Find the mass of $\ce{NaBrO3}$ required to prepare $\pu{150 ml}$ of $\pu{0.75 N}$ solution based on the reaction $$\ce{BrO3- + 6 H+ + 6e- -> Br- + 3 H2O}$$

My analysis:

$\pu{150 ml}$ solution of $\pu{0.75 N}$ means $0.1125$ equivalents of $\ce{NaBrO3}$ are present in the solution. Since, according to the given equation, n-factor (of equivalence) of $\ce{NaBrO3}$ is $6$. So $0.1125$ equivalents means $0.01875$ moles of $\ce{NaBrO3}$. The molecular mass of $\ce{NaBrO3}$ is $151$. Thus, I get the mass as $\pu{2.83 g}$ approximately. Am I correct?

Since the book says the answer should be $\pu{1.42 g}$


1 Answer 1


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...).

  • 3
    $\begingroup$ 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. $\endgroup$ Nov 17, 2013 at 16:52
  • $\begingroup$ @NicolauSakerNeto yes I think is quite useless... $\endgroup$
    – G M
    Nov 17, 2013 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, 2013 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, 2013 at 20:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.