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**Hint:**This question requires us to complete the statement that is established in Raoult's Law. Since this law is related to vapour pressure, it can be closely related to the ideal gas law as well. Raoult's rule is based on the assumption that the forces present amongst the molecules for each molecule are identical.

**Complete answer:**

To explain the answer to this question first we must state Raoult's Law, then proceed to understand the law in terms of the given options.

The pressure that is applied or produced by vapours on the surface of a liquid in conditions that are equilibrium, at a fixed temperature is referred to as vapour pressure.

According to the law in consideration, that is Raoult’s Law:

The partial vapour pressure of the solvent that is involved in a solution (or sometimes a mixture) is either identical or equivalent to the product of pure solvent's vapour pressure and the mole fraction of that solvent involved in the solution.

The equation for this would be:

$ \Rightarrow $ $ {P_{solution}} = {P^ \circ }_{solvent} \times {x_{solvent}} $

where ‘ $ {P_{solution}} $ ’ is the vapour pressure of solution, ‘ $ {P^ \circ }_{solvent} $ ’ is the vapour pressure of pure solvent and ‘ $ {x_{solvent}} $ ’ is the mole fraction of the solvent in solution.

During the procedure of adding a solute to solvent, there is a decrease in vapour pressure. This is caused by the fact that once the solute is inserted into the pure solvent (liquid), the surface of the solvent will then contain ions or molecules from both ‘solute’ as well as ‘pure solvent’.

The number of solvent molecules that convert into the vapour phase decreases, and as a consequence, the vapour phase exerts a pressure that is again lesser than the original pressure. This is referred to as ‘relative lowering of vapour pressure’.

We can form a derivation to see the relative lowering of vapour pressure:

Consider that the solution formed is binary so then the mole fractions of the solvent and solute be $ {x_1},\;{x_2} $ respectively.

Let the vapour pressures of solvent and pure solvent be $ P,\;{P^ \circ } $ respectively.

By Raoult’s law we can say: $ P = {P^ \circ }{x_1} $

Now we can say that the reduction in the solvent’s vapour pressure ( $ \Delta P $ ) is represented as:

$ \Rightarrow \Delta P = {P^ \circ } - P $

Replacing the vapour pressure of solvent using Raoult’s law:

$ \Rightarrow \Delta P = {P^ \circ } - {P^ \circ }{x_1} $

$ \Rightarrow \Delta P = {P^ \circ } \times (1 - {x_1}) $

The solution was considered as binary so this is true: $ {x_2} = 1 - {x_1} $

So we can write $ \Delta P = {P^ \circ } \times {x_2} $

Rearranging we get:

$ \Rightarrow {x_2} = \dfrac{{\Delta P}}{{{P^ \circ }}} $ , here the relative lowering of vapour pressure is represented by $ \dfrac{{\Delta P}}{{{P^ \circ }}} $ .

This implies that from the above equation we can say that the mole fraction of the solute controls the relative lowering of vapour pressure.

Relative lowering of vapour pressure can be considered as a 'colligative property', since the relative lowering relies only upon the concentration or amount of the particles of solute that contain in it. Hence we can clearly say that relative lowering is determined by the mole fraction of solute.

So it is evident from the above derivation and statement that the options: (A) ‘mole fraction of solvent’, (C) ‘independent of mole fraction of solute’, (D) ‘molality of solution’ are all incorrect options.

**Therefore the correct answer is clearly option (B) mole fraction of solute.**

**Note:**

Although Raoult’s law is widely used, it does have its limitations like:

- Raoult's law is well suited to explaining optimal (ideal) solutions. Such kind of ideal compounds, on the contrary, are difficult to come by and occur very rarely. Different chemical elements must be chemically identical in the same way.

- There is no assurance that the components in the substance are uniform when their attractive forces are considered, these substances differ or vary from the established law.

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