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

Explanation

Let the vapour pressure of pure octane be `p_(1)^(0)`
`(80)/(100)p_(1)^(0)=0.8 p_(1)^(0)`
Molar mass of solute, `M_(2)=40 g mol^(-1)`
Mass of octane, `w_(1)=114 g`
Molar mass of octane, `(C_(3)H_(18))`,
`= M_(1)=8xx12+18xx1=114 g mol^(-1)`
Applying the reaction,
`(p_(1)^(0)-p_(1))/(p_(1)^(0))=(w_(2)xxM_(1))/(M_(2)xx w_(1))`
`therefore (p_(1)^(0)-0.8 p_(1)^(0))/(p_(1)^(0))=(w_(2)xx114)/(40xx114)`
`therefore (0.2 p_(1)^(0))/(p_(1)^(0))=(w_(2))/(40)`
`therefore w_(2)=8 g`