Calculate the molality of potassium carbonate solution which is formed by dissolving 2.5 gm of it in one litre solution. (density of solution of 0.85 g/mL).

Explanation

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`