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2.5 g of the carbonate of a metal war treated with 100 ml of `1 N H_(2)SO_(4)`. After the completion of the reaction, the solution was boiled off to expel `CO_(2)` and was then titrated against 1 N NaOH solution. The volume of alkali that would be consumed, if the equivalent weight of the metal is 20.A. (a)`50`B. (b)`25`C. (c )`75`D. (d)`100` |
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Answer» Correct Answer - d Equivalent weight of metal carbonate `=20+30=50` `2.5 g` of metal carbonate `=2.5/50=0.05 eq.` Number of equivalent of `H_(2)SO_(4)` would have reached `=0.05` Number of equivalent of `H_(2)SO_(4)` taken `=(100xx1)/1000=0.1` `:.` Number of equivalent of `H_(2)SO_(4)` remains unreached `=0.1-0.05=0.05 eq`. `:.` Number of equivalent of alkali consumed `=0.05 eq.` Milli eq. =Normality `xx` Volume in mL. `:. 1.0xxV=0.05xx1000` `V=(0.05xx1000)/1.0=50 mL` |
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