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A solution prepared by dissolving 1.25 g...

A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 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)`

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The correct Answer is:
Mass of solute `(W_(B))=1.25g`
Mass of solvent `(W_(A))=99g`
Elevation in boiling point `(DeltaTb)=8031-80.10^(@)C`
`K_(b)=2.53^(@)Ckg "mol"^(-1)`
Now, `DeltaT_(b)=K_(b)(W_(b)xx100)/(W_(A)xxM_(B))`
`0.21=(2.53xx1.25xx1000)/(99xxM_(B))` or `M_(B)=(2.53xx1.25xx1000)/(99xx0.21)` ir `M_(B)=152.116g`
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