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JEE Main & Advanced Sample Paper JEE Main Sample Paper-6

  • question_answer
    5.39 g of a mixture of \[FeS{{O}_{4}}.7{{H}_{2}}O\]and anhydrous ferric sulphate requires 80 mL of 0.125 N permanganate solution for complete conversion to the ferric sulphate. The individual weight of ferric sulphate in the original mixture is

    A)  2.61 g                                  

    B)  5.22 g

    C)  1.305 g                                

    D)  2.78 g

    Correct Answer: A

    Solution :

    Equivalent weight of \[FeS{{O}_{4}}.7{{H}_{2}}O=Mol.wt.=278\] 80 mL 0.125 (N) permanganate solution = (80 x 0.125) N solution = meq. of \[FeS{{O}_{4}}.7{{H}_{2}}O\] =10 Weight\[=\frac{10\times 278}{1000}=2.78g\] Weight of anhydrous \[F{{e}_{2}}{{(S{{O}_{4}})}_{3}}\] \[=5.39-2.78=2.61g\]


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