The limiting molar conductivities - for NaCl-KBr and KCl are 126- 152 and 150- S- cm-2- mol-1respectively- The - for NaBr is278 S cm2mol-1176-S cm2mol-1302-S cm2mol-1128-S cm2mol-1

The correct option is D 128 S cm2mol 1(126 scm^2)∧ ^∘_NaCl = ∧ ^∘_Na++∧ ^∘_Cl^ .......(1) (152 scm^2)∧ ^∘_KBr =∧ ^∘_K^++∧ ^∘_Br^ ...

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Standard X

Physics

Ideal Gas Equation

The limiting ...

Question

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 $N a B r$ is

278 S cm²mol^-1

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176 S cm²mol^-1

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128 S cm²mol^-1

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302 S cm²mol^-1

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Solution

The correct option is C 128 S cm²mol^-1
$(126 s c m^{2}) \land_{N a C l}^{\circ} = \land_{N a +}^{\circ} + \land_{C l^{-}}^{\circ}$ $. . . . . . . (1)$
$(152 s c m^{2}) \land_{K B r}^{\circ} = \land_{K^{+}}^{\circ} + \land_{B r^{-}}^{\circ}$ $. . . . . . . (2)$
$(150 s c m^{2}) \land_{K C l}^{\circ} = \land_{K^{+}}^{\circ} + \land_{C l^{-}}^{\circ}$ $. . . . . . . (3)$
By equation $(1) + (2) - (3)$
$∵ \land_{N a B r}^{\circ} = \land_{N a +}^{\circ} + \land_{B r^{-}}^{\circ} = 126 + 152 - 150 = 128 S c m^{2} m o l^{- 1}$

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Similar questions

Q. The limiting molar conducticities

\land^{0}

for NaCl,KBr and KCl are 126,152 and

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

respectively. The

\land_{m}^{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 :

The limiting molar conductivity for NaCl,KBr and KCl are 126,152 & 150 S.cm^2mol^-1 respectively.then molar conductivity for NaBr 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

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The limiting molar conductivity for KCl, NaBr and NaCl are 100, 120, 130 S cm^-2mol^-1 respectively. Calculate the limiting molar conductivity for KBr?

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The limiting molar conductivities λ∘ for NaCl,KBr and KCl are 126, 152 and 150 S cm2 mol−1respectively. The λ∘ for NaBr is

The correct option is C 128 S cm2mol-1(126 scm2)∧∘NaCl=∧∘Na++∧∘Cl− .......(1) (152 scm2)∧∘KBr=∧∘K++∧∘Br− .......(2) (150 scm2)∧∘KCl=∧∘K++∧∘Cl− .......(3) By equation (1)+(2)−(3) ∵ ∧∘NaBr=∧∘Na++∧∘Br−=126+152−150=128 Scm2mol−1

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 $N a B r$ is