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If the density of a 1-ltre solution of `98% H_(2)SO_(4) (wt.//vol.)` is `1.88 g mol^(-1)`, the molality of the solution will beA. `13.13`B. `10.10`C. `11.11D. `12.12` |
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Answer» Correct Answer - 3 `98% H_(2)SO_(4) (wt.//vol.)` implies that every `100 mL` of solution contains `98 g` of `H_(2)SO_(4)`. Thus `n_(H_(2)SO_(4))=(mass_(H_(2)SO_(4)))/(molar mass_(H_(2)SO_(4)))` `=(98 g)/(98 g//mol)=1 mol` Mass of solution =`(Volume)_("soln")xx("density")_("soln")` `=(100 mL)(1.88 g mL^(-1))` `=188 g` Mass of solvent =(Mass of solution) - (Mass of solute) `=(188 g) - (98 g)` `=90 g` Molality (m)`= n_(H_(2)SO_(4))/g_("solvent")xx(1000 g)/(kg)` `=(1 mol)/(90 g)xx(1000 g)/(kg)` `=11.11 m` |
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