Question

The vapour pressure of pure liquid solvent A is 0.80 atm. When a nonvolatile substance B is added to the solvent its vapour pressure drops to 0.60 atm. What is the mole fraction of compound B in the solution? (if your answer is x, then write$x\times 100$)

Answer 1

When a nonvolatile substance is added to volatile solvent then lowering of vapour pressure occurs. This is called Raoult’s law. Also, this relative lowering of vapour pressure is directly proportional to the mole fraction of solute.

Formula used:
Relative lowering of vapour pressure = $\dfrac{{{P}_{A}}^{0}-{{P}_{A}}}{{{P}_{A}}^{0}}={{X}_{B}}$

Complete answer:
We have been given vapour pressure of pure solvent A to be 0.80 atm, and that of substance B (solute) to be 0.60 atm. Relative lowering of vapour pressure is observed according to Raoult’s law, as substance B is nonvolatile. So, the mole fraction of solute B is calculated by the formula for mole fraction, according to Raoult's law that states that lowering of vapour pressure is directly proportional to mole fraction of solute. So, $\dfrac{{{P}_{A}}^{0}-{{P}_{A}}}{{{P}_{A}}^{0}}={{X}_{B}}$, where ${{P}_{A}}^{0}$ is vapour pressure of pure solvent A, ${{P}_{A}}$ is vapour pressure of solution, ${{X}_{B}}$ is mole fraction of substance B (solute). So, putting the respective given values in this formula we have,
${{X}_{B}}=\dfrac{0.8-0.6}{0.8}$
${{X}_{B}}=0.25$
${{X}_{B}}=0.25\times 100$
${{X}_{B}}=25$
Hence, the mole fraction of solute is 0.25, and 25 when multiplied by 100.

Note:
Mole fraction does not have any unit, as it is a fraction, which is given by ${{X}_{A}}=\dfrac{{{n}_{A}}}{{{n}_{A}}+{{n}_{B}}}$ , where number of moles of component A divided by total number of moles in both component gives the mole fraction of component A. The relation between mole fractions of both components of a solution is ${{X}_{A}}+{{X}_{B}}=1$.