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?

for my answer I got 0.370 C6H6.
please help!

You're right for C6H6.

XC2H5OH is 1.00 - 0.370 = ?

To find the mole fraction of benzene in the mixture, we need to calculate the amount of moles of benzene and ethanol first. Then, we can find the mole fraction by dividing the moles of benzene by the total moles in the mixture.

1. Determine the molar masses of benzene (C6H6) and ethanol (C2H5OH):
Molar mass of benzene (C6H6) = 6(12.01 g/mol) + 6(1.01 g/mol) = 78.11 g/mol
Molar mass of ethanol (C2H5OH) = 2(12.01 g/mol) + 6(1.01 g/mol) + 16.00 g/mol = 46.07 g/mol

2. Calculate the moles of benzene and ethanol using the equal masses provided:
Moles of benzene = mass of benzene / molar mass of benzene
= mass of benzene / (78.11 g/mol)
Moles of ethanol = mass of ethanol / molar mass of ethanol
= mass of ethanol / (46.07 g/mol)

3. Since equal masses are used, the masses of benzene and ethanol are the same. Let's assume the mass to be 'm', and substitute in the above equations:
Moles of benzene = m / (78.11 g/mol)
Moles of ethanol = m / (46.07 g/mol)

4. Calculate the total moles in the mixture:
Total moles = Moles of benzene + Moles of ethanol

5. Calculate the mole fraction of benzene:
Mole fraction of benzene = Moles of benzene / Total moles

Substituting the values and evaluating the expression will give you the mole fraction of benzene in the mixture.
Let's calculate it step by step using the values you provided:
Mole fraction of benzene = (0.370 mol) / (0.370 mol + 0.370 mol) = 0.370

Therefore, the mole fraction of benzene in the mixture is 0.370.