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JEE Main & Advanced Sample Paper JEE Main Sample Paper-17

  • question_answer
    What is the mass of sodium bromate solution necessary to prepare \[\text{85}\text{.5}\,\text{c}{{\text{m}}^{3}}\]of \[\text{0}\text{.672 N}\]solution when the half-cell reaction is \[\text{Br}{{\text{O}}^{-}}_{3}+6{{H}^{+}}+6e\to B{{r}^{-}}+3{{H}_{2}}O?\]

    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 \[\left( \frac{0.112}{1000}\times 85.5 \right)\times 151=1.446\,gm\]


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