Calculate the molarity of sulphuric acid solution in which mole fracti

1.	Calculate the molarity of sulphuric acid solution in which mole fraction water is 0.85.
Answer» Solution :Mole fraction of water = 0.85, Mole fraction of `H_(2)SO_(4)=1 -0.85 = 0.15` Let `n_(B)` moles of `H_(2)SO_(4)` be dissolved in 1000 g of water to represent the molality of the solution `:.` No. of moles of water `(n_(A))=((1000G))/(("18 g mol"^(-1)))=55.55` mol No. of moles of `H_(2)SO_(4)=n_(B)` `(n_(B))/(n_(A)+n_(B))=0.15or(n_(B))/(n_(B)+55.55)=0.15` `n_(B)=0.15 n_(B)+55.5xx0.15or` `n_(B)=(55.5xx0.15)/(0.85)=9.8`.

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