a mixture is prepared using equal masses of two volatile liquid benzene, C6H6 and ethanol C2H5OH. what is the mole fraction of benzene in this mixture?
I would pick a number for the masses.
The convert masses to mols.
mols fraction X = molsX/total # of mols.
Post your work if you get stuck.
DrBob222 you are the best thanks for all you help...you great. thanks
ok, so I assumed the masses to equal 5.00 g
molar mass of benzene is =78.12 g/mol
molar mass of ethanol is =46.08g/mol
moles of C6H6=
5.00g/78.12g/mol=0.064004moles
moles of C2H5OH=
5.00 g/48.08g/mol=0.10851g/mol
so mole fraction of benzene should equal to=
0.064004mol/(0.064004+0.10851)=0.371mol
so the mole fraction of benzene is 0.371 mol
Is my approach to the problem correct?
To find the mole fraction of benzene in the mixture, we need to calculate the moles of benzene and ethanol separately, and then divide the moles of benzene by the total moles of the mixture.
Here's how you can do it:
1. Determine the molar mass of benzene (C6H6) and ethanol (C2H5OH):
- Molar mass of benzene (C6H6) = 12.01 g/mol (Carbon) × 6 + 1.01 g/mol (Hydrogen) × 6 = 78.11 g/mol
- Molar mass of ethanol (C2H5OH) = 12.01 g/mol (Carbon) × 2 + 1.01 g/mol (Hydrogen) × 6 + 16.00 g/mol (Oxygen) + 1.01 g/mol (Hydrogen) = 46.07 g/mol
2. Calculate the moles of benzene and ethanol:
- Moles of benzene = Mass of benzene / Molar mass of benzene
- Since equal masses of benzene and ethanol are used, the moles of benzene and ethanol will be the same.
3. Calculate the total moles of the mixture:
- Total moles of the mixture = Moles of benzene + Moles of ethanol
4. Calculate the mole fraction of benzene:
- Mole fraction of benzene = Moles of benzene / Total moles of the mixture
By following these steps, you can find the mole fraction of benzene in the mixture.