A sample of argon gas at STP occupies 56.2 liters. Find the number of moles of argon and the mass in the sample
PV = nRT
n = PV/RT @ STP T= 0+273k; P = 1atm R = 0.821 L- atm/mol-K
n = 1atm(56.2L)/0.821 L- atm/mol-K(273k)
n = 2.50 mole
2.50 moles of Ar =39.984g of Ar/1mole of Ar
Solution:
1) Rearrange PV = nRT to this:
n = PV / RT
2) Substitute:
n = [ (1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K) ]
n = 2.50866 mol (I'll keep a few guard digits)
3) Multiply the moles by the atomic weight of Ar to get the grams:
2.50866 mol times 39.948 g/mol = 100. g (to three sig figs)
1 mol occupies 22.4 L at STP. Solve for n, then n = grams/molar mass. You know molar mass and n, solve for grams.
OR you may use PV = nRT to solve for n.
Ah, argon, the noble gas. It's quite important to know its number of moles and mass, otherwise we might find ourselves in an argon-t situation! Let's crunch some numbers.
First, at STP (Standard Temperature and Pressure - not to be confused with Stupid Terrible Puns), we have the following conditions:
- Temperature: 273.15 Kelvin (K)
- Pressure: 1 atmosphere (atm)
We can use the ideal gas law, which states: PV = nRT, where:
- P is the pressure
- V is the volume
- n is the number of moles
- R is the ideal gas constant (0.0821 L·atm/(mol·K))
- T is the temperature
Now, we already know V is 56.2 liters, P is 1 atm, and T is 273.15 K. Plugging these values in, we can rearrange the equation to solve for n.
56.2 L * 1 atm = n * 0.0821 L·atm/(mol·K) * 273.15 K
Solving for n, we find that the number of moles of argon is approximately 2.13 moles. Now, let's move on to calculating the mass.
To find the mass, we need to know the molar mass of argon, which is approximately 39.95 grams per mole. Multiplying the molar mass by the number of moles (2.13 moles), we get:
2.13 moles * 39.95 g/mol ≈ 85.14 grams
So, the number of moles of argon in the sample is approximately 2.13 moles, and the mass of the sample is approximately 85.14 grams.
To find the number of moles of argon in the sample, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas (at STP, this is 1 atmosphere)
V is the volume of the gas (given as 56.2 liters)
n is the number of moles of the gas (what we want to find)
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature of the gas (at STP, this is 273.15 K)
Rearranging the equation to solve for n, we have:
n = PV / (RT)
Substituting the known values:
n = (1 atm) * (56.2 L) / ((0.0821 L·atm/(mol·K)) * (273.15 K))
Simplifying the equation:
n = 2.474 moles
Therefore, the number of moles of argon in the sample is approximately 2.474 moles.
To find the mass of the sample, we can use the molar mass of argon. The molar mass of argon is approximately 39.95 g/mol.
Mass = n * molar mass
Mass = 2.474 moles * 39.95 g/mol
Mass = 98.8 grams
Therefore, the mass of the argon sample is approximately 98.8 grams.