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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`) |
Answer» Correct Answer - 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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