Question

Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6. at 25 degrees Celsius, the vapor pressure of pure benzene is 100.84 torr. what is the vapor pressure of a solution made from dissolving 10.1 g of biphenyl in 28.5 g of benzene?

Answer 1

Concept and reason

The concept used is to calculate the vapor pressure of the solution.

Fundamentals

A solute is a substance that is dissolved in another substance. It is present in smaller quantities.

A solvent is a liquid in which the solute is dissolved. It is present in larger amounts.

According to Raoult’s law, when a non-volatile solute is mixed in a solvent, the vapor pressure of a solution is equal to the vapor pressure of the pure solvent multiplied by its mole fraction.

It is expressed as follows:

${P_{{\rm{solution}}}} = {x_{{\rm{solvent}}}} \times {P_{{\rm{solvent}}}}$

Moles fraction of a solvent is the number of moles of solvent divided by the total moles of solute and solvent.

It is expressed as follows:

${x_{{\rm{solvent}}}} = \frac{{{n_{{\rm{solvent}}}}}}{{{n_{{\rm{solvent}}}} + {n_{{\rm{solute}}}}}}$

Here in this question, benzene is the solvent and biphenyl is the solute. So, the vapor pressure of the solution is as follows:

${P_{{\rm{solution}}}} = {x_{{\rm{benzene}}}} \times {P_{{\rm{benzene}}}}$

1.

Mass of biphenyl = $10.1{\rm{ g}}$

Molar mass of biphenyl = $154{\rm{ g/mol}}$

$\begin{array}{l}\\{\rm{Moles of biphenyl}} = \frac{{10.1{\rm{ g}}}}{{154{\rm{ g/mol}}}}\\\\{\rm{ = 0}}{\rm{.066 mol}}\\\end{array}$

Mass of benzene = $28.5{\rm{ g}}$

Molar mass of benzene = $78{\rm{ g/mol}}$

$\begin{array}{l}\\{\rm{Moles of benzene}} = \frac{{28.5{\rm{ g}}}}{{78{\rm{ g/mol}}}}\\\\{\rm{ = 0}}{\rm{.37 mol}}\\\end{array}$

The mole fraction of benzene is calculated as follows:

$\begin{array}{l}\\{\rm{Mole fraction of benzene}}\left( {{x_{{\rm{benzene}}}}} \right) = \frac{{0.37{\rm{ mol}}}}{{\left( {0.37 + 0.066} \right){\rm{ mol}}}}\\\\{\rm{ = 0}}{\rm{.85}}\\\end{array}$

Vapor pressure of pure benzene = $100.84{\rm{ torr}}$

The vapor pressure of the solution is calculated as follows:

$\begin{array}{l}\\{P_{{\rm{solution}}}} = \left( {0.85} \right) \times \left( {100.84{\rm{ torr}}} \right)\\\\{\rm{ = 85}}{\rm{.71 torr}}\\\end{array}$

Ans: Part 1

Therefore, the vapor pressure of the solution is $85.71{\rm{ torr}}$ .