Question

What does Raoult’s law state?

Answer 1

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.

Answer 2

Step-by-step 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) * x_(A)`

`p_(B) = p^(o)_(B) * x_(B)`

P =
`p_(A) + p_(B)`

=
`p^(o)_(A) * x_(A) + p^(o)_(B) * x_(B)`

Since,
`x_(A) + x_(B)` = 1

`x_(A)` = 1 -
`x_(B)`

hence, P =
`p^(o)_(A) * (1 - x_(B)) + p^(o)_(B) * 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,
`(p^(o) - P_(s))/(P^(o))` =
`(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