contestada

Biphenyl, c12h10, is a nonvolatile, nonionizing 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 13.6 g of biphenyl in 26.4 g of benzene?

Respuesta :

Edufirst
One form of Raoult's Law states that the vapor pressure of a solution of a non-volatile solute at certain temperature is equal to the vapor pressure of the pure solvent at the same temperature multiplied by the mole fraction of the solvent, this is:

p = X solvent * P pure solvent,

X solvent = number of moles of solvent / total number of moles.

Here the solute is 13.6 g of C12 H10 and the solvent is 26.4 g C6H6.

=>

moles of solvent = mass in grams / molar mass

molar mass of C6H6 = 6 * 12 g/mol + 6 * 1g/mol = 78 g/mol

moles of solvent = 26.4 g / 78 g/mol = 0.33846 mol

molar mass of C12H10 = 12 * 12g/mol + 10*1g/mol = 154 g/mol

moles of solute = 13.6 g / 154 g/mol = 0.08831 mol

=> X solvent = 0.33846 / (0.33846 + 0.08831) = 0.793

=> p = 0.793 * 100.84 torr = 79.97 torr ≈ 80.0 torr

Answer: 80.0 torr

The vapor pressure of the solution of biphenyl and benzene has been 80.672 torr.

Raoult's law can be used for the determination of the vapor pressure of the solute in the solution.

The law can be given as:

Vapor pressure of solution = mole fraction of solvent [tex]\times[/tex] Vapor pressure of the solvent

Mole fraction of solvent = moles of solvent to moles of solution

Moles of solvent (Benzene) = [tex]\rm \dfrac{26.4}{78}[/tex]

Moles of solvent = 0.33 mol

Moles of solute = [tex]\rm \dfrac{13.6}{154}[/tex]

Moles of solute = 0.08 mol

Moles of solution = solvent + solute

Moles of solution = 0.33 + 0.08 = 0.41 mol

Moles of solvent = [tex]\rm \dfrac{0.33}{0.41}[/tex]

Moles of solvent = 0.80

Vapor pressure of solution = 0.80 [tex]\times[/tex] 100.84

Vapor pressure of solution = 80.672 torr.

For more information about vapor pressure, refer to the link:

https://brainly.com/question/14466591

Otras preguntas