Calculate the vapor pressure of a solution of 0.99 mol of cholesterol in 5.4 mol of toluene at 32°c. pure toluene has a vapor pressure of 41 torr at 32°c. (assume ideal behavior.)

Respuesta :

Explanation:

No. of moles of cholesterol solute, [tex]n_{1}[/tex] = 0.99 mol

No. of moles of toulene solvent, [tex]n_{2}[/tex] = 5.4 mol

Hence, total number of moles of solution will be as follows.

                  n = [tex]n_{1} + n_{2}[/tex]

                     = 0.99 mol + 5.4 mol

                     = 6.39 mol

Therefore, mole fraction of cholestrol solute ([tex]\chi_{1}[/tex]) is as follows.

                = [tex]\frac{\text{no. of moles of chloesterol}}{\text{Total number of moles of solution (n)}}[/tex]  

     [tex]\chi_{1}[/tex] = [tex]\frac{0.99 mol}{6.39 mol}[/tex]

                      = 0.154

Vapor pressure of pure toulene solvent ([tex]p^{o}[/tex]) = 41 torr

Vapor pressure of solution = P

Hence, formula to calculate relative lowering of vapor pressure is as follows.

     Relative lowering of vapor pressure = [tex]\frac{p^{o} - P}{p^{o}}[/tex]

As per relative lowering of vapor pressure colligative property, the relative lowering of vapor pressure is equal to the mole fraction of solute.

Hence,     [tex]\frac{p^{o} - P}{p^{o}}[/tex] = [tex]\chi_{1}[/tex]

               [tex]\frac{41 torr - P}{41 torr}[/tex] = 0.154

                         41 torr - P = 6.314

                                      P = 34.686 torr

or,                                      = 35 torr

Therefore, we can conclude that vapor pressure of the solution is 35 torr.