derive an expression for relative lowering in vapour pressure
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let the vapour pressure of pure solvent is P° the vapour pressure of the solution formed on adding solute to it will be P.
As per the result law we know that the relative lowering of vapour pressure is equals to the mole fraction of the solute added .
P°-P/P°=x solute
there are 'n' mole dissolved in 'N' moles of solvent.
so,the moles fraction of solute X solute
=n/n+N
so P°-P/P°=n/n+N
since, n =wt/m.wt
putting this is n place of 'n' and 'N' we get
P°-P/P°=wt/m.wt/wt/m.wt+wt/m.wt
for a very dilute solution P°-P/P°=n/N
the above is the relations of relative lowering of vapour pressure on adding solute
As per the result law we know that the relative lowering of vapour pressure is equals to the mole fraction of the solute added .
P°-P/P°=x solute
there are 'n' mole dissolved in 'N' moles of solvent.
so,the moles fraction of solute X solute
=n/n+N
so P°-P/P°=n/n+N
since, n =wt/m.wt
putting this is n place of 'n' and 'N' we get
P°-P/P°=wt/m.wt/wt/m.wt+wt/m.wt
for a very dilute solution P°-P/P°=n/N
the above is the relations of relative lowering of vapour pressure on adding solute
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