1.

Molarity of H_(2)SO_(4) is 0.8 M and its density os 1.06 g//cm^(3). What will be the concentration in terms of molality and molar fraction?

Answer»


Solution :`"Molarity of "H_(2)SO_(4)=0.8 M = 0.8" MOL L"^(-1)`
`"Mass of "H_(2)SO_(4)="No. of MOLES"xx"Moles mass"`
`=(0.8"mol")xx(98.0" G mol"^(-1))=78.4 g`
`"Mass of 1 L solution"=(1000cm^(3))xx(1.06" g cm"^(-3))=1060 g`
`"Mass of water in solution"=(1060-78.4)=981.6 g= 0.9816 kg`
`"Molality of solution (m)"=("No. of moles of "H_(2)SO_(4))/("Mass of water in kg")=((0.8"mol"))/((0.9816" kg"))`
`=0.815" mol kg"^(-1)=0.815 m`
`"Mole fraction of"H_(2)SO_(4)=n_(H_(2)SO_(4))/(n_(H_(2)SO_(4))+n_(H_(2)O))`
`=((0.8mol))/((0.8mol)+((98.6g))/(18" g mol"^(-1)))=0.8/(0.8+54.5)=0.8/55.3=0.014`


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