A sample of oxygen gas has a volume of 8.00 L at 20.0 ° C and 700. torr. How many moles does the sample contain?
There is no point in posting the same question many times
Use PV=nRT
is answer near 3.15* 10^-2
To determine the number of moles in the sample of oxygen gas, you 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, let's convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 20.0 °C + 273.15
T(K) = 293.15 K
Next, let's convert the given pressure from torr to atm:
P(atm) = P(torr) / 760
P(atm) = 700 torr / 760
P(atm) ≈ 0.92105 atm
Now, we have all the values needed to calculate the number of moles:
P = 0.92105 atm
V = 8.00 L
T = 293.15 K
R = 0.0821 L·atm/mol·K
Plugging these values into the ideal gas law equation:
(0.92105 atm)(8.00 L) = n(0.0821 L·atm/mol·K)(293.15 K)
Simplifying the equation:
7.3684 L·atm = n(24.084315 L·atm/mol·K)
Dividing both sides by (24.084315 L·atm/mol·K):
n = (7.3684 L·atm) / (24.084315 L·atm/mol·K)
Calculating the right side of the equation:
n ≈ 0.305 moles
Therefore, the sample of oxygen gas contains approximately 0.305 moles.