3.7 g of a gas at 25°C occupies the same volume as 0.184 g of hydrogen at 17°C and at the same pressure. What is the molecular mass of the gas ?

Answer

To find the molecular mass of the gas, we can use the relationship between the masses, temperatures, and molecular weights of the gases involved, given that they occupy the same volume at the same pressure.

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin:**
- The temperature of the gas (T1) is given as 25°C. To convert this to Kelvin:
\[ T1 = 25 + 273 = 298 \, \text{K} \]
- The temperature of hydrogen (T2) is given as 17°C. To convert this to Kelvin:
\[ T2 = 17 + 273 = 290 \, \text{K} \]

2. **Identify Given Values:**
- Mass of the gas (W1) = 3.7 g
- Mass of hydrogen (W2) = 0.184 g
- Molecular weight of hydrogen (m2) = 2 g/mol

3. **Use the Ideal Gas Law Relationship:**
- Since the pressure and volume are constant, we can use the relationship:
\[ \frac{n1}{T1} = \frac{n2}{T2} \]
- Where \( n1 \) and \( n2 \) are the number of moles of the gas and hydrogen, respectively.

4. **Express Number of Moles:**
- The number of moles can be expressed as:
\[ n1 = \frac{W1}{m1} \quad \text{and} \quad n2 = \frac{W2}{m2} \]
- Substituting these into the relationship gives:
\[ \frac{W1/m1}{T1} = \frac{W2/m2}{T2} \]

5. **Rearranging the Equation:**
- Rearranging the equation to solve for \( m1 \) (the molecular mass of the gas):
\[ \frac{W1}{m1} \cdot \frac{T2}{T1} = \frac{W2}{m2} \]
- Cross-multiplying gives:
\[ W1 \cdot T2 \cdot m2 = W2 \cdot T1 \cdot m1 \]

6. **Substituting Known Values:**
- Substitute the known values into the equation:
\[ 3.7 \cdot 290 \cdot 2 = 0.184 \cdot 298 \cdot m1 \]

7. **Calculating:**
- Calculate the left side:
\[ 3.7 \cdot 290 \cdot 2 = 2146.0 \]
- Calculate the right side:
\[ 0.184 \cdot 298 = 54.872 \]
- Now, substituting back gives:
\[ 2146.0 = 54.872 \cdot m1 \]
- Solving for \( m1 \):
\[ m1 = \frac{2146.0}{54.872} \approx 39.05 \, \text{g/mol} \]

8. **Final Answer:**
- The molecular mass of the gas is approximately:
\[ m1 \approx 39.05 \, \text{g/mol} \]