Detailed kinetic explanation for vapor pressure reduction by dissolved solute?
I'd like to specifically commment on this:
According to the NCERT for Class XII, Part I, [pg. 46, para 3][1],

In  a  pure  liquid  the  entire  surface  is occupied by the molecules of the liquid. If a non-volatile solute is added to a solvent to give  a  solution  [Fig.  2.4.(b)],  the  vapor pressure  of  the  solution  is  solely  from  the solvent  alone.  This  vapor  pressure  of  the solution at a given temperature is found to be  lower  than  the  vapor  pressure  of  the pure  solvent  at  the  same  temperature.  In the solution, the surface has both solute and solvent molecules; thereby the fraction of the surface covered by the solvent molecules gets reduced.  Consequently,  the  number  of solvent molecules escaping from the surface is correspondingly reduced, thus, the vapor pressure is also reduced.
[emphasis mine]

NCERT's explanation for why vapor pressure is lowered due to the presence of a dissolved solute is incorrect, because vapor pressure is independent of the surface area accesible to the solvent (assuming it's non-zero).  Consider these two examples:

You have two sealed containers, both of which contain an open beaker of identical solvent.  In container A, the beaker is low and wide.  In container B, the beaker is tall and narrow.  The surface area of solvent in container B is thus smaller.  Yet, assuming the conditions in the respective containers are identical, the equilibrium vapor pressures will be the same.

You have two sealed containers, both of which contain an open beaker of identical solvent.  The beakers are identical, so the surface area of liquid is the same. The conditions in the containers are also identical.  Glucose is dissolved into the solvent in container A, while sucrose is dissolved into the solvent in container B.  The final concentrations are the same.  Sucrose is larger than glucose.  Consequently, the fraction of the surface area covered by solvent in container A is larger than in container B.  Yet, ignoring non-ideality, the vapor pressures are the same; they depend only on the concentration of solute, not its nature.


The surface area only matters for kinetics, i.e., for how quickly the solvent can escape the container to reach equilibrium.  Clearly, with a larger surface area, the equilibrium vapor pressure will be reached more rapidly.
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Here's the thermodynamic explanation: The addition of solute lowers the chemical potential of the solvent, due to the entropy of mixing.  [There can also be energetic effects, in either direction, but with a soluble solute the entropy term is dominant.]
At equilibrium, the chemical potential of the solvent in the liquid and vapor phases must be equal.  Thus, since the chemical potential of the solvent in the liquid phase has lowered, that of the the solvent in the vapor phase must lower as well.
And since the chemical potential of a gas increases with its partial pressure, the vapor pressure of the solvent in the gas phase will decrease until its chemical potential reaches the new (lowered) chemical potential of the solvent in the liquid phase.
As far as the question itself goes, I'm uncertain, because the question is asking for a kinetic, rather than a thermodynamic, explanation.  But I can offer two alternative pictures:
Let's represent the movement between the liquid and vapor phases of the solvent (&quot;X&quot;), as follows:
$$\ce{X_{(l)}&lt;=&gt;X_{(g)}}$$
Picture (I): Both &quot;B&quot; and &quot;C&quot; are correct.
Let's start with the vapor in equilibrium with pure solvent.  At equilibrium, the rates of the forward and backward reactions are equal.
Now suppose we add solute to the liquid phase.  This shifts the reaction to the left, which means that the rate of the forward reaction initially decreases (which is answer &quot;B&quot;) and the rate of the backward reaction initially increases (which is answer &quot;C&quot;).  Eventually the system reaches its new equilibrium point, at which point the rates of the forward and reverse reactions are again equal.
Picture (II): &quot;B&quot; is correct.
Again, let's start with the vapor in equilibrium with pure solvent.  At equilibrium, the rates of the forward and backward reactions are equal.
Now suppose we add solute to the liquid phase. The rate at which gas enters the liquid phase depends only on the concentration of gas, so the backwards reaction will be unaffected by the presence of the solute in the liquid.  However, the rate of the forward reaction will be reduced, since it is now more favorable to for the solvent to remain in the liquid state.  Hence the answer is &quot;B&quot;.
As I mention in the Comments:
In Picture (I), the change in relative chemical potentials would raise the energetic barrier in the forward direction and lower it in the reverse direction, which would in turn increase the % of collisions that are succesful in the rev. direction, thus increasing the rate of the rev. reaction even though the collision frequency of the vapor with the liquid woudn't be changed by the presence of the dissolved solute.
But: If there isn't any barrier in the rev. direction, such that 100% of vapor molecules that collide with the liquid phase become liquid, then Picture (I) would not apply.
I'm afraid I don't know enough about the microscopic kinetics to have certainty on this.  Perhaps someone that has done simulation work on this could provide an answer.  [But to get this, you would want to vote to reopen the question.]