Problem: A gaseous mixture of O2 and Kr has a density of 1.181 g/L at 485 torr and 300. K. What is the mole percent O2 in the mixture?
Attempt: I have tried everything possible that I know and I can't get anywhere close to the solution. d=m/v=pM/RT... nothing works. Help please?
It seems to me you could do the following:
PV=nRT
PV = (grams/molar mass)RT
P= gRT/molar mass*V
P*molar mass = (g/v)RT
molar mass = density*RT/P
I plugged in the numbers and obtained about 45 something. Check my thinking. Check my arithmetic.
So the 45 something molar mass is made up of some percentage (or fraction) of 83.80 Kr (check that number) and some percentage (fraction) of 32.00 oxygen. Let x = fraction of oxygen, then 1-x = fraction of Kr.
32.00(x) + 83.80(1-x) = 45 something.
Solve for x. You can convert to grams and to mols and to mol fractions and mol percent from there. I kept looking for a way to obtain partial pressure of oxygen and Kr and from there the mol percent but that didn't happen. I think there may be an easier way to do this but I think the above will get it. Let me know.
To solve this problem, you need to use the ideal gas law equation, which relates the pressure, volume, temperature, and the number of moles of a gas.
The ideal gas law equation is given as:
PV = nRT
Where:
P = pressure of the gas (in atm or torr)
V = volume of the gas (in L)
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/mol·K or 62.36 L·torr/mol·K)
T = temperature of the gas (in Kelvin)
The given information is the density, pressure, and temperature. To solve the problem, we'll need to convert the density to molar mass and then use the ideal gas law to find the number of moles.
1. Convert density to molar mass:
Density (d) = mass (m) / volume (v)
Given that density (d) = 1.181 g/L, we need to determine the molar mass of the mixture. Assume there are x moles of O2 and y moles of Kr in 1 mole of the mixture.
The molar mass of the mixture (M) can be expressed as:
M = (x * M(O2)) + (y * M(Kr))
Where:
M(O2) = molar mass of oxygen
M(Kr) = molar mass of krypton
Since we are interested in mole percent of O2, we'll use only O2 in the molar mass calculation. Oxygen's molar mass (M(O2)) is 32.00 g/mol.
Substituting the given density into the equation:
d = m/v => 1.181 g/L = (x * M(O2)) + (y * M(Kr))
2. Use the ideal gas law equation to calculate the number of moles:
Rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Given that pressure (P) = 485 torr and temperature (T) = 300 K, we can now substitute the values:
n = (485 torr * 1 L) / (62.36 L·torr/mol·K * 300 K) = 0.026 moles
3. Substitute the number of moles into the previous equation:
1.181 g/L = (0.026 moles * 32.00 g/mol) + (y * M(Kr))
Now you have two equations with two variables. Simplify and solve for x, y, or M(Kr).
You can rearrange the equations and solve them manually, or you can use an equation solver or software like Microsoft Excel or Wolfram Alpha to find the values of x (moles of O2), y (moles of Kr), or M(Kr) (molar mass of krypton).
Once you have these values, you can calculate the mole percent of O2 using the formula:
Mole Percent O2 = (moles of O2 / total moles of gas) * 100
By following these steps, you should be able to determine the mole percent of O2 in the mixture.