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JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    Excess of NaOH (aq) was added to 100 mL of \[FeC{{l}_{3}}\](aq) resulting into 2.14 g of \[Fe{{(OH)}_{3}}.\] The molarity of \[FeC{{l}_{3}}\](aq) is : (Given molar mass of \[Fe=56gmo{{l}^{-1}}\]and molar mass of \[Cl=35.5gmo{{l}^{-1}}\]) [JEE Online 08-04-2017]

    A)  0.3 M                   

    B)  0.2 M

    C)  0.6 M                   

    D)  1.8 M

    Correct Answer: B

    Solution :

    \[\underset{\begin{smallmatrix}  100ml \\  m=? \end{smallmatrix}}{\mathop{3NaOH+Fec{{l}_{3}}}}\,\to \underset{2.14gm}{\mathop{Fe{{(CH)}_{3}}+NaCl}}\,\] Moles of \[Fe(C{{H}_{3}})=\frac{2.14}{107}=2\times {{10}^{-2}}mol\] moles \[FeC{{l}_{3}}=2\times {{10}^{-2}}mol\] \[M=\frac{2\times {{10}^{-2}}}{100}\times 1000=0.2M\]


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