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Calculate the boiling point of a solution containing 0.456 g of camphor (mol. Mass = 152) dissolved in 31.4 g of acetone (b.p. = 56.30^(@)C), if the molecular elevation constant per 100 g of acetone is 17.2^(@)C. |
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Answer» SOLUTION :Here, we have `w_(2)=0.456 g, M_(2)=152, w_(1)=31.4 g, T_(0)=56.30^(@)C, K_(b)=17.2^(@)C//100 g` `Delta T_(b)=(100 K_(b).w_(2))/(w_(1)M_(2))=(100xx17.2xx0.456)/(31.4xx152)=0.16^(@)C` `therefore` BOILING point of solution `(T_(b))=T_(b)^(@)+Delta T_(b)=56.30+0.16 = 56.46^(@)C` |
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