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 the vapour pressure of solvent be `p^0`.
Then, vapour pressure of solution = `0.80p^0`
Let the mass of solute = w g
Number of moles of the solute ` = (w)/(40)`
Number of moles of solvent (octane) =` (114 g)/(114 g "mol"^(-1))` = 1 mol [ molar mass of `C_8H_18 =114 g "mol"^(-1)` ]
Applying Raoult.s law and substituting the values in the equation, we have
`(p^0 -p_s)/(p^0) = x_2`
` (p^0 - 0.80 (p^0)/(p^0)) = (w/40)/(w // 40 +1)`
`0.2 ((w)/(40) + 1) = (w)/(40) rArr (0.8 w)/(40) = 0.2`
` w = 10 g `