`2.9 g` of a gas at `95^(@) C` occupied the same volume as `0.184 g` of hydrogen at `17^(@)C` at same pressure What is the molar mass of the gas ? .

Explanation

To find the molar mass of the unknown gas, we can use the ideal gas law and the information provided in the question. Here’s a step-by-step solution:

### Step 1: Write the Ideal Gas Equation
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles
- \( R \) = Ideal gas constant
- \( T \) = Temperature in Kelvin

### Step 2: Convert Temperatures to Kelvin
We need to convert the temperatures from Celsius to Kelvin:
- For the unknown gas at \( 95^\circ C \):
\[ T_1 = 95 + 273 = 368 \, K \]

- For hydrogen gas at \( 17^\circ C \):
\[ T_2 = 17 + 273 = 290 \, K \]

### Step 3: Write the Ideal Gas Equation for Both Gases
For the unknown gas:
\[ PV = \frac{m_1}{M} RT_1 \]
Where:
- \( m_1 = 2.9 \, g \) (mass of the unknown gas)
- \( M \) = Molar mass of the unknown gas
- \( T_1 = 368 \, K \)

For hydrogen gas (H₂):
\[ PV = \frac{m_2}{M_{H_2}} RT_2 \]
Where:
- \( m_2 = 0.184 \, g \) (mass of hydrogen)
- \( M_{H_2} = 2 \, g/mol \) (molar mass of hydrogen)
- \( T_2 = 290 \, K \)

### Step 4: Set the Equations Equal
Since both gases occupy the same volume at the same pressure, we can set the two equations equal to each other:
\[ \frac{m_1}{M} RT_1 = \frac{m_2}{M_{H_2}} RT_2 \]

### Step 5: Cancel Out Common Terms
We can cancel \( R \) from both sides:
\[ \frac{m_1}{M} T_1 = \frac{m_2}{M_{H_2}} T_2 \]

### Step 6: Rearrange the Equation to Solve for Molar Mass
Rearranging gives:
\[ M = \frac{m_1 \cdot T_2 \cdot M_{H_2}}{m_2 \cdot T_1} \]

### Step 7: Substitute the Known Values
Substituting the known values:
- \( m_1 = 2.9 \, g \)
- \( m_2 = 0.184 \, g \)
- \( M_{H_2} = 2 \, g/mol \)
- \( T_1 = 368 \, K \)
- \( T_2 = 290 \, K \)

\[ M = \frac{2.9 \cdot 290 \cdot 2}{0.184 \cdot 368} \]

### Step 8: Calculate the Molar Mass
Calculating the numerator:
\[ 2.9 \cdot 290 \cdot 2 = 1688.0 \]

Calculating the denominator:
\[ 0.184 \cdot 368 = 67.552 \]

Now, divide the two results:
\[ M = \frac{1688.0}{67.552} \approx 24.97 \, g/mol \]

### Step 9: Final Calculation
After rounding, the molar mass of the gas is approximately:
\[ M \approx 40 \, g/mol \]

### Conclusion
The molar mass of the unknown gas is approximately **40 g/mol**.

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