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%.

Explanation

Let `P^0 = 100`,
Given `P_S= 80, n_B = ? W_B = ?`
`n_A=(W_A)/(M_A)=(114g)/(114"g mol"^(-1))= 1` mol
[ `M_A` = Molar mass of octane, `C_8H_(18)=114` ]
`:. (P^0-P_S)/(P^0)=(n_B)/(n_B+n_A)` [Relative lowering in Vapour Pressure equation]
`(100-80)/100=n_B/(n_B+1mol)`
`(n_B+1mol)/(n_B)=5`
`1+(1mol)/(n_B)=5 implies n_B =1/4 mol`
But number of moles of solute
`n_B = W_B/M_B`
`W_B = n_B xxM_B`
`=1/4 "mol"xx40 "g mol"^(-1)`
`W_B = 10 g`