The vapour pressure (VP) of a dilute sol...

The vapour pressure `(VP)` of a dilute solution of non-volatile solute is `P` and the `VP` of a pure solvent is `P^(@)`. The lowering of the `VP` is

`+ve`

`-ve`

`P//P^(@)`

`P^(@)//P`

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The correct Answer is:

Lowering is always positive

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Vapour pressure of the solution of a non- volatile solute is always __.

When non-volatile solute is assed to a pure solvent, the:

Partial pressure of solvent in solution of non-volatile solute is given by equation.

Addition of non-volatile solute to solvent lowers its vapoure pressure. Therefore, the vapour pressure of a solution (i.e, V.P. of solvent in a solution) is lower than that of pure solvent in a solution) is lower than that of pure solvent, at the same temperature. A higher temperature is needed to raise the vapour pressure upto one atmosphere pressure, when boiling point is attined. However, increase in b.pt. is small . for example, 0.1 molal aqueous sucrose solution boils at 10.05^(@)C Sea water, an aqueous solution, which is rich in Na^(+) and Cl^(-) ions, freezes about 1^(@)C lower than frozen water . At the freezing point of a pure-solvent, the reates at which two molecule stick together to form the solid and leave it to return to liquid are equal when solute is present. Few solvent molecules are in contact with surface of solid. However, the rate at which the solvent molecules leave, surface of solid remains unchanged. That is why, temperature is lowered to restore the equalibrium. The freezing depression in a dilute solution is proportional to molality of the solute. The freezing point of benzene solution was 5.4^(@)C . The osmotic pressure of same solution at 10^(@)C is (freezing point of benzene = 5.5^(@)C ). Assume solution to be dilute. [ K_(f) for C_(6)H_(6) is 4.9 K "molality"^(-1) ].

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