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Calculate mole fraction of solute in an aqueous 4m solution assuming the density of the solution as 1.0g//mL |
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Answer» `:.` wt of the solution `=` density `XX` volume (mL) `=1.25xx1000=1250g` Wt. of `Na_(2)S_(2)O_(3)` present in 1L of the solution `=` molarity `xx` mol.wt. `=3xx158=474g` Wt `%` of `Na_(2)S_(2)O_(3)=(474)/(1250)xx100=37.92%` (ii) Wt. of SOLUTE `(Na_(2)S_(2)O_(3))=474g` Moles of solute `=(474)/(158)=3` Wt. of SOLVENT `(H_(2)O)=1250-474=776g` Moles of solvent `=(776)/(18)=43.11` `:.` mole fraction of `Na_(2)S_(2)O_(3)=(3)/(3+43.11)=0.063` (iii) Molality of `Na_(2)S_(2)O_(3)= ("moles of " Na_(2)S_(2)O_(3))/("wt.of solvent in grams")xx1000` `=(3)/(776)xx1000=3.865m` `:.1` mole of `Na_(2)S_(2)O_(3)` contains 2 moles of `Na^(+)` ions and 1 mole of `S_(2)O_(3)^(2-)` ions `:.` molality of `Na^(+)=2xx3.865=7.73m` Molality of `S_(2)O_(3)^(2-)=3.865m` |
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