Question

The Vapor pressure of benzene is 100.0 mmHg at 26.1 degrees celsius. Calculate the vapor pressure of a solution containing 23.5 g of camphor (C10H16O) dissolved in 97.2 g of benzene. Camphor is a low votality solid

Answer 1

@Photon336

Answer 2

Moles C10H16O = 23.5 g/ 153.23 g/mol = 0.161
Moles benzene = 97.2 g / 78.1121 g/mol = 1.26
Mole fraction benzene = 1.24 / 1.24 + 0.153 = .959
vapor pressure = 100 x 0.959 = 95.945 mm Hg

Answer 3

@Photon336 I'll be back in a few hours. Sorry I didn't get back to you earlier.

Answer 4

@tatianagomezb

Answer 5

\[P_{vapor~pressure} = P_{V/pressure~pure~benzene}*X_{mol fraction

Benzene}+P_{V/Ppressure pure camphor}*X_{mol fraction camphor}\]

Answer 6

Moles of benzene = 1.26
Moles of camphor = 0.161

Total #moles
\[Benzene+Camphor = (1.26+0.161) = 1.421_{moles~total}\]

Mole fraction of benzene
\[X_{BENZENE} = (\frac{ 1.26 }{ 1.421 }) = 0.887 \]

Mole fraction of Camphor
\[X_{CAMPHOR} = (\frac{ 0.161 }{ 1.421 }) = 0.113\]


\[P_ {vapor~pressure} = P^{0}X_{BENZENE} +P^{0}X_{CAMPHOR} \]

Answer 7

\[VaporPressure~Camphor~25^{0}C = 0.65mmHG\]


\[VaporPressure~of~Benzene 100~mmHG~26.1^{0}C\]

\[(100~mmHG)*(0.887)+(0.65mmHG)(0.113) = 88.77~mmHG \]

Answer 8

I saw what you did and you were correct when you had to calculate the mole fraction, you needed to do that. But I also think that you needed to take into account the amount of camphor you put in too as well as the vapor pressure of pure camphor. You have to dig around to find the numbers but this reaction is happening around approximately 25C

Answer 9

remember vapor pressure of pure substances changes as temperature changes so you would need the value of the vapor pressure at that particular temperature.

Answer 10

@Photon336 I knew I was close but I got tripped up towards the end. Thanks for the help.