Given that 1 3 _m ^ ( F e^ 3 + ) = 68 ,oh m^ - 1 ,c m^ - 1 ,e q^ … - Vidyadip

MCQ

Given that $\frac{1}{3}\mathop {{\lambda _m}}\limits^\infty \left( {F{e^{3 + }}} \right) = 68\,oh{m^{ - 1}}\,c{m^{ - 1}}\,e{q^{ - 1}}$ and $\frac{1}{2}\mathop {{\lambda _m}}\limits^\infty \left( {SO_4^{2 - }} \right) = 80\,oh{m^{ - 1}}\,c{m^{ - 1}}\,e{q^{ - 1}}$ What will be value of $\mathop {{\lambda _{eq}}}\limits^\infty \left( {F{e_2}{{\left( {S{O_4}} \right)}_3}} \right)$ ? ............ ${\rm{oh}}{{\rm{m}}^{ - 1}}{\mkern 1mu} {\rm{c}}{{\rm{m}}^2}$ $\mathrm{eq}^{-1}$

✓
$148$
B
$880$
C
$364$
D
$130$

✓

Answer

Correct option: A.

$148$

a
${{\overset{\infty }{\mathop{\lambda }}\,}_{\text{eg}{{\left( A{{l}_{2}}\text{(S}{{\text{O}}_{4}}) \right)}_{3}}}}=\frac{1}{3}\Lambda _{\text{M}}^{\infty }$ $\left( \text{A}{{\text{l}}^{+3}} \right)+\frac{1}{2}\Lambda _{\text{M}}^{\infty }\left( \text{SO}_{4}^{2-} \right)$

$=68+80$

$=148\, \Omega^{-1} \,\mathrm{cm}^{-1} \,\mathrm{eq}^{-1}$

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