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

Explanation

Raoult's law formula `(P_(0)-P_(S))/(P_(0))=x_(s)`
`=(n_(s))/(n_(0)+n_(s))`
`n_(s) lt lt n_(0)`
`:. (P_(0)-P_(s))/(P_(0))=(n_(s))/(n_(0))=(w)/(m)xx(M)/(W)`
Given vapour pressure reduced to `80%`, when non-volatile solute is dissolved in octane.
`P_(0)=1" atm"`
`P_(s)=0.8" atm"`
`W=114" gms"`
M = 114 gm/mole
w = ? , m = 40
`(P_(0)-P_(s))/(P_(0))=(n_(s))/(n_(0))=(w)/(m)xx(M)/(W)`
`(1-0.8)/(1)=(w)/(40)xx(114)/(114)`
w = 8 gms