1.

Calculate the molality of sulphuric acid solution acid solution with mole fraction of 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"`.
`THEREFORE"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_(B)+n_(B))=0.15orn_(B)/(n_(B)+55.55)=0.15`
`n_(B)=0.15""n_(B)+55.5xx0.15orn_(B)=(55.5xx0.15)/(0.85)=9.8`.


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