What is the volume occupied by 2.20 mol of argon at STP? Round to the nearest hundredth. Don't forget the units.
One (1) moles of an ideal gas occupies 22.4 L at STP.
8l
To find the volume occupied by a given amount of a gas at STP (Standard Temperature and Pressure), we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (at STP, it is 1 atmosphere)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (at STP, it is 273.15 Kelvin)
We need to rearrange the equation to solve for V:
V = (nRT) / P
Given the following values:
n = 2.20 mol
R = 0.0821 L·atm/(mol·K)
P = 1 atm
T = 273.15 K
Substituting the values into the equation:
V = (2.20 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
Calculating the expression:
V = 57.48 L
Rounding to the nearest hundredth:
V ≈ 57.48 L
Therefore, the volume occupied by 2.20 mol of argon at STP is approximately 57.48 L.