A) \[1.146\,g\]
B) \[1.246\,g\]
C) \[1.346\,g\]
D) \[1.446\,g\]
Correct Answer: D
Solution :
Since 6e are involved in the reaction, we have \[\text{Molarity = }\frac{\text{Normality}}{\text{6}}=\frac{0.672}{6}=0.112\,M\] Molar mass of \[NaBr{{O}_{3}}=151\,\text{g}\,\text{mo}{{\text{l}}^{-1}}\] Mass of\[\text{NaBr}{{\text{O}}_{\text{2}}}\]needed to prepare the required solution \[\left( \frac{0.112}{1000}\times 85.5 \right)\times 151=1.446\,gm\]You need to login to perform this action.
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