The lowering of vapour pressure of the solvent takes place on dissolving a non-volatile solute because

Question

The lowering of vapour pressure of the solvent takes place on dissolving a non-volatile solute because (A) the density of the solution increases (B) the surface tension of the solution decreases (C) the molecules of the solvent on the surface are replaced by the molecules of the solute (D) the mole fraction of solvent is less than 1

Tardigrade Admin · Accepted Answer

Vapour pressure does not depend upon surface area of a volatile liquid (p/p°)= xsolvent In case of a solution rsoivem < 1. ∴ p < p°

Q. The lowering of vapour pressure of the solvent takes place on dissolving a non-volatile solute because

the density of the solution increases

the surface tension of the solution decreases

the molecules of the solvent on the surface are replaced by the molecules of the solute

the mole fraction of solvent is less than 1

Vapour pressure does not depend upon surface area of a volatile liquid $\frac{p}{p ^{\circ}} = x_{so l v e n t}$ In case of a solution rsoivem $< 1. ∴ p < p^{\circ}$

Q. The lowering of vapour pressure of the solvent takes place on dissolving a non-volatile solute because

the density of the solution increases

the surface tension of the solution decreases

the molecules of the solvent on the surface are replaced by the molecules of the solute

the mole fraction of solvent is less than 1

Vapour pressure does not depend upon surface area of a volatile liquid p∘p​=xsolvent​ In case of a solution rsoivem <1.∴p<p∘

Vapour pressure does not depend upon surface area of a volatile liquid $\frac{p}{p ^{\circ}} = x_{so l v e n t}$ In case of a solution rsoivem $< 1. ∴ p < p^{\circ}$