The volume of a gas X and chlorine diffusing during the same taime are 35 ml and 29 ml respectively. If the molar mass of chlorine is 71, calculate the molar mass of gas(X).

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. **Identify the Given Information:**
- Volume of gas X (Vx) = 35 ml
- Volume of chlorine (VCl) = 29 ml
- Molar mass of chlorine (MCl) = 71 g/mol

2. **Write the Formula for Graham's Law:**
According to Graham's law, the ratio of the rates of diffusion of two gases is equal to the inverse ratio of the square roots of their molar masses:
\[ \frac{Rate_X}{Rate_{Cl}} = \frac{\sqrt{M_{Cl}}}{\sqrt{M_X}} \]

3. **Express the Rates in Terms of Volume:**
Since the time taken for both gases to diffuse is the same, we can express the rates in terms of the volumes:
\[ \frac{V_X}{V_{Cl}} = \frac{\sqrt{M_{Cl}}}{\sqrt{M_X}} \]

4. **Substitute the Known Values:**
Plugging in the values we have:
\[ \frac{35}{29} = \frac{\sqrt{71}}{\sqrt{M_X}} \]

5. **Square Both Sides to Eliminate the Square Roots:**
\[ \left(\frac{35}{29}\right)^2 = \frac{71}{M_X} \]

6. **Calculate \(\left(\frac{35}{29}\right)^2\):**
\[ \frac{35^2}{29^2} = \frac{1225}{841} \approx 1.45 \]

7. **Set Up the Equation:**
\[ 1.45 = \frac{71}{M_X} \]

8. **Rearrange to Solve for \(M_X\):**
\[ M_X = \frac{71}{1.45} \]

9. **Calculate \(M_X\):**
\[ M_X \approx 48.97 \text{ g/mol} \]

### Final Answer:
The molar mass of gas X is approximately **49 g/mol**.