Calculate the molar mass of an unknown gas which diffuses 1.117 times faster than oxygen oxygen gas through same aperture under the same conditions of temperature and pressure.

Explanation

To calculate the molar mass of an unknown gas that diffuses 1.117 times faster than oxygen gas, we can use Graham's law of diffusion. Here’s a step-by-step solution:

### Step 1: Understand Graham's Law of Diffusion
Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be expressed as:

\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]

Where:
- \( R_1 \) = rate of diffusion of gas 1 (unknown gas)
- \( R_2 \) = rate of diffusion of gas 2 (oxygen)
- \( M_1 \) = molar mass of gas 1 (unknown gas)
- \( M_2 \) = molar mass of gas 2 (oxygen)

### Step 2: Set Up the Equation
Given that the unknown gas diffuses 1.117 times faster than oxygen, we can express this as:

\[ R_1 = 1.117 \times R_2 \]

Substituting this into Graham's law gives us:

\[ \frac{1.117 \times R_2}{R_2} = \sqrt{\frac{M_2}{M_1}} \]

This simplifies to:

\[ 1.117 = \sqrt{\frac{M_2}{M_1}} \]

### Step 3: Substitute Molar Mass of Oxygen
The molar mass of oxygen (O₂) is approximately 32 g/mol. Therefore, we can substitute \( M_2 \) with 32 g/mol:

\[ 1.117 = \sqrt{\frac{32}{M_1}} \]

### Step 4: Square Both Sides
To eliminate the square root, we square both sides of the equation:

\[ (1.117)^2 = \frac{32}{M_1} \]

Calculating \( (1.117)^2 \):

\[ 1.117^2 \approx 1.250 \]

So we have:

\[ 1.250 = \frac{32}{M_1} \]

### Step 5: Solve for Molar Mass of Unknown Gas
Now, we can rearrange the equation to solve for \( M_1 \):

\[ M_1 = \frac{32}{1.250} \]

Calculating this gives:

\[ M_1 \approx 25.6 \text{ g/mol} \]

### Conclusion
The molar mass of the unknown gas is approximately **25.65 g/mol**.

---