To find the molar mass of the gas, you can use the ideal gas law equation:
\[ PV = nRT \]
Where:
- \( P \) is the pressure of the gas (in atm)
- \( V \) is the volume of the gas (in liters)
- \( n \) is the number of moles of the gas
- \( R \) is the ideal gas constant (\(0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K}\))
- \( T \) is the temperature of the gas (in Kelvin)
First, let's convert the given values to appropriate units:
- Pressure: \( 738 \, \text{mmHg} = 0.971 \, \text{atm} \) (since 1 atm = 760 mmHg)
- Volume: \( 110 \, \text{mL} = 0.110 \, \text{L} \) (since 1 L = 1000 mL)
- Temperature: \( 30 \, \text{°C} = 273 \, \text{K} \) (temperature in Kelvin)
Substituting these values into the ideal gas law equation, we get:
\[ (0.971 \, \text{atm}) \times (0.110 \, \text{L}) = n \times (0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K}) \times (273 \, \text{K}) \]
Solving for \( n \) (the number of moles), we get:
\[ n = \frac{(0.971 \, \text{atm}) \times (0.110 \, \text{L})}{(0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K}) \times (273 \, \text{K})} \]
\[ n \approx 0.0051 \, \text{moles} \]
Next, to find the molar mass (\( M \)) of the gas, you can use the formula:
\[ M = \frac{\text{mass}}{\text{moles}} \]
Given the mass of the gas is \( 0.170 \, \text{g} \), substitute the values:
\[ M = \frac{0.170 \, \text{g}}{0.0051 \, \text{moles}} \]
\[ M \approx 33.33 \, \text{g/mol} \]
Therefore, the molar mass of the gas is approximately \( 33.33 \, \text{g/mol} \).