How many grams of CO would occupy 22.414 Liters at 0 °C and 1 atm?

To solve this problem, we need to use the ideal gas law equation, which is PV = nRT.

Step 1: Identify the knowns

• Pressure (P) = 1 atm
• Volume (V) = 22.414 L
• Temperature (T) = 0 °C = 273.15 K (converted from Celsius to Kelvin)
• R (ideal gas constant) = 0.0821 L.atm/K.mol

Step 2: Solve for n (number of moles) Rearrange the ideal gas law equation to solve for n: n = PV/RT Substitute the known values into the equation: n = (1 atm * 22.414 L) / (0.0821 L.atm/K.mol * 273.15 K) = 1 mol

Step 3: Convert moles to grams The molar mass of CO (carbon monoxide) is approximately 28.01 g/mol. Therefore, 1 mol of CO is equivalent to 28.01 g.

So, 22.414 liters of CO at 0 °C and 1 atm would weigh approximately 28.01 grams.