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0.7g of `(NH_(4))_(2)SO_(4)` sample was boiled with 100mL of 0.2 N NaOH solution was diluted to 250 ml. 25mL of this solution was neutralised using 10mL of a 0.1 N `H_(2)SO_(4)` solution. The percentage purity of the `(NH_(4))_(2)SO_(4)`sample is :A. 94.3B. 50.8C. 47.4D. 79.8 |
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Answer» Correct Answer - A m.eq of `(NH_(4))_(2)SO_(4)+` m.eq of `H_(2)SO_(4)`=m.eq of NaOH `("m.molea"xx2)++(0.1xx10xx(250)/(25))=0.2xx100` `:. " m.mole of " (NH_(4))_(2)SO_(4)=5` wt. of `(NH_(4))_(2)SO_(4)=(5)/(1000)xx132=0.66 g` `:. " % of " (NH_(4))_(2)SO_(4)=(0.66)/(0.7)xx100=94.28% ~~94.3%` |
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