Ionic conductance at infinite dilution of Al3 + and SO42 - ions are 189ohm- 1cm2eq- 1 and 160ohm- 1cm2eq- 1 respectively. Calculate the λ eq ∞ of Al 2( SO 4)3:-

Question

Ionic conductance at infinite dilution of Al3 + and SO42 - ions are 189ohm- 1cm2eq- 1 and 160ohm- 1cm2eq- 1 respectively. Calculate the λ eq ∞ of Al 2( SO 4)3:- (A) 349ohm- 1cm2eq- 1 (B) 143ohm- 1cm2eq- 1 (C) 720ohm- 1cm2mole- 1 (D) 120ohm- 1cm2eq- 1

Tardigrade Admin · Accepted Answer

Lambda eq ∞ Al 2( SO 4)3=∧ eq ∞( Al +3)+λ eq ∞( SO 4-2) 189+160 =349ohm- 1cm2eq- 1

Q. Ionic conductance at infinite dilution of $A l^{3 +}$ and $S O_{4}^{2 -}$ ions are $189 o h m^{- 1} c m^{2} e q^{- 1}$ and $160 o h m^{- 1} c m^{2} e q^{- 1}$ respectively. Calculate the $λ_{e q}^{\infty}$ of $A l_{2} (S O_{4})_{3} : -$

$349 o h m^{- 1} c m^{2} e q^{- 1}$

$143 o h m^{- 1} c m^{2} e q^{- 1}$

$720 o h m^{- 1} c m^{2} m o l e^{- 1}$

$120 o h m^{- 1} c m^{2} e q^{- 1}$

$Λ_{e q}^{\infty} A l_{2} (S O_{4})_{3} = \land_{e q}^{\infty} (A l^{+ 3}) + λ_{e q}^{\infty} (S O_{4}^{- 2})$
$189 + 160$
$= 349 o h m^{- 1} c m^{2} e q^{- 1}$

Q. Ionic conductance at infinite dilution of Al3+ and SO42−​ ions are 189ohm−1cm2eq−1 and 160ohm−1cm2eq−1 respectively. Calculate the λeq∞​ of Al2​(SO4​)3​:−

349ohm−1cm2eq−1

143ohm−1cm2eq−1

720ohm−1cm2mole−1

120ohm−1cm2eq−1