Question

What is the weight of sodium bromine and molarity of solution necessary to prepare 85.5 ml of 0.672 N solution when tha half-cell reaction is
BrO^​-₃ + 6H⁺ + 6e ----> Br ^- + 3H₂​O

Answer 1

$$\begin{gathered} \text{Dear Student,} \\ \mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{NaBrO}}_3=151\mathrm{g}/\mathrm{mol} \\ \mathrm{Equivalent}\mathrm{mass}\mathrm{of}{\mathrm{NaBrO}}_3=\frac{\mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{NaBrO}}_3}{6} \\ {\mathrm{N}}_{{\mathrm{NaBrO}}_3}=6{\mathrm{M}}_{{\mathrm{NaBrO}}_3} \\ {\mathrm{M}}_{{\mathrm{NaBrO}}_3}=\frac{{\mathrm{N}}_{{\mathrm{NaBrO}}_3}}{6}=\frac{0.672}{6}=0.112\mathrm{M} \\ \mathrm{Moles}\mathrm{of}{\mathrm{NaBrO}}_3\mathrm{in}85.5\mathrm{ml}\mathrm{of}0.112\mathrm{M}\mathrm{solution}=\mathrm{Molarity}\times \mathrm{Volume}\mathrm{in}\mathrm{liters}=0.112\mathrm{M}\times \frac{85.5}{1000}=0.00957 \\ \mathrm{Weight}\mathrm{of}{\mathrm{NaBrO}}_3=\mathrm{moles}\times \mathrm{Molecular}\mathrm{mass}=0.00957\times 151=1.446\mathrm{g} \end{gathered}$$