JEE Main & Advanced Sample Paper JEE Main Sample Paper-1

  • question_answer
    Directions: Questions No. 64 are based on the following paragraph. When ammonium vanadate is heated with oxalic add solution, a compound Z is formed. A sample of Z was titrated with\[KMn{{O}_{4}}\] solution in hot acidic solution. The resulting liquid was reduced with \[S{{O}_{2}}\], the excess \[S{{O}_{2}}\]boiled off, and the liquid again titrated with \[KMn{{O}_{4}}\]. The ratio of the volumes of \[KMn{{O}_{4}}\] used in the two titrations was 5 : 1. \[KMn{{O}_{4}}\] oxidises all oxidation state of vanadium to Vanadium (+V) and \[S{{O}_{2}}\] reduces vanadium (+V) to vanadium (+IV). Read the above experiment and answer the following questions. Consider following redox reaction\[VO_{3}^{2-}+MnO_{4}^{-}\to M{{n}^{2+}}+VO_{4}^{3-}\]1 mole of\[VO_{3}^{2-}\] is oxidised by x mole of\[MnO_{4}^{-}\] Thus, x is

    A)  0. 2                                       

    B)  0. 4

    C)  0. 8                                       

    D)  1. 0

    Correct Answer: A

    Solution :

     \[\underset{+4}{\mathop{VO_{3}^{2-}}}\,\xrightarrow[{}]{{}}\underset{+5}{\mathop{VO_{4}^{3-}}}\,\underset{\text{1}\,\text{unit}}{\mathop{Change}}\,\] \[\underset{+7}{\mathop{MnO_{4}^{-}}}\,\xrightarrow[{}]{{}}\underset{+2}{\mathop{Mn_{{}}^{2+}}}\,5\,\text{unit}\] \[\therefore \]  \[5VO_{3}^{2}\equiv 1MnO_{4}^{-}\] \[1\,VO_{3}^{2-}\,\equiv \,\frac{1}{5}\,MnO_{4}^{-}\,=0.2\]


You need to login to perform this action.
You will be redirected in 3 sec spinner