A piece of `Al` wieghing `2.7 g` is titrated with `75.0 mL` of `H_(2) SO_(4)` (specific gravity `1.8 mL^(-1)` and 24.7% `H_(2) SO_(4)` by weight). After the metal is completely dissolved, the solution is diluted to `400 mL`. Calculate the molarity of free `H_(2) SO_(4)` solution.
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Mass of `H_(2)SO_(4)=(24.7)/(100)xx75xx1.18` =21.8595 g Reaction : `underset(2xx27)(2Al)+underset(3xx98)(3H_(2)SO_(4)) to Al_(2)(SO_(4))_(3)+3h_(2)` `H_(2)SO_(4)` required for dissolving 2.7 g Al `=(3xx98)/(2xx27)xx2.7=14.7 g` `H_(2)SO_(4)` left unreacted =(21.895-14.7)g=7.1595 g 7.1595 g `H_(2)SO_(4)` is present in 400 mL Amt. of `H_(2)SO_(4)` present in one litre`=(7.1595)/(400)xx1000 g` =17.898 g No. of g moles of `H_(2)SO_(4)=(17.898)/(98)=0.1826` Hence, molartiy of `H_(2)SO_(4)=0.1826 M`
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