1.

Calculate the molality of potassium carbonete solution formed by dissolving 2.5 g of it in one litre of solution (density of solution=0.85 g mL^(-1)).

Answer»


Solution :`"MASS of 1000 ML (one litre ) of solution"=Vxxd=(1000mL)xx(0.85"g mL"^(-1))=850g`
`"Mass of solvent (water)"=850g-2.5g)=847.5 g=0.8475 KG`
`"Molar mass of"K_(2)CO_(3)=2xx39+12+3xx16=138" g mol"^(-1)`
`"Molality of solution (m)"("Mass of" K_(2)CO_(3)//"Mola mass")/("Volume of solution in litres")`
`"Molality of solution (m)"("Mass of" K_(2)CO_(3)//"Mola mass")/("Mass of solvent in kg")=((2.6g)//138" g mol"^(-1))/((0.8475kg))=0.021 m`.


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