Find the value of Lambda eq ° for potash alum. λ°m(K+) =73.5 Ω-1 cm2 mol -1 λ°m (Al3+) = 189 Ω-1 cm2 mol-1 λ°m (SO2-4) = 160Ω-1 cm2 mol-1

Question

Find the value of Lambda eq ° for potash alum. λ°m(K+) =73.5 Ω-1 cm2 mol -1 λ°m (Al3+) = 189 Ω-1 cm2 mol-1 λ°m (SO2-4) = 160Ω-1 cm2 mol-1 (A) 145.6 Ω-1 cm2 eq-1 (B) 1165 Ω-1cm2 eq-1 (C) 532 Ω-1 cm2 eq-1 (D) 195.5 Ω-1cm2 eq-1

Tardigrade Admin · Accepted Answer

K2SO4 ⋅ Al2 (SO4)3 ⋅ 24H2O ⇒ 2K+(aq) +2Al3+(aq) +4SO2-4(aq) ∧°m (potash alum) = 2λ°m (K+) +2λ°m (Al3+) +4λ°m (SO2-4) = 2 ×73.5 +2 ×189 +4 ×160 = 1165 m-1 cm2 mol-1 n for potash alum = 8 (total +ve charge) ∧°eq = (∧°m/n) = (1165/8) = 145.6Ω-1 cm2 eq-1

Q. Find the value of $Λ_{e q}^{\circ}$ for potash alum.
$λ_{m}^{\circ} (K^{+}) = 73.5 Ω^{- 1} c m^{2} m o l^{- 1}$
$λ_{m}^{\circ} (A l^{3 +}) = 189 Ω^{- 1} c m^{2} m o l^{- 1}$
$λ_{m}^{\circ} (S O_{4}^{2 -}) = 160 Ω^{- 1} c m^{2} m o l^{- 1}$

$145.6 Ω^{- 1} c m^{2} e q^{- 1}$

$1165 Ω^{- 1} c m^{2} e q^{- 1}$

$532 Ω^{- 1} c m^{2} e q^{- 1}$

$195.5 Ω^{- 1} c m^{2} e q^{- 1}$

Q. Find the value of Λeq∘​ for potash alum. λm∘​(K+)=73.5Ω−1cm2mol−1 λm∘​(Al3+)=189Ω−1cm2mol−1 λm∘​(SO42−​)=160Ω−1cm2mol−1

145.6Ω−1cm2eq−1

1165Ω−1cm2eq−1

532Ω−1cm2eq−1

195.5Ω−1cm2eq−1

Q. Find the value of $Λ_{e q}^{\circ}$ for potash alum.
$λ_{m}^{\circ} (K^{+}) = 73.5 Ω^{- 1} c m^{2} m o l^{- 1}$
$λ_{m}^{\circ} (A l^{3 +}) = 189 Ω^{- 1} c m^{2} m o l^{- 1}$
$λ_{m}^{\circ} (S O_{4}^{2 -}) = 160 Ω^{- 1} c m^{2} m o l^{- 1}$

$145.6 Ω^{- 1} c m^{2} e q^{- 1}$

$1165 Ω^{- 1} c m^{2} e q^{- 1}$

$532 Ω^{- 1} c m^{2} e q^{- 1}$

$195.5 Ω^{- 1} c m^{2} e q^{- 1}$