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The boling point of benzone is 353.23K. When 1.80 g of a non-valatioe solute is mixed in 90 g of benzene, the boling point id raised to 354.11 Calculate the molar mass of the solute. Given that K_(b) benzene is 2.53 K kg mol^(-1) |
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Answer» Solution :`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))` `W_(B)=1.80 KG,k_(b)=2.53K Kg mol^(-1)` `DeltaT_(b)=(354.11-353.23)=0.88 K` `M_(B)=((1.80)XX(2.53 kg mol^(-1)))/((0.88)xx(0.090 kg))=9.5 g mol^(-1)` |
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