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NEET Chemistry Analytical Chemistry Question Bank Self Evaluation Test - Analytical Chemistry

  • question_answer
    0.5 g mixture of \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] and \[KMn{{O}_{4}}\] was treated with excess of KI in acidic medium. \[{{I}_{2}}\]liberated required \[100c{{m}^{3}}\] of 0.15 N \[N{{a}_{2}}{{S}_{2}}{{\text{O}}_{3}}\] solution for titration. The percentage amount of  \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] in the mixture is

    A) 85.36%            

    B) 14.64%

    C) 58.63%            

    D) 26.14%

    Correct Answer: B

    Solution :

    [b] Let the amount of the \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] in the mixture be x g, then amount of \[KMn{{O}_{4}}\] will be (0.5-x)g \[\therefore \left( \frac{x}{49}+\frac{0.5-x}{31.6} \right)=\frac{100\times 0.15}{1000}\] Where 49 is Eq. wt. of \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] and 31.6 is Eq. wt. of \[KMn{{O}_{4}}.\] On solving, we get x = 0.073 g Percentage of \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] \[=\frac{0.0732\times 100}{0.5}=14.64\]%


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