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Calculate the molality of 1 litre solution of 93% H_(2)SO_(4) (weight/volume). The density of the solution is "1.84 g mL"^(-1) |
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Answer» SOLUTION :`93%H_(2)SO_(4)(w//v)=93gH_(2)SO_(4)` in `100CM^(3)` of the solution = 93 g in 184 g of the solution `therefore" SOLVENT (WATER )"=184-93=91g=0.091kg" ,Molality"=(93//98mol)/(0.091kg)=10.43m.` |
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