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What would be the molar mas of a compound if 6.21 g of it dissolved in 24.0 g of chloroform a solution that has a boiling point of `68.04^(@)C`? Given that the boiling point of pure chloroform is `61.7^(@)C` and `K_(b) for chloroform= 3.63^(@)C/m`. |
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Answer» `M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))` `K_(b)=3.63^(@)C//m=3.63 K kg mol^(-1),W_(B)=6.21g,W_(A)=24.0g=0.024 kg,` `DeltaT_(b)=(68.04-61.7)^(@)C=6.34K` `M_(B)=((3.63K kg mol^(-1))xx(6.21))/((6.34 K)xx(0.024 kg))=148.15 g mol^(-1)` |
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