A solution prepared by dissolving `1.25g` of oil of winter green (methyl sallicylate) in `99.0g` of benzene has a boiling point of `80.31^(@)C`. Determine the molar mass of this compound. (`B.P.` of pure benzene `=80.10^(@)C` and `K_(b)` for benzene `=2.53^(@)C kg "mol".1`)

Explanation

Here` w_(2)=1.25 g, w_1= 99.0 g, M_(2)78.08g mol_(-1)`
` Delta` ` t_(b)=(80.31-80.10) ^(@)C=0.21 ^(@)C or0.21 K `
`therefore` `Delta``t_(b)=K_(b)m=K_(b)xx(w_(2)xx1000)/(M_(2)xxw_(1))`
`M_(2)=k_(bxxw_(2)xx1000)/(Delta T_(b)xxw_(1))=(2.53xx1.25xx1000)/(0.21xx99)`
`(3162.5)/20.79= `M_(2)(3162.5)/20.79= 152.11approx152 g" "mol^(-1)`