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"Rate of vapourisation is reduced by presence of non-volatile solute "-Explain. |
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Answer» Solution :When a nonvolatile solute is dissolved in a pure solvent ,the VAPOUR pressure of the pure solvent will decrease .In such solutions , the vapour pressure of the solution will depend only on the solvent molecules as the solute is nonvolatile. For example when sodium chloride is added to the water the vapour pressure of the solution is determined by the number of molecules of the solvent PRESENT in the surface at any time and is proportional to the MOEL fraction of the solvent ` P_("solution ")prop x_A` Where `X_A ` is the mole fraction of the solvent ` P_("solution ")"" =kX_A ` When `X_A"" ="" 1, K =P^(@) _("solvent ") ` ` (P_("solvent ")^(@) ` is the PARTIAL pressure of pure solvent ) ` P_("solution ")""= ""P^(@) _("solvent ") X_A ` `(P_"solution "))/( P_("solvent ")^(@) ) ` ` 1- (P_("solution "))/( P_("solvent ")^(@)) "" ="" 1-X_A ` ` (P_("solvent" )^(@) -P_("solution "))/( P_("solvent ")^(@)) "" =""X_B ` Where `X_B ` is the fraction of the solute ` (therefore X_A +X_B =1, X_B =1-X_A ) ` The above expression gives the relative lowering of vapour pressure.Based on this expression , Raoult''s LAW can also be stated as " the relative lowering of vapour pressure of an ideal solution containing the nonvolatile solute is equal to the mole fraction of the solute at a given temperature ". |
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