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

Explanation

Let the vapour pressure of pure octane = `p_0`
Vapour pressure of the solution = `0.8p_0`
Lowering of vapour pressure = `0.2p_0`
Relative lowering ` = (0.2p_0)/(p_0)`
According to Raoult.s law, Relative lowering of vapour pressure of a solution is equal to the mole fraction of the solute.
Let the mass of solute be = w g
Its molecular mass is given as = 40 u
No. of moles of the solute = `(w)/(40)`
No. of moles of the solvent ` = 114/114 = 1` [ mol. mass of octane = 114]
Mole fraction of the solute ` = (w//40)/(1 + w//40)`
According to Raoult.s law,
`(0.2p_0)/(p_0) = (w/40)/(1 + w//40) ` on simplification
`0.2 = (w//40)/((40 + w)/(40)) `
`0.2 = (w)/(40 + w)`
or 0.2(40 + w) = w or 8 + 0.2w - w = 0
or - 0.8w = -8 or w = 10 g