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On dissolving 3.24 g of sulphur in 40 g of benzene, the boiling point of the solution was higher than sulphur? (K_(b) for benzene = 2.53 K kg mol^(-1) , atomic mass of sulphur = 32 g mol^(-1)). |
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Answer» Solution :The given VALUES are: `W_(B) = 3.24 g`, `W_(A) = 40 g` `DeltaT_(b) = 0.81 K`, `K_(b) = 2.53 KG mol^(-1)` Using formula, `Mw_(B) = (K_(b) xx 1000 xx W_(B)) /(DeltaT_(b) xx W_(A))` On substituting all the values, we get `:. Mw_(B) = (2.53 xx 1000xx 3.24) / (0.81 xx 40) = 253` Let the molecular formula of sulphur = `S_(x)` Atomic mass of sulphur = `32` Molecular mass = `32 xx x` `:. 32X = 253` `x= (253)/(32) = 7.91 ~~ 8` `:.` Molecular formula of sulphur `= S_(8)` |
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