An oxygen sample has a volume of 4.50 L at 27°C and 800.0 torr. How many oxygen molecules does it contain?

Answer 1

To find the number of oxygen molecules in the sample, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure in atm
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L * atm / (mol * K))
T = temperature in Kelvin

First, we need to convert the given temperature from Celsius to Kelvin:

T(K) = T(°C) + 273.15
T(K) = 27°C + 273.15
T(K) = 300.15 K

Next, we need to convert the given pressure from torr to atm:

P(atm) = P(torr) / 760
P(atm) = 800.0 torr / 760
P(atm) = 1.0526 atm

Now we have all the values we need to solve for the number of moles (n):

PV = nRT
n = PV / RT

n = (1.0526 atm) * (4.50 L) / (0.0821 L * atm / (mol * K) * 300.15 K)
n ≈ 0.2336 mol

Finally, we can use Avogadro's number (6.022 x 10^23 molecules/mol) to convert from moles to molecules:

Number of molecules = n * Avogadro's number
Number of molecules ≈ 0.2336 mol * (6.022 x 10^23 molecules/mol)
Number of molecules ≈ 1.405 x 10^23 molecules

Therefore, the oxygen sample contains approximately 1.405 x 10^23 oxygen molecules.