15. Which of the following (all at STP) will contain the most atoms?
(a) 454 L of helium gas
(b) 80.00 g of argon gas
(c) 454 L of argon gas
(d) 80.00 g of helium gas
which is greater?
a. 454/22.4
b. 80/40
c. 454/40
d. 80/4
To determine which substance will contain the most atoms, we need to use the concept of Avogadro's law, which states that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules.
To apply Avogadro's law, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since the substances are at STP (standard temperature and pressure), we can assume the temperature (T) and pressure (P) are constant for all the options. Therefore, we can rewrite the equation as V = nRT/P.
Let's calculate the number of moles (n) for each substance:
(a) 454 L of helium gas:
Assuming we are using STP conditions, the molar volume of any gas at STP is 22.4 L/mol. Therefore, we can calculate the number of moles of helium gas:
n = V / (22.4 L/mol) = 454 L / (22.4 L/mol) = 20.27 mol
(b) 80.00 g of argon gas:
To find the number of moles of argon gas, we need to use its molar mass. The molar mass of argon is 39.95 g/mol. Therefore, we can calculate the number of moles:
n = mass / molar mass = 80.00 g / 39.95 g/mol = 2.00 mol
(c) 454 L of argon gas:
Using the same process as in (a), we calculate the number of moles of argon gas:
n = V / (22.4 L/mol) = 454 L / (22.4 L/mol) = 20.27 mol
(d) 80.00 g of helium gas:
Using the same process as in (b), we calculate the number of moles of helium gas:
n = mass / molar mass = 80.00 g / 4.00 g/mol = 20.00 mol
Now, let's determine which option contains the most atoms. To do this, we need to convert the number of moles to the number of atoms for each substance.
To convert moles to atoms, we need to use Avogadro's number, which is 6.022 × 10^23 atoms/mol.
(a) 454 L of helium gas:
atoms = moles * Avogadro's number = 20.27 mol * 6.022 × 10^23 atoms/mol = 1.221 × 10^25 atoms
(b) 80.00 g of argon gas:
atoms = moles * Avogadro's number = 2.00 mol * 6.022 × 10^23 atoms/mol = 1.204 × 10^24 atoms
(c) 454 L of argon gas:
atoms = moles * Avogadro's number = 20.27 mol * 6.022 × 10^23 atoms/mol = 1.221 × 10^25 atoms
(d) 80.00 g of helium gas:
atoms = moles * Avogadro's number = 20.00 mol * 6.022 × 10^23 atoms/mol = 1.204 × 10^25 atoms
Therefore, option (c), which is 454 L of argon gas, will contain the most atoms.