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Calculate the molefraction of each component in a solution made by mixing 52.0 g CH_(3)COOH (acetic acid) and 100 g water. |
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Answer» Solution :`n_(1)=(w_(1))/(M_(1))=(100)/(18)` mol for water, M (water)=18 G `mol^(-1)` `n_(2)=(w_(2))/(M_(1))=(52.0g)/(60" g "mol^(-1))=(13)/(15)mol`(CHECK that `x_(1)+x_(2)=1`) `therefore x_(2)=(n_(2))/(n_(1)+n_(2))=((13)/(15))/((13)/(15)+((100)/(18)))=0.135 and x_(1)=1-x_(2)=1-0.135=0.865` |
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