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Find the mass of $\ce{NaBrO_3}$ required to prepare $150 \text{ml}$ of $0.75N$ solution based on the reaction $$\ce{BrO_3^- +6H^+ +6e^- \bond{->}Br^- +3H_2O}$$

My analysis: $150 \text{ml}$ solution of $0.75N$ means $0.1125$ equivalents of $\ce{NaBrO_3}$ are present in the solution. Since, according to the given equation, n-factor (of equivalence) of $\ce{NaBrO_3}$ is $6$. So $0.1125$ equivalents means $0.01875$ moles of $\ce{NaBrO_3}$. The molecular mass of $\ce{NaBrO_3}$ is $151$. Thus, I get the mass as $2.83g$ approximately. Am I correct? (Because the book says the answer should be $1.42g$)

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To me your answer is correct. It seems that the book result have a '2' dividing the final result. Could you post all the text of the book exercise? Two possibilities, the result from the book is wrong (could be) or there is a mistake in one of the values (75 ml instead 150, 0.375 N instead 0.75 N...).

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    $\begingroup$ The lesson to learn here is, of course: don't use normality. It's based on gram-equvialents, an old ambiguous concept from the 1800s when we still figuring out substance formulae. It's about time to bury it. $\endgroup$ – Nicolau Saker Neto Nov 17 '13 at 16:52
  • $\begingroup$ @NicolauSakerNeto yes I think is quite useless... $\endgroup$ – G M Nov 17 '13 at 16:55
  • $\begingroup$ The question is exactly as given here. So, probably, the result given in the book is wrong. $\endgroup$ – Tejas Nov 18 '13 at 13:34
  • $\begingroup$ @TejasAdsul I've try to solve the problem I have your same result...I think Rauru is right $\endgroup$ – G M Nov 18 '13 at 20:11

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