Calculate the limiting molar conductivity of NaCl and KBr from the above table IonΛ∘S cm2mol−1IonΛ∘S cm2mol−1H+349.6OH-199.1Na+50.1CI-76.3K+73.5Br-78.1Ca2+119.0CH3COO-40.9Mg2+106.0SO42-160.0

ΛNaCl∘=50.1+76.3=126.4S cm2mol−1ΛKBr∘=73.5+78.1=151.6S cm2mol−1

Calculate the limiting molar conductivity of NaCl and KBr from the above table IonΛ∘S cm2mol−1IonΛ∘S cm2mol−1H+349.6OH-199.1Na+50.1CI-76.3K+73.5Br-78.1Ca2+119.0CH3COO-40.9Mg2+106.0SO42-160.0

ΛNaCl∘=50.1+76.3=126.4S cm2mol−1ΛKBr∘=73.5+78.1=151.6S cm2mol−1

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Calculate the limiting molar conductivity of NaCl and KBr from the above table

Ion $Λ^{\circ} (S {cm}^{2} {mol}^{- 1})$ Ion $Λ^{\circ} (S {cm}^{2} {mol}^{- 1})$
H⁺ 349.6 OH^- 199.1
Na⁺ 50.1 CI^- 76.3
K⁺ 73.5 Br^- 78.1
Ca²⁺ 119.0 CH³COO^- 40.9
Mg²⁺ 106.0 ${SO}_{4}^{2 -}$ 160.0

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NaCl -150.6
KBr -126.4

NaCl -151.6
KBr -126.4

NaCl -126.4
KBr -151.6

NaCl -126.4
KBr -150.6

answer is B.

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Detailed Solution

$\begin{aligned} Λ_{NaCl}^{\circ} = 50 .1 + 76 .3 = 126 .4 S {cm}^{2} {mol}^{- 1} \\ Λ_{KBr}^{\circ} = 73 .5 + 78 .1 = 151 .6 S {cm}^{2} {mol}^{- 1} \end{aligned}$

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Ion	$Λ^{\circ} (S {cm}^{2} {mol}^{- 1})$	Ion	$Λ^{\circ} (S {cm}^{2} {mol}^{- 1})$
H⁺	349.6	OH^-	199.1
Na⁺	50.1	CI^-	76.3
K⁺	73.5	Br^-	78.1
Ca²⁺	119.0	CH³COO^-	40.9
Mg²⁺	106.0	${SO}_{4}^{2 -}$	160.0