State Raoult law and obtain expression for lowering of vapour pressure when nonvolatile solute is dissolved in solvent.

State Raoult law and obtain expression for lowering of vapour pressure

1.	State Raoult law and obtain expression for lowering of vapour pressure when nonvolatile solute is dissolved in solvent.
Answer» Solution :Ranoult.s law: This law states that "in the case of a solution of volatile liquids the partial vapour pressure of each component (A & B) of the solution is directly proporational to its mole fraction. `P _(A)prop X _(A)` when `x _(A) =1,` then `k = P^(@)_(A) ""(P_(A)^(@)=` vapour pressure of pure component) `therefore P _(A) = P _(A)^(@) x a` ` P _(B) = P _(B) ^(@) . x _(b)` when a non volatile is dissolved in pure WATER, the vapour pressure of the pure solvent will decrease. In such solution, the vapours pressure of the solution will depend only on the solvent molecules as the solute in non-volatile. `P _("solution") prop x _(A)` `P _("solution") =k. x _(A)` when `x _(A) =1,k = p^(@)_("solvent")` `therefore P _("solution") =p_("solvent"^(@) . x _(A)` Lowering of vapour pressure `=P _("solvent")^(@)-P_("solution")` RELATIVE lowering of vapour pressure `=(P^(@)-P)/(P ^(@))=x _(B)` where `x _(B)=` Mole fraction of solute.