Question

What does Raoult’s law state?

Answer 1

Explanation:

Raoult's law for volatile solutes states that at a given temperature vapor pressure of a component is equal to the mole fraction of that component component in the solution multiplied to the vapor pressure of the component in the pure state.

Mathematically,     $p_{A} = p^{o}_{A} \times x_{A}$

                              $p_{B} = p^{o}_{B} \times x_{B}$

                              P = $p_{A} + p_{B}$

                                 = $p^{o}_{A} \times x_{A} + p^{o}_{B} \times x_{B}$

Since,  $x_{A} + x_{B}$ = 1

           $x_{A}$ = 1 - $x_{B}$

hence,    P = $p^{o}_{A} \times (1 - x_{B}) + p^{o}_{B} \times x_{B}$

              P = $(p^{o}_{B} - p^{o}_{A})x_{B} + p^{o}_{A}$

On the other hand, Raoult's law for non-volatile solutes states that relative lowering of vapor pressure of a solution that has non-volatile solute is equal to the mole fraction of the solute in the solution.

Mathematically,       $\frac{p^{o} - P_{s}}{P^{o}}$ = $\frac{n_{2}}{n_{1} + n_{2}}$ = $x_{2}$

where,    $x_{2}$ = mole fraction of solute

               $n_{1}$ = moles of solvent

               $n_{2}$ = moles of solute

               $P_{s}$ = vapor pressure of the solution

               $p^{o}$ = vapor pressure of pure solvent

Answer 2

Raoult's Law basically states that the vapor pressure of a solution equals the sum of the vapor pressures of each volatile component, if it was multiplied by that mole fraction of the component in solution.