Vapour Pressure of solutions MCQ - Practice Questions with Answers

It is the pressure exerted by vapours when in equilibrium with the liquid at a given temperature. It depends upon nature of liquid and temperature. Pure liquid has always a vapour pressure greater than its solution.
Vapour pressure of a liquid helps us to have an idea of forces of attraction amongst the molecules of liquid that is, more the force of attraction, lower is the vapour pressure and vice versa.
Vapour pressure of a liquid increase with an increase in temperature due to an increase in kinetic energy of solvent molecules that is, increase in evaporation however it is independent of the nature of the vessel.

Vapour Pressure of a Solution
When a miscible solute is added to a pure solvent, it results in the formation of solution. As some molecules of solute will replace the molecules of the solvent from the surface, therefore, escaping tendency of solvent molecules decreases. This causes a lowering of vapour pressure.

• The vapour pressure of a solution is less than that of pure solvent.
• If vapour pressure of solvent is P and that of solution is Ps then lowering of V.P = P - Ps.
• Vapour pressure of solution decreases as surface area occupied by solvent molecule decreases and density increases.

Vapour pressure depends on the following factors:

• Vapour pressure does not depend on the shape or size of the container.
•  Vapour pressure depends on the nature of substance/liquid.
• Vapour pressure is inversely proportional to the force of attraction between molecules.
  
• Vapour pressure depends upon temperature. As temperature increases, evaporation increases and thus vapour pressure increases.

This is the solution when it has constituents i.e, solute and solvent as volatile. Lets represent solvent as "A" and solute as "B".
Now before mixing, vapour pressure of A is P^o_A and vapour pressure of B is P^o_B.
After mixing of solute and solvent, vapour pressure of solvent A and solute B will be partial pressures i.e, P_A and P_B.

Now according to Raoult's law, vapour pressure of liquid A is proportional to the mole fraction of liquid A. 
Thus, P_A = KX_A and P_B = KX_B
Now, when we have only liquid A, then partial pressure of A is equal to P^o_A, thus K = P^o_A.
Thus, we can write:
P_A = P^o_AX_A and P_B = P^o_BX_B

Now according to Dalton's law of partial pressure, we have:
Total pressure(P_T) = P_A + P_B 

Thus, P_T = P^o_AX_A + P^o_BX_B

The non-volatile solute is the solute which is present in solid-state. The solution is prepared by mixing of this solute in the liquid solvent.
Let's consider the solvent as "A" and solute as "B". Now, when no solute is present in the solvent, then the vapour pressure of the solvent is represented as P^o_A. Now when the solute is dissolved in the solvent then the vapour pressure of the solvent decreases and is represented as P_A. 

According to Raoult's law, we know:
P_A = P^o_AX_A       .............(i)

Now, total mole fraction of solvent(X_A) and solute(X_B) is equal to 1.
Thus, X_A + X_B = 1
On putting the value of X_A in equation (i), we get:
P_A = P^o_A [1-X_B]
P_A = P^o_A - P^o_A X_B
Therefore, P^o_A X_B = P^o_A - P_A

Thus, (P^o_A - P_A) is also known as the lowering of vapour pressure

This equation is also known as 'Relative lowering of vapour pressure'.