Derive the relationship between relative lowering of vapour pressure and molar mass of solute.

Explanation

When a non-voltile solute is added to a solvent, the vapour of the solution decreases.
The vapour pressure of a solvent `(P_(1))` in a solution cvontaining a non- volatile solute is given by Raoult's law, Vapout pressure of the pure solvent
`=P_(1)^(@)`
Vapour pressure of the solvent in solution
`=P_(1)`
`P_(1)=x_(1)P_(1)^(@)`
`DeltaP_(1)=P_(1)^(@)-P_(1)`
`=P_(1)^(@)-x_(1)P_(1)^(@)`
`=P_(1)^(@)(1-x_(1))`
In a binary solution,
`1-x_(1)=x_(2)`
`DeltaP_(1)=P_(1)^(@)x_(2)`
`DeltaP_(1)//P_(1)^(@)=(P_(1)^(@)-P_(1))//P_(1)^(@)=x_(2)`
The lowering of vapour pressure relative to the vapour pressure of pure solvent is called relative lowering of vapour pressure.
`DeltaP_(1)//P_(1)^(@)to` Relative lowering of vapour pressure Thus, the relative lowering in vapour presure depends only on the concentration of solute particles and is independent of their identity.
If the solution contains more than one non-volatile solute, then the relative lowering in vapour pressure of a solvent is equal to the sum of the mole fractions of all the non-volatile solutes.
If `n_(1)and n_(2)` are respectively the number of moles of the solvent and solute in a binary solution, then the relative lowering in the vapour pressure of the solvent,
`(P_(1)^(@)-P_(1))//P_(1)^(@)=x_(1)+x_(2)+x_(3)+...+x_(n)`
if `n_(1) and n_(2)` are the number of moles of the solvent and solute,
`(P_(1)^(@)-P_(1))//P_(1)^(@)=n_(2)//(n_(1)+n_(2))`
For dilute solutions `n_(2)lt lt n_(1)`
`n_(1)=W_(1)//M_(1),n_(2)=W_(2)//M_(2)`
`(P_(1)^(@)-P_(1))//P_(1)^(@)=(W_(2)xxM_(1))//(W_(1)xxM_(2))`
or `(P_(1)^(@)-P_(1))/(P_(1)^(@))=(W_(2)M_(1))/(W_(1)M_(2))`
or ` (DeltaP_(1))/(P_(1)^(@))=(W_(2)M_(1))/(W_(1)M_(2))`
`W_(1)` = Mass of solvent
`W_(2)=` Mass of solute
`M_(1)=` Molar mass of solvent
`M_(2)=` Molar mass of solute