Questions: Gas A was collected in a 1.312 L container at 356.18 K. The pressure inside the container was 790.91 torr. The molar mass of gas A is 110.75 g / mol. Recall that PV = nRT, R = 0.08206 L* atm /(mol* K), and 1 atm = 760 torr. How many grams of gas A were collected? SHOW ALL WORK

Solution Steps

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

\[ P = \frac{790.91 \, \text{torr}}{760 \, \text{torr/atm}} = 1.0407 \, \text{atm} \]

The ideal gas law is given by \(PV = nRT\). We need to solve for \(n\), the number of moles.

\[ n = \frac{PV}{RT} \]

Substitute the known values:

• \(P = 1.0407 \, \text{atm}\)
• \(V = 1.312 \, \text{L}\)
• \(R = 0.08206 \, \text{L atm/mol K}\)
• \(T = 356.18 \, \text{K}\)

\[ n = \frac{(1.0407 \, \text{atm})(1.312 \, \text{L})}{(0.08206 \, \text{L atm/mol K})(356.18 \, \text{K})} \]

\[ n = \frac{1.3661}{29.2303} = 0.04674 \, \text{mol} \]

The mass of the gas can be found by multiplying the number of moles by the molar mass.

\[ \text{Mass} = n \times \text{Molar Mass} = 0.04674 \, \text{mol} \times 110.75 \, \text{g/mol} \]

\[ \text{Mass} = 5.176 \, \text{g} \]

Final Answer