Raoult law states that. "in the case of a solution of volatile liquids, the partial vapour pressure of each component (A & B) of the solution is directly proportional to its mole fraction.
According to Raoult's law,
P_A \(\alpha\) X_A 
p_A = K X_A
When xA = 1, k = \({ p }_{ A }^{ 0 }\)
Where \({ p }_{ A }^{ 0 }\) is the vapour pressure of pure component 'A' at the same temperature
"The relative lowering of vapour pressure of an ideal solution containing the nonvolatile solute is equal to the mole fraction of the solute at a given temperature".

Derivation:
when sodium chloride is added to the water, the vapour pressure of the salt solution is lowered. The vapour Pressure of the solution is determined by the number of molecules of the solvent present in the surface at any time and is proportional to the mole fraction of the solvent.
P_solution \(\infty \) X_A
Where X_A is the mole fraction of the solvent
P_solution  = K X_A
When X_A = 1, K = P⁰ solvent
\({ P }_{ solvent }^{ 0 }\) is the partial pressure of pure solvent
P_solution = \({ P }_{ solvent }^{ 0 }\) X_A
\(\frac { { P }_{ solution } }{ { P }_{ solvent }^{ 0 } } ={ X }_{ A }\)
\(1-\frac { { P }_{ solution } }{ { P }_{ solvent }^{ 0 } } 1-{ X }_{ A }\)
\(\frac { { P }_{ solvent }^{ 0 }-{ P }_{ solution } }{ { P }_{ solvent }^{ 0 } } ={ x }_{ B }\)
Where X_B the mole fraction of the solute
( \(\because\) X_A + A_B = 1, X_B = 1 - X_A)
The above expression gives the relative lowering of vapour pressure. Based on this expression, Raoult's law can also be stated as "the relative lowering of vapour pressure of an ideal solution containing, the nonvolatile solute is equal to the mole fraction of the solute at a given temperature".

(i) The pressure of the vapour in equilibrium with its liquid is called vapour pressure of the liquid at the given temperature.
(ii) The ratio of the difference between the vapour pressure of pure solvent and the vapour pressure of a solution to the vapour pressure of pure solvent is called the relative lowering of vapour pressure.
Relative lowering of vapour pressure = \(\frac{\mathbf{P}^{0} \text { solvent }-\mathbf{P}_{\text {soultion }}}{\mathbf{P}_{\text {solvent }}^{0}}\).

\(10\%{W\over V}\) means that 10g of solute in 100 ml solution
\(\therefore\) amount of iodopovidone in 1.5 ml = \({10\ g\over 100\ ml}\times 1.5\ ml\)
= 0.15 g.

Molar mass = \({mass\ of\ unknown\ solute × RT\over osmotic\ pressure\ ×\ volume\ of\ solution}\)
\(= {1.5\times 8.314\times 10^{-2}\times 400\over 0.3\times 1.5}\)
= 110.85 gram mol^-1.