A sample of oxygen at 24.0 degrees celcius and 99.3 kpa was found to have a volume of 455 ml. How many grams of O2 were in the sample?
Answer
Answer for the question
To determine the number of grams of oxygen (O2) in the sample, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in kilopascals)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K) or 8.3145 J/(mol·K))
T = temperature (in Kelvin)
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 24.0°C + 273.15 = 297.15 K
Next, we need to convert the volume from milliliters to liters:
V(L) = V(mL) / 1000
V(L) = 455 mL / 1000 = 0.455 L
Now we have all the values needed to substitute into the ideal gas law equation:
PV = nRT
(99.3 kPa)(0.455 L) = n(8.3145 J/(mol·K))(297.15 K)
Simplifying the equation, we can solve for n (number of moles of O2):
n = (99.3 kPa * 0.455 L) / (8.3145 J/(mol·K) × 297.15 K)
n ≈ 0.0205 mol
Finally, to determine the mass of the oxygen, we need to use the molar mass of O2, which is approximately 32 g/mol:
Mass = n × Molar mass
Mass = 0.0205 mol × 32 g/mol
Mass ≈ 0.656 g
Therefore, there are approximately 0.656 grams of O2 in the given sample.
Use the gas law and do some conversion:
PV=nRT
Where
P=99.3kpa
V=455mL=0.455L
R=0.0821 L·atm/mol·K
T=24.0 C
and
n=???
Convert kpa to atm and C to K
1Atm = 101.325 kPa
So,
99.3 kPa*(1 atm/101.325kPa)= 0.980 atm
K=C + 273
So,
24.0 + 273=297K
Solve for n:
n=PV/RT
n*(32.00g/mole)= O2 in grams
Answer contains three significant figures.