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When `1L` of `0.1 M` sulphuric acid solution is allowed to react with `1 L` of `0.1 M` sodium hydroxide then the molarity of sodium sulphate formed is `(H_(2)SO_(4) + 2NaOH rarr Na_(2)SO_(4) + 2H_(2)O)`:A. `0.1 M`B. `0.05 M`C. `0.025 M`D. `0.2 M` |
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Answer» Correct Answer - C `NaOH` is `LR` `H_(2)SO_(4) + 2NaOH rarr Na_(2)SO_(4) + 2H_(2)O` `[Na_(2)SO_(4)] = (0.05)/(2) = 0.025 M` |
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