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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» Correct Answer - 0.014 `"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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