A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system.A liquid vapourises to form a more disordered gas.When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness.As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas.Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution.
Similarly, the greater randomness of the solution opposes the tendency to freeze.In consequence, a lower the temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution.Elevation of `B.Pt.(DeltaT_b)` and depression of `F.Pt.(DeltaT_f)` of a solution are the colligative properties which depend only on the concentration of particles of the solute, not their identify.For dilute solutions. `DeltaT_b and DeltaT_f` are proportional to the molality of the solute in the solution.
`DeltaT_b=K_bm , K_b`=Ebullioscopic constant=`(RT_(b)^(@^(2))M)/(1000 DeltaH_(vap))`
And `DeltaT_f=K_fm , K_f`=Cryoscopic constant=`(RT_(f)^(@^(2))M)/(1000 DeltaH_(fus))` (M=molecular mass of the sovent)
The values of `K_b and K_f` do depend on the properties of the solvent.For liquids, `(DeltaH_(vap))/T_b^@` is almost constant.
[Troutan's Rule, this constant for most of the unassociated liquids (not having any strong bonding like Hydrogen bonding in the liquid state ) is equal to `90 J//"mol"`.]
For solutes undergoing changes of molecular state is solution (ionization or association), the observed `DeltaT` values differ from the calculated ones using the above relations In such situations, the relationships are modified as `DeltaT_b=i K_bm, DeltaT_f=i K_fm`
where i=Van't Hoff factor, greater than unity for ionization and smaller than unity for association of the solute molecules.
To aqueous solution of Nal increasing amounts of solid `HgI_2` is added.The vapour pressure of the solution

Options

A. decrease to a constant value
B. increase to a constant value
C. increases first and then decreases
D. remains constant because `HgI_2` is sparingly soluble in water

Answer

B

Explanation

`2Na^+(aq)+2I^(-)(aq)+HgI_2(s)to2Na^+(aq)+HgI_4^(2-)(aq)`
The number of mole particle decreases from `4(2Na^+ + 2I^(-)) "to" 3(2Na^+ + HgI_4^(2-))` Hence, the colligative property will decrease or, the vapour pressure will increase to a constant value until NaI is completely consumed.