Calculate the mass of a non-volatile solute ( molecular mass `40`) which should be dissolved in `114 g` octane to reduce its vapour pressure to `80%`.

Explanation

Reduction of vapour pressure to `80` means that if `p^(@)=" 100 mm, then "p_(s)="80 mm."`
Applying complete formula
`(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1)+n_(2))=(w_(2)//M_(2))/(w_(1)//M_(1)+w_(2)//M_(2))`
`=(100-80)/(100)=(w_(2)//40)/(114//114+w_(2)//40)" "("Mol. mass of octane "C_(8)H_(18)="114 g mol"^(-1))`
`"or "(20)/(100)=(w_(2)//40)/(1+w_(2)//40)" or "(1)/(5)(1+(w_(2))/(40))=(w_(2))/(40)" or "w_(2)=10g`
Note that complete formula is required because concentration of solution is greater than `5%`.
Complete formula can also be applied in the form
`(p^(@)-p_(s))/(p_(s))=(w_(2)//M_(2))/(w_(1)//M_(1))" or "(100-80)/(80)=(w_(2)//40)/(114//114)" or "(1)/(4)=(w_(2))/(40)" or "w_(1)=10g`
Alternatively, suppose mass of solute dissolved = 2 g
Moles of solute `=(w)/(40 )g`
Moles of solvent (octane) `=(114)/(114)="1 mole "("Mol. mass of "C_(8)H_(18)="114 g mol"^(-1))`
`therefore" Mole fraction of solvent "=(1)/(1+w//40)`
For a non-volatile solution,
Vapour pressure of solution = Mole fraction of solvent in the solution `xx` Vapour pressure of pure solvent
`p_(s)=x_(1)xxp^(@)`
`therefore" "80=(1)/(1+w//40)xx100" or "1+(w)/(40)=(100)/(8)" or "(w)/(40)=(10)/(8)-1=(2)/(8)=(1)/(4)" or "w=10 g`