The limiting molar conducticities - 0 for NaCl-KBr and KCl are 126-152 and 150Scm2mol-1 respectively- The - m0 for NaBr is-

The correct option is C 128 S cm^2mol^ 1 ∧_NaBr^o=∧_NaCl^o+∧_kBr^o ∧_KCl^o =126+152 150 =128 s cm^2 mol^ 1 Hence, the correct option is C. ...

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Standard XII

Chemistry

Kohlrausch Law

The limiting ...

Question

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:

278 S

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

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176 S

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

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128 S

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

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302 S

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

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Solution

The correct option is C 128 S $c m^{2} m o l^{- 1}$
$\land_{N a B r}^{o} = \land_{N a C l}^{o} + \land_{k B r}^{o} - \land_{K C l}^{o}$

$= 126 + 152 - 150$

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

Hence, the correct option is C.

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Similar questions

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. Calculate

Λ_{m}^{0}

for

C a C l_{2}

and

M g S O_{4}

from the data

λ_{C a^{2 +}}^{0}

= 119.0 S

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

λ_{C l^{⊖}}^{0}

= 76.3 S

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

λ_{M g^{2 +}}^{0}

= 106.0 S

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

λ_{S O_{4}^{2 -}}^{0}

= 160.0 S

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

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

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. Molar conductivities

(\land_{m})

at infinite dilution of NaCl, HCl and

C H_{3} C O O N a

are 126.4, 425.9 and 91.0 S

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

respectively.

(\land_{m})

for

C H_{3} C O O H

will be:

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The limiting molar conducticities ∧0 for NaCl,KBr and KCl are 126,152 and 150Scm2mol−1 respectively. The ∧0m for NaBr is:

The correct option is C 128 Scm2mol−1∧oNaBr=∧oNaCl+∧okBr−∧oKCl=126+152−150=128 s cm2 mol−1Hence, the correct option is C.

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 correct option is C 128 S $c m^{2} m o l^{- 1}$
$\land_{N a B r}^{o} = \land_{N a C l}^{o} + \land_{k B r}^{o} - \land_{K C l}^{o}$

$= 126 + 152 - 150$

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

Hence, the correct option is C.