On adding a non-volatile solute to a solvent, the vapour pressure of solvent decreases and becomes z $ \times $[vapour pressure of solvent] where z is
1) Mole fraction of solvent
2) Mole fraction of solute
3) Molality
4) Molarity

Hint: On addition of the non- volatile solute to solvent, the relationship between the vapour pressure and the mole fraction of the solvent is formed which is given by $p = {p^*}x$, here x is the mole fraction of the solvent and p* is the vapour pressure of solvent.

Complete step by step answer:
This is based on the principle of Raoult’s law.
Raoult’s law states that the partial vapour pressure of a component of an ideal mixture is the vapour pressure of the solvent multiplied by its mole fraction of solvent. In an ideal mixture it is presumed that there are equal intermolecular interactions between all molecules inside the mixture.
Therefore according to Raoult’s law
$p = {p^*}x$
Where,
${p^*}$ = represents vapour pressure of solvent
$x$ = represents mole fraction of the solvent
Therefore for the given question when a non-volatile solute is added to a solvent, the z is the Mole fraction of solvent

So, the correct answer is Option 1.

Additional Information:
In chemistry when solute is dissolved in solvent it forms the solution which is homogeneous in nature.
Solvent: liquid in which solute is dissolved.
Solute: the dissolved substances which are added into the solvent

Note: Vapour pressure is defined as measure of the ability of a substance or material to change its state into the vapour or the gaseous state, and it is directly dependent on the temperature, with increases in temperature it increases.
Boiling point of liquid can be defined using vapour pressure: It is the temperature at which its surface vapour pressure gets in equilibrium with the surroundings pressure.