under similar conditions diffuse in 30 s. What is the molecular mass of the gas ?

Answer

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. **Identify the Given Data:**
- Volume of unknown gas (V1) = 20 dm³
- Time for unknown gas (T1) = 60 s
- Volume of O₂ (V2) = 14.1 dm³
- Time for O₂ (T2) = 30 s
- Molar mass of O₂ (M2) = 32 g/mol

2. **Calculate the Rate of Diffusion:**
- Rate of diffusion (R) can be expressed as:
\[ R = \frac{V}{T} \]
- For the unknown gas:
\[ R1 = \frac{V1}{T1} = \frac{20 \, \text{dm}^3}{60 \, \text{s}} = \frac{1}{3} \, \text{dm}^3/\text{s} \]
- For O₂:
\[ R2 = \frac{V2}{T2} = \frac{14.1 \, \text{dm}^3}{30 \, \text{s}} = 0.47 \, \text{dm}^3/\text{s} \]

3. **Apply Graham's Law:**
- According to Graham's law:
\[ \frac{R1}{R2} = \sqrt{\frac{M2}{M1}} \]
- Rearranging gives:
\[ \frac{R1}{R2} = \sqrt{\frac{32}{M1}} \]

4. **Substitute the Values:**
- Substitute the calculated rates into the equation:
\[ \frac{\frac{1}{3}}{0.47} = \sqrt{\frac{32}{M1}} \]

5. **Calculate the Left Side:**
- Calculate the left side:
\[ \frac{1}{3} \div 0.47 \approx 0.71 \]
- Therefore:
\[ 0.71 = \sqrt{\frac{32}{M1}} \]

6. **Square Both Sides:**
- Squaring both sides gives:
\[ 0.71^2 = \frac{32}{M1} \]
- Calculate \(0.71^2\):
\[ 0.5041 = \frac{32}{M1} \]

7. **Rearrange to Find M1:**
- Rearranging gives:
\[ M1 = \frac{32}{0.5041} \approx 63.4 \, \text{g/mol} \]

8. **Final Result:**
- The molecular mass of the unknown gas is approximately **63 g/mol**.