# How do I calculate the normality of NaBrO3?

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 $1.42g$)