What volume will each of the following occupy at STP?
(a) 1.05 × 1023 molecules of CH4
(b) 1.8 mol CH4
(c) 33.3 g argon
1 mole of a gas occupies 22.4 L @ STP and it will contain 6.02E23 molecules and it will have a mass equal to the atomic/molecular mass.
a. 22.4 x 1.05E23/6.02E23 = ?
b. 22.4 L/mol x 1.8 mol = ?
c. 33.3 g Ar/39.9 = mols. Then follow b.
To calculate the volume occupied by each of the given substances at STP, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (at STP, it is 1 atm)
V = volume
n = number of moles of gas
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature (at STP, it is 273.15 K)
Let's go step by step to find the volume for each substance:
(a) 1.05 × 10^23 molecules of CH4:
First, we need to convert the number of molecules to moles.
1 mole of any substance contains Avogadro's number (6.022 × 10^23) of molecules.
So, the number of moles of CH4 is:
1.05 × 10^23 molecules CH4 / (6.022 × 10^23 molecules/mole) = 0.174 moles CH4
Now, we can substitute the values in the ideal gas law equation:
PV = nRT
(1 atm) * V = (0.174 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K)
Simplifying the equation:
V = (0.174 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm)
Calculating:
V = 3.83 L
So, 1.05 × 10^23 molecules of CH4 will occupy approximately 3.83 liters at STP.
(b) 1.8 mol CH4:
Given that we already have the number of moles, we can directly substitute the value in the ideal gas law equation:
PV = nRT
(1 atm) * V = (1.8 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K)
Simplifying the equation:
V = (1.8 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm)
Calculating:
V = 39.5 L
So, 1.8 moles of CH4 will occupy approximately 39.5 liters at STP.
(c) 33.3 g argon:
First, we need to convert grams to moles using the molar mass of argon.
The molar mass of argon (Ar) is 39.9 g/mol.
The number of moles of argon is:
33.3 g Ar / 39.9 g/mol = 0.834 mol Ar
Now, we can substitute the values in the ideal gas law equation:
PV = nRT
(1 atm) * V = (0.834 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K)
Simplifying the equation:
V = (0.834 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm)
Calculating:
V = 18.3 L
So, 33.3 grams of argon will occupy approximately 18.3 liters at STP.
To calculate the volume of a gas at STP (Standard Temperature and Pressure), you can use the ideal gas law equation, which states that PV = nRT. In this equation, P represents the pressure, V represents the volume, n represents the number of moles, R is the ideal gas constant, and T represents the temperature (which is 273.15 Kelvin at STP).
To solve for V, rearrange the equation to V = (nRT) / P.
(a) To find the volume occupied by 1.05 × 10^23 molecules of CH4 at STP, we need to convert the number of molecules to moles. One mole of any substance contains 6.022 × 10^23 particles (Avogadro's number). Therefore, we can calculate the number of moles using the formula:
Number of moles = (Number of molecules) / (Avogadro's number)
Plugging in the values, we get:
Number of moles = (1.05 × 10^23) / (6.022 × 10^23) ≈ 0.174
Now, we can use the ideal gas law equation to find the volume:
V = (nRT) / P
V = (0.174 * 0.0821 * 273.15) / 1
V ≈ 4.33 L
Therefore, 1.05 × 10^23 molecules of CH4 will occupy approximately 4.33 liters at STP.
(b) To calculate the volume occupied by 1.8 moles of CH4 at STP, we can directly substitute the value of n into the ideal gas law equation:
V = (nRT) / P
V = (1.8 * 0.0821 * 273.15) / 1
V ≈ 41.2 L
Therefore, 1.8 moles of CH4 will occupy approximately 41.2 liters at STP.
(c) To calculate the volume occupied by 33.3 grams of argon at STP, we first need to convert the mass of argon to moles using its molar mass. The molar mass of argon is approximately 39.95 g/mol.
Number of moles = (Mass of argon) / (Molar mass of argon)
Number of moles = 33.3 g / 39.95 g/mol ≈ 0.834
Now, we can use the ideal gas law equation to find the volume:
V = (nRT) / P
V = (0.834 * 0.0821 * 273.15) / 1
V ≈ 17.2 L
Therefore, 33.3 grams of argon will occupy approximately 17.2 liters at STP.