Henry's law state that

the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution” and is expressed as:
p = KH x 

Here KH is the Henry’s law constant

Answer:

for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.
p_A ∝ x_A

p_A x x_A where p_A is vapour pressure of solvent having mole fraction x_A.
PA = P⁰A x A
But x_A + x_B = L
∴ x_A = 1 – x_B
When x_B is mole fraction of non-voltile solute B
Pa = P⁰_A (1–x_B)
= p⁰A – p⁰_A x _B
Total vapour of solution is equal to p_A as nonvolatile solute does not have any vapour pressure.
i.e., ${\mathrm{p}}_{\mathrm{B}} = 0$   Total vapour pressure,
               $$\begin{gathered} \mathrm{p} = {\mathrm{p}}_{\mathrm{A}}+{\mathrm{p}}_{\mathrm{B}} = {{\mathrm{p}}^{\mathrm{o}}}_{\mathrm{A}}- {{\mathrm{p}}^{\mathrm{o}}}_{\mathrm{B}}\times \mathrm{B} \\ {\mathrm{x}}_{\mathrm{B}} = \frac{ {{\mathrm{p}}^{\mathrm{o}}}_{\mathrm{A}} - {\mathrm{p}}_{\mathrm{A}}}{{{\mathrm{p}}^{\mathrm{o}}}_{\mathrm{A}}} \end{gathered}$$

The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The ideal solutions have two other important properties. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero, i.e.,

ΔmixH = 0, ΔmixV = 0 

Amount of mole of a constituent divided by the total amount of moles of all constituent in a mixture.

Mole fraction of a component =
$\frac{\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{the} \mathrm{component}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{component}}$