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Calcuate the molality of a sulphuric acid solution in which the mole fraction of water is 0.85. |
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Answer» Solution :Mole fraction of water = 0.85 Mole fraction of `H_(2)SO_(4)` in the solution `=1-0.85=0.15,"i.e., "(n_(2))/(n_(1)+n_(2))=0.15` where `n_(2)` is the number of mole of `H_(2)SO_(4)` and `n_(1)` is the number of moles of `H_(2)O` in the solution. Molality of `H_(2)SO_(4)` solution means the number of moles of `H_(2)SO_(4)` present in 1000 g of `H_(2)O`. Thus, we have to FIND `n_(2)` when `w_(1)=1000`, i.e., `n_(1)=(1000)/(18)=55.55" moles."` Substituting the value of `n_(1)` in equation (i), we GET `(n_(2))/(55.55+n_(2))=0.15"or"n_(2)=0.15n_(2)+8.3325 or 0.85n_(2)=8.3325` `n_(2)=(8.3325)/(0.85)="9.8 moles, i.e, Molality"= 9.8m.` Alternatively, `(n_(1))/(n_(1)+n_(2))=0.85"...(i)"THEREFORE""(n_(2))/(n_(1)+n_(2))=1-0.85=0.15"...(II)"` Dividing (ii) by (i), we get `(n_(2))/(n_(1))=(0.15)/(0.85)or(n_(2))/(1000//18)=(0.15)/(0.85)orn_(2)=(0.15)/(0.85)XX(1000)/(18)=9.8" moles"` `"Hence,molality= 9.8 m"` |
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