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 \[{{\overset{{}^\circ }{\mathop{\Lambda }}\,}_{m}},\] \[\left( x+\frac{y}{4}+\frac{3z}{4} \right)=\left( \frac{4x'+y'+2z'}{8} \right)\]You need to login to perform this action.
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