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Calculate the molality of `K_(2)CO_(3)` solution which is formed by dissolving `2.5g` of it in one litre of solution. Density of solution is `0.85g ml^(-1)`. (At. wt. of `K=39,C=12,O=16)` |
Answer» Given Mass of solute `(K_(2)CO_(3))=2.5g` Volume of solution `=1000ml` Density of solution `=0.85g ml^(-1)` Asked Molality of solution `(m)=?` Formulae: `m=(W_(B))/(M_(B))xx(1000)/(W_(A))` Explanation: `m=` molality of solution, `M_(B)=` Molecular weight solute `W_(A)=` Mass of solvent in grams Calculation: Mass of solute `=2.5g` Mass of solution`="volume"xx"density"=1000mlxx0.85g ml^(-1)=850g` Mass of solvent `=(850-2.5)847.5g`. Substitution in formulae: `m=(2.5)/(138)xx(1000)/(847.5)` `m=0.0213` |
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