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KVPY Sample Paper KVPY Stream-SX Model Paper-31

  • question_answer
    For the decomposition of \[{{H}_{2}}{{O}_{2}}\] (aq) it was found that \[{{V}_{{{O}_{2}}}}\] (t = 15 min.) was 100 mL (at \[0{}^\circ C\]and 1 atm) while \[{{V}_{{{O}_{2}}}}\] (maximum) was 200 mL (at \[0{}^\circ C\] and 2 atm) If the same reaction had been followed by the titration method and if \[V_{KMn{{O}_{4}}}^{(CM)}\] (t = 0) had been 40 mL, what would \[V_{KMn{{O}_{4}}}^{(cM)}\](t = 15 min) have been?

    A) 30 mL

    B) 25 mL

    C) 20 mL

    D) 15 mL

    Correct Answer: A

    Solution :

    \[\frac{1}{4}th\] reaction has completed upto 15 min. Hence \[{{V}_{KMn{{O}_{4}}}}\] will be \[\frac{3}{4}\times 40=30\,mL\]


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