The vapour pressure of `C Cl_(4)" at "30^(@)C` is 143 mm of Hg.0.5 gm of a nonvolatile non electrolyte substance with molar mass 65 is dissolved in 100 ml of `C Cl_(4)`. What will be the vapour pressure of the solution. Density of `C Cl_(4)" at "30^(@)C = "1.58 gm per cc."`

Explanation

Vapour pressure of pure solvent, `P^(@)=143`mm of Hg.
Vapour pressure of solution, `p = ?`
Weight of solute, `W_(2) = 0.5 gm`
Mol. wt of solute, `M_(2) = 65`
Mol. wt of solvent `(C Cl_(4)), M_(1) = 154`
Weight of solvent, `W_(1) = 100xx1.58 =158 gm`
`("mass" =" density"xx "volume")`
`(p^(@)-p)/(p^(@))=(W_(2))/(M_(2)).(M_(1))/(W_(1))" (By Raoult.s law)"`
`therefore (143-p)/(143)=(0.5)/(65)xx(154)/(158)`
`therefore" p = 141.93 mm of Hg."`