Write Raoult's law for non - volatile solute and volatile solvent and

1.	Write Raoult's law for non - volatile solute and volatile solvent and explain it.
Answer» Solution :The vapour pressure of a solvent in solution is less than that of the pure solvent because the lowering of vapour PRESURE depends only on the concentration of the solute particles. A relation between vapour pressure of the solution, mole fraction and vapour pressure of the solvent, i.e., `p_(1)=x_(1).p_(1)^(0)` The REDUCTION in the vapour pressure of solvent `(Delta p_(1))` is given as : `p_(1)=x_(1).p_(1)^(0)` It should be decrease in vapour presure = `p_(1)^(0)-p_(1)=p_(1)^(0)-p_(1)^(0)x_(1)` `1-x_(1)=x_(2)` `Delta p_(1)=x_(2).p_(1)^(0)` In a solution containing several non - volatile solutes, the lowering of the vapour pressure depends on the sum of the mole fraction of different solutes. `(Delta p_(1))/(p_(1)^(0))=(p_(1)^(0)-p_(1))/(p_(1)^(0))=x_(2)` So, relative lowering of vapour presure `((p_(1)^(0)-p_(1))/(p_(1)^(0)))` is equal to the mole fraction `(X_(2))` of the solute. `(p_(1)^(0)-p_(1))/(p_(1)^(0))=(n_(2))/(n_(1)+n_(2))` where, `n_(1)=` moles of solvent and `n_(2)=` moles of solute For dilute solutions `n2 lt lt n1`, hence neglecting n2 in the denominator we have, `(p_(1)^(0)p_(1))/(p_(1)^(0))=(n_(2))/(n_(1))` `(p_(1)^(0)-p_(1))/(p_(1)^(0))=(w_(2)xx M_(1))/(M_(2)xx w_(1))` where, `w_(1)=` WEIGHT of solvent `w_(2)=` weight of solute `M_(1)=`MOLAR weight of solvent `M_(2)=` molar weight of solute.

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