Questions: A 0.02 gram sample of an unknown gas is found to occupy a volume of 0.025 L at a pressure of 98.7 mm Hg and a temperature of 33 °C. The molecular weight of the unknown gas is

Solution Steps

First, we need to convert the given units to standard units for use in the ideal gas law equation.

• Pressure: Convert from mm Hg to atm. \[ P = \frac{98.7 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 0.1299 \, \text{atm} \]

• Temperature: Convert from Celsius to Kelvin. \[ T = 33 + 273.15 = 306.15 \, \text{K} \]

The ideal gas law is given by: \[ PV = nRT \] where:

• \( P \) is the pressure in atm,
• \( V \) is the volume in liters,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant, \( R = 0.0821 \, \text{L atm/mol K} \),
• \( T \) is the temperature in Kelvin.

Rearrange the equation to solve for \( n \): \[ n = \frac{PV}{RT} \]

Substitute the known values: \[ n = \frac{(0.1299 \, \text{atm})(0.025 \, \text{L})}{(0.0821 \, \text{L atm/mol K})(306.15 \, \text{K})} = 0.0001295 \, \text{mol} \]

The molecular weight \( M \) is calculated using the formula: \[ M = \frac{\text{mass}}{n} \]

Substitute the known values: \[ M = \frac{0.02 \, \text{g}}{0.0001295 \, \text{mol}} = 154.5 \, \text{g/mol} \]

Final Answer