The limiting molar conductivities - 0 for NaCl- KBr and KCl are 126- 152 and 150 Scm2 mo l-1 respectively- The - 0 for NaBr is-

The limiting molar conductivities $\land^{0}$ for $N a C l, K B r$ and $K C l$ are $126, 152$ and $150$ $S$ $c m^{2} m o l^{- 1}$ respectively. The $\land^{0}$ for $N a B r$ is :

Q. The limiting molar conductivities

λ^{\circ}

for

N a C l, K B r

and

K C l

are

126, 152

and

150 S c m^{2} m o l^{- 1}

respectively. The

λ^{\circ}

for

N a B r

Q. The limiting molar conductivities

λ^{\circ}

for

N a C l, K B r

and

K C l

are

126, 152

and

150 S c m^{2} m o l^{- 1}

respectively. The

λ^{\circ}

for

N a B r

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The limiting molar conductivities ∧0 for NaCl, KBr and KCl are 126, 152 and 150 Scm2 mol−1 respectively. The ∧0 for NaBr is:

The correct option is A 128 Scm2mo1−1As we know,Λ0NaBr=Λ0NaCl+Λ0KBr−Λ0KCl=Λ0Na++Λ0Br−=126+152−150 = 128 Scm2mol−1Option A is correct answer.

The limiting molar conductivities $\land_{0}$ for $N a C l$ , $K B r$ and $K C l$ are 126, 152 and 150 $S c m^{2}$ mo $l^{- 1}$ respectively. The $\land_{0}$ for $N a B r$ is:

The correct option is A $128 S c m^{2} m o 1^{- 1}$
As we know,

$Λ_{N a B r}^{0} = Λ_{N a C l}^{0} + Λ_{K B r}^{0} - Λ_{K C l}^{0}$

= $Λ_{N a^{+}}^{0} + Λ_{B r^{-}}^{0} = 126 + 152 - 150$

= $128 S c m^{2} m o l^{- 1}$

Option A is correct answer.