`3.7 gm` of gas at `25^(@)C` occupied the same volume as `0.184 gm` of hydrogen at `17^(@)C ` and at the same pressure. What is the molecular mass of the gas ?

Options

A. `82.66`
B. 41.33
C. 20.67
D. 10.33

Explanation

To find the molecular mass of the gas, we can use the ideal gas law and the fact that both gases occupy the same volume under the same conditions of pressure. Here’s a step-by-step solution:

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

Since both gases occupy the same volume at the same pressure, we can set up a relationship between the two gases.

### Step 2: Convert Temperatures to Kelvin
Convert the temperatures from Celsius to Kelvin:
- For the gas at \( 25^\circ C \):
\[ T_1 = 25 + 273 = 298 \, K \]
- For hydrogen at \( 17^\circ C \):
\[ T_2 = 17 + 273 = 290 \, K \]

### Step 3: Write the Equation for Both Gases
Using the ideal gas law, we can express the relationship for both gases:
For the unknown gas:
\[ \frac{W_1}{M_1} \cdot R \cdot T_1 \]
For hydrogen:
\[ \frac{W_2}{M_2} \cdot R \cdot T_2 \]

Since \( R \) and \( P \) are the same for both gases, we can equate them:
\[ \frac{W_1}{M_1} \cdot T_1 = \frac{W_2}{M_2} \cdot T_2 \]

### Step 4: Substitute Known Values
Substituting the known values:
- \( W_1 = 3.7 \, g \) (mass of the unknown gas)
- \( W_2 = 0.184 \, g \) (mass of hydrogen)
- \( M_2 = 2 \, g/mol \) (molecular mass of hydrogen)

The equation now looks like:
\[ \frac{3.7}{M_1} \cdot 298 = \frac{0.184}{2} \cdot 290 \]

### Step 5: Solve for \( M_1 \)
Rearranging the equation to solve for \( M_1 \):
\[ M_1 = \frac{3.7 \cdot 298 \cdot 2}{0.184 \cdot 290} \]

### Step 6: Calculate \( M_1 \)
Now, calculate \( M_1 \):
1. Calculate the numerator:
\[ 3.7 \cdot 298 \cdot 2 = 2205.2 \]
2. Calculate the denominator:
\[ 0.184 \cdot 290 = 53.36 \]
3. Divide the numerator by the denominator:
\[ M_1 = \frac{2205.2}{53.36} \approx 41.3 \, g/mol \]

### Final Answer
The molecular mass of the gas is approximately \( 41.3 \, g/mol \).

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