Calculate the mass of a non-volatile solute (molar mass "40 g mol"^(-1)) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Calculate the mass of a non-volatile solute (molar mass "40 g mol"^(-1

1.	Calculate the mass of a non-volatile solute (molar mass "40 g mol"^(-1)) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Answer» Solution :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`