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The molality of a sulphuric acis solution in which the mole fraction of water is 0.86 is |
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Answer» Solution :Mole FRACTION of `H_(2)SO_(4)` in the solution `=1-0.86=0.14` `(n_(2))/(n_(1)+n_(2))=0.14` MOLALITY is `n_(2)` in 1000 g of water, i.e., in `n_(1)` `=1000//18=55.55` moles `THEREFORE (n_(2))/(55.55+n_(2))=0.14"or"n_(2)=0.14n_(2)+7.777` `"or"0.86n_(2)=7.777 or n_(2)=9.0` |
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