JEE Main & Advanced Sample Paper JEE Main Sample Paper-19

  • question_answer \[0.5\text{ }g\] 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 \[0.15N\]. \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\] 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 :

     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 \[3.16\] is Eq. wt. of \[KMn{{O}_{4}}\]. On solving, we get \[x=0.073\text{ }g\] % age of \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}=\frac{0.0732\times 100}{0.5}=14.64%\]


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