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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 44

  • question_answer
    For pseudo alum, the ionic equivalent and ionic molar conductivities of their ions are given. \[\lambda {{{}^\circ }_{eq}}(S{{O}_{4}}^{2-})'\lambda {{{}^\circ }_{eq}}(F{{e}^{2+}})\] and \[\lambda {{{}^\circ }_{m}}(A{{l}^{3+}})\] are x, y and z respectively: \[\lambda {{{}^\circ }_{m}}(S{{O}_{4}}^{2-})'\lambda {{{}^\circ }_{eq}}(F{{e}^{2+}})\]and \[\lambda {{{}^\circ }_{m}}(A{{l}^{3+}})\]are \[{{x}^{1}},{{y}^{1}}\] and \[{{z}^{1}}\] respectively For pseudo alum, the correct relation between \[\wedge {{{}^\circ }_{eq}}\] and \[\wedge {{{}^\circ }_{m}}\] is

    A) \[\left( x+\frac{y}{4}+\frac{3z}{4} \right)=\left( \frac{{{x}^{1}}+{{y}^{1}}+{{z}^{1}}}{2} \right)\]

    B)   \[\left( x+\frac{y}{2}+\frac{z}{2} \right)=\left( \frac{4{{x}^{1}}+{{y}^{1}}+2{{z}^{1}}}{8} \right)\]

    C) \[\left( x+\frac{y}{4}+\frac{3z}{4} \right)=\left( \frac{4{{x}^{1}}+{{y}^{1}}+2{{z}^{1}}}{8} \right)\]

    D) \[\left( x+\frac{y}{2}+\frac{z}{2} \right)=\left( \frac{4{{x}^{1}}+{{y}^{1}}+2{{z}^{1}}}{8} \right)\]

    Correct Answer: C

    Solution :

    [c] Pseudo alum \[(\overset{+2}{\mathop{Fe}}\,\overset{-2}{\mathop{S{{O}_{4}}}}\,.\overset{+2\times 3}{\mathop{A{{l}_{2}}(}}\,\overset{-2\times 3}{\mathop{S{{O}_{4}}{{)}_{3}}.}}\,24{{H}_{2}}O)\] Total charge = (\[+8\] or \[-8\]) Eq. of \[F{{e}^{+2}}=\frac{+2}{+8}=\frac{1}{4}Eq.=\left( \frac{y}{4} \right)\] Eq. of \[A{{l}^{+3}}=\frac{+2\times 3}{+8}=\frac{3}{4}\,Eq.\left( \frac{3z}{4} \right)\] Eq. of \[S{{O}_{4}}^{2-}=\frac{-8}{-8}=1\,\,Eq.(x)\] \[{{\overset{{}^\circ }{\mathop{\Lambda }}\,}_{eq}}=\left( x+\frac{y}{4}+\frac{3z}{4} \right)\] \[FeS{{O}_{4}}.A{{l}_{2}}{{(S{{O}_{4}})}_{3}}.24{{H}_{2}}O\to F{{e}^{+2}}+2A{{l}^{+3}}+4S{{O}_{4}}^{2-}+24{{H}_{2}}O\]             \[=(y'+2z'+4x')\]                         \[{{\overset{{}^\circ }{\mathop{\Lambda }}\,}_{m}}=(4x'+y'+2z')\] Comparing the value of \[{{\overset{{}^\circ }{\mathop{\Lambda }}\,}_{eq}}\] and \[{{\overset{{}^\circ }{\mathop{\Lambda }}\,}_{m}},\]             \[\left( x+\frac{y}{4}+\frac{3z}{4} \right)=\left( \frac{4x'+y'+2z'}{8} \right)\]


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