Biphenyl, C12H10, is a nonvolatile, non ionizing solute that is soluble in benzene, C6H6. At 25 °C, the vapor pressure of pure benzene is 100.84 torr. What is the vapor pressure of a solution made from dissolving 14.3 g of biphenyl in 25.8 g of benzene?
Didn't I do this one too? Where are you stuck on this?
No i never put this one up because i thought i knew how to do it but i keep getting the wrong answer so i guess im doing something wrong
Show me what you're doing and I'll try to find the error.
To find the vapor pressure of the solution, we can use Raoult's law. Raoult's law states that the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the solvent.
First, we need to convert the given masses of biphenyl and benzene into moles. We can use the molar mass of each compound to do this.
The molar mass of biphenyl (C12H10) can be calculated as follows:
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.008 g/mol
Molar mass of biphenyl = (12.01 * 12) + (1.008 * 10) = 154.22 g/mol
Next, we can calculate the number of moles of biphenyl:
moles of biphenyl = mass of biphenyl / molar mass of biphenyl
moles of biphenyl = 14.3 g / 154.22 g/mol
Similarly, we can calculate the number of moles of benzene:
moles of benzene = mass of benzene / molar mass of benzene
moles of benzene = 25.8 g / 78.11 g/mol
Now, we can calculate the mole fractions of biphenyl and benzene in the solution:
mole fraction of biphenyl = moles of biphenyl / (moles of biphenyl + moles of benzene)
mole fraction of benzene = moles of benzene / (moles of biphenyl + moles of benzene)
Finally, we can use Raoult's law to find the vapor pressure of the solution:
vapor pressure of solution = mole fraction of benzene * vapor pressure of benzene
vapor pressure of solution = mole fraction of benzene * 100.84 torr
By plugging in the calculated values, you should be able to find the vapor pressure of the solution.