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

Explanation

Here, `p = 80%` of `p_(1)^(@) = 0.80 p_(1)^(@)`
Number of moles of solute `(n_(2)) = (W_(2))/(40) mol`
Number of moles of solvent (octane)
`(n_(1)) = (114 g)/(114 g mol^(-1)) = 1 mol`
(`therefore` As molar mass of `C_(8)H_(18) = 114 mol^(-1)`)
Now, `(p_(1)^(@) - p)/(p_(1)^(@)) = chi_(2) = (n_(2))/((n_(1) + n_(2)))`
`therefore (p_(1)^(@) - 0.80 p_(1)^(@))/(p_(1)^(@)) = ((W_(2)//40))/((W_(2)//40+1))`
`0.2((W_(2))/(40)+1) = (W_(2))/(40)` or `(0.8 W_(2))/(40) = 0.2`
or `W_(2) = 10 g`