The mass of 3.01 x 1023 molecules of gas is 18.0 grams. What volume does 12.0 grams of the gas occupy at STP
1 mol of a gas is 6.02E23 molecules so moles you have is
(3.01E23/6.02E23) = 0.5 mol and that has a mass of 18 g so 1 mol has a mass of 36 g. Then 36 g will have a volume of 22.4 L @ STP
22.4 L x 12/36 = volume occupied by 12 g
Well, if the mass of 3.01 x 10^23 molecules is 18.0 grams, we can calculate the molar mass of the gas.
To do that, we divide the total mass (18.0 grams) by the number of molecules (3.01 x 10^23).
So, the molar mass would be 18.0 grams / 3.01 x 10^23 = a BIG number that I can't even pronounce. Let's call it "Molar Mass Hocus Pocus".
Now, we can use the molar mass Hocus Pocus and the given mass of 12.0 grams to calculate the number of moles of the gas. We divide the given mass by the molar mass to get:
12.0 grams / Molar Mass Hocus Pocus = Moles of Gas Hokus Pocus
Now, since it's STP (Standard Tons of Pranks), we know that at STP one mole of any gas occupies about 22.4 liters of space.
So, we can multiply the moles of gas Hokus Pocus by 22.4 liters to find the volume of 12.0 grams of the gas at STP.
Voilà! You have your answer in liters, or as I like to call it, "Laughter Units."
To find the volume of 12.0 grams of the gas at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation:
PV = nRT,
where:
P = pressure (at STP = 1 atmosphere)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (at STP = 273.15 Kelvin)
First, we need to calculate the number of moles of the gas. We know that the mass of 3.01 x 10^23 molecules of the gas is 18.0 grams. To find the molar mass of the gas, we can divide the mass by the number of molecules:
Molar mass = mass / number of molecules
Molar mass = 18.0 g / (3.01 x 10^23 molecules)
This gives us the molar mass. Let's calculate it.
Molar mass = 18.0 g / (3.01 x 10^23 molecules)
Molar mass ≈ 5.98 x 10^-24 g/molecule
Next, we can find the number of moles of the gas in 12.0 grams by dividing the mass by the molar mass:
Number of moles = mass / molar mass
Number of moles = 12.0 g / (5.98 x 10^-24 g/molecule)
Let's calculate the number of moles.
Number of moles = 12.0 g / (5.98 x 10^-24 g/molecule)
Number of moles ≈ 2.01 x 10^24 moles
Now that we have the number of moles, we can substitute this value along with the other known values into the ideal gas law equation to find the volume (V):
PV = nRT
V = (nRT) / P
V = (2.01 x 10^24 moles x 0.0821 L·atm/(mol·K) x 273.15 K) / 1 atm
Let's calculate the volume.
V = (2.01 x 10^24 moles x 0.0821 L·atm/(mol·K) x 273.15 K) / 1 atm
V ≈ 457 L
Therefore, 12.0 grams of the gas occupies approximately 457 liters at STP.
To find the volume of gas at STP (Standard Temperature and Pressure) given its mass, we need to use the concept of molar mass and the ideal gas law.
1. Determine the molar mass of the gas:
The molar mass of a compound is the mass of one mole of that substance. In this case, we are given the mass of 3.01 x 10^23 molecules, which is equal to the mass of one mole.
Since the mass of 3.01 x 10^23 molecules is given as 18.0 grams, we can calculate the molar mass as follows:
Molar mass = Mass / Number of moles
Molar mass = 18.0 g / 1 mol
Molar mass = 18.0 g/mol
2. Calculate the number of moles of gas:
To find the number of moles of gas, we can divide the given mass (12.0 grams) by the molar mass we calculated in step 1:
Number of moles = Mass / Molar mass
Number of moles = 12.0 g / 18.0 g/mol
Number of moles ≈ 0.667 mol
3. Apply the ideal gas law to determine the volume:
The ideal gas law equation is written as:
PV = nRT
At STP conditions, the temperature (T) is 273.15 K, the pressure (P) is 1 atm, and the gas constant (R) is 0.0821 L·atm/mol·K. Rearranging the equation, we get PV = nRT → V = (nRT) / P.
Substituting the values we have:
V = (0.667 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm
Calculating this gives us approximately:
V ≈ 15.7 L
Therefore, 12.0 grams of the gas would occupy a volume of approximately 15.7 liters at STP.