Questions: What is the molecular weight of a gas if a 21.0 g sample has a pressure of 836 mm Hg at 25.0°C in a 2.00 L flask? ( R=0.0821 L atm / mol K) 11.1 amu 243 amu 1.89 amu 234 amu none of the above

Solution Steps

The pressure is given in mm Hg, so we need to convert it to atmospheres. The conversion factor is \(1 \, \text{atm} = 760 \, \text{mm Hg}\).

\[ P = \frac{836 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 1.1053 \, \text{atm} \]

The temperature is given in degrees Celsius, so we need to convert it to Kelvin. The conversion formula is \(T(K) = T(^{\circ}C) + 273.15\).

\[ T = 25.0 + 273.15 = 298.15 \, \text{K} \]

The ideal gas law is \(PV = nRT\), where \(n\) is the number of moles. We can solve for \(n\):

\[ n = \frac{PV}{RT} = \frac{(1.1053 \, \text{atm})(2.00 \, \text{L})}{(0.0821 \, \text{L atm/mol K})(298.15 \, \text{K})} \]

\[ n = \frac{2.2106}{24.4662} = 0.0903 \, \text{mol} \]

The molecular weight \(M\) is given by the formula \(M = \frac{\text{mass}}{n}\).

\[ M = \frac{21.0 \, \text{g}}{0.0903 \, \text{mol}} = 232.678 \, \text{g/mol} \]

Final Answer