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Calculate the molality of a 1-litre solution of 93% H_(2)SO_(4) (wt./vol). The density of the solution is 1.84 g.mL |
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Answer» Solution :The solution is `1000mL` containing `930g` of `H_(2)SO_(4)` The weight of the solution will be `1840g`. The weight of the SOLVENT `(H_(2)O)` will therefore be `(1840-930)` i.e.910g `:.` molality `=("mole of" H_(2)SO_(4))/("wt.of"H_(2)O(G))xx1000` `=(930//98)/(910)xx1000=10.428m` |
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