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When 1.0 g of a non-volatie solute was d...

When 1.0 g of a non-volatie solute was dissolved in 10 g benzene, elevation in boiling point temperature was found to be `1^(@)C`. Calculate the molar mass of the dissoved substance. `(K_(b) "for benzene = 2.53 K kg" mol^(-1))`.

Text Solution

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The correct Answer is:
253 h `mol^(-1)`
`W_(B)=1.0g, W_(A)=10g=0.01 kg`
`DeltaT_(b)=1^(@)C or 1 K, K_(b)=2.53" K kg mol"^(-1)`
`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((2.53" K kg mol"^(-1))xx(1.0 g))/((1 K)xx(0.01" kg"))=253" g mol"^(-1)`
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The boiling point of benzene is 353 .23 K. When 1.80 gram of non- volatile solute was dissolved in 90 gram of benzene , the boiling point is raised to 354.11 K. Calculate the molar mass of solute . [K_(b) for benzene = 2.53 K kg mol^(-1)]