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Molarity of H(2)SO(4) is 0.1 M and its d...

Molarity of `H_(2)SO_(4)` is 0.1 M and its density is `1.06 g cm^(-3)`. What will be concentration of the solution in terms of molarity and mole fraction ?

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Mass of one litre of solution `= V xx d = 1000 "cm"^(3) xx 1.06 "g cm"^(-3) = 1060 g`
Mass of `H_(2)SO_(4)` in 0.8 M solution `= (98 g)xx((0.8M))/((1.0M))=78.4g`
Mass of solvent `(H_(2)O)=(1069-78.4)=981.6g`
Molality of solution `(m)=("Mass of "H_(2)SO_(4)//"Molar mass")/("Mass of solvent in kg")`
`=((78.4g)//(98"g mol"^(-1)))/(981.6//1000kg)=0.815"mol kg"^(-1)=0.815` m
Mole fraction of solute `= ("No. of moles of "H_(2)SO_(4))/("No. of moles of "H_(2)SO_(4)+" No. of moles of water")`
`=((0.8 "mol"))/((0.8+981.6//18)"mol")=((0.8 "mol"))/((0.8+54.53)"mol")=0.0145`
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