The partial pressure of fluorine gas (F2) in a mixture of gases where the total pressure is 1.00 atm is 300. torr. What percent of the mixture is made of fluorine gas
I worked this problem below.
the ratio of the presure is directly proportional to the mole ratio.
molesF2=.3 total moles
total moles=RT/PV
percent of Moles F2 is 30 percent.
percent by mass depends on what the mass of the other gases is.
To determine the percentage of the mixture made of fluorine gas, we need to calculate the mole fraction of fluorine gas and then convert it to a percentage.
First, let's convert the given partial pressure of fluorine gas from torr to atm. Since 1 atm is equal to 760 torr, 300 torr is equivalent to 300/760 = 0.3947 atm.
Next, using Dalton's Law of partial pressures, we know that the sum of the partial pressures of all the gases in a mixture is equal to the total pressure. In this case, the total pressure is 1.00 atm.
Therefore, the mole fraction of fluorine gas (X_F2) can be calculated by dividing its partial pressure by the total pressure:
X_F2 = Partial pressure of F2 / Total pressure
= 0.3947 atm / 1.00 atm
= 0.3947
Now, to convert the mole fraction of fluorine gas into a percentage, we multiply it by 100:
Percentage of F2 = X_F2 * 100
= 0.3947 * 100
= 39.47%
So, approximately 39.47% of the mixture is made of fluorine gas.