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Standard XII

Chemistry

Kohlrausch Law

Calculate the...

Question

Calculate the molar conductivity of $H N O_{2}$ at infinite dilution (in Scm $^{2}$ mol $^{- 1}$ ).
Given that $λ_{H^{+}}^{\infty} = 349.6 S c m^{2} m o l^{- 1}$ and $λ_{N O_{2}^{-}}^{\infty} = 71.7 S c m^{2} m o l^{- 1}$ .

421.3

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321.3

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421.3 \times 10^{- 4}

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321.3 \times 10^{- 4}

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Solution

The correct option is B $421.3$
The molar conductivity of $H N O_{2}$ at infinite dilution (in $S c m^{2} m o l^{- 1}$ ) is:

$Λ_{H N O 2}^{o} = λ_{H^{+}}^{0} + λ_{N O 2^{-}}^{0} = 349.6 + 71.7 = 421.3$ .

It is equal to the sum of the molar conductivity of hydrogen ions and the molar conductivity of nitrite ions at infinite dilution.

Option A is correct.

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Similar questions

Calculate the molar conductivity of water at infinite dilution (in Sm $^{2}$ mol $^{- 1}$ ).

(molar conductivity of hydrogen ions and the molar conductivity of hydroxide ions at infinite dilution are $349.6 \times 10^{- 4}$ Sm $^{2}$ mol $^{- 1}$ and $198.3 \times 10^{- 4}$ Sm $^{2}$ mol $^{- 1}$ )

Q. The molar conductivities at infinite dilution of

A g N O_{3}

and

N a C l

are 116.0 and 110.0

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

respectively. The conductivity of

A g C l

in water is

8.42 \times 10^{- 6} S c m^{- 1}

and of water used is

1.16 \times 10^{- 6} S c m^{- 1}

. The solubility of

A g C l

6.0 \times 10^{- 5} m o l / L

. Find the molar conductivity at infinite dilution of

N a N O_{3}

(

i n S c m^{- 1} m o l^{- 1}

) in the solution.

Electrolyte	$K C l$	$K N O_{3}$	$H C l$	$N a O A c$	$N a C l$
$Λ^{\infty} {(Scm}^{2} {mol}^{- 1})$	$149.9$	$145$	$426.2$	$91$	$126.5$

Calculate

Λ_{H O A c}^{\infty}

using appropirate molar conductances of the electrolytes listed above at infinite dilution in

H_{2} O

25^{o} C

Q. The molar conductivities of

A g N O_{3}, N a C l

and

N a N O_{3}

at infinite dilution are 116.0, 110.0 and

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

, respectively. The conductivity of

A g C l

in water is

8.42 \times 10^{- 6} S c m^{- 1}

and of water used is

1.16 \times 10^{- 6} S c m^{- 1}

. Find the solubility of AgCl in

m o l l i t r e^{- 1}

Q. Given the following molar conductivity at infinite dilution and

25^{o}

C
HCl:

\land_{m}^{\infty}

426.2

{c m}^{2}

{m o l}^{- 1}

KCl:

\land_{m}^{\infty}

= 114.42 S

{c m}^{2}

{m o l}^{- 1}

C H_{3} C O O K

\land_{m}^{\infty}

149.86

{c m}^{2}

{m o l}^{- 1}

The molar conductance at infinite dilution and

25^{o}

C, for acetic acid solution is:

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Calculate the molar conductivity of HNO2 at infinite dilution (in Scm2 mol−1 ).Given that λ∞H+=349.6 Scm2mol−1 and λ∞NO−2=71.7 Scm2mol−1.

The correct option is B 421.3The molar conductivity of HNO2 at infinite dilution (in Scm2mol−1 ) is: ΛoHNO2=λ0H++λ0NO2−=349.6+71.7=421.3. It is equal to the sum of the molar conductivity of hydrogen ions and the molar conductivity of nitrite ions at infinite dilution.Option A is correct.

Calculate the molar conductivity of $H N O_{2}$ at infinite dilution (in Scm $^{2}$ mol $^{- 1}$ ).
Given that $λ_{H^{+}}^{\infty} = 349.6 S c m^{2} m o l^{- 1}$ and $λ_{N O_{2}^{-}}^{\infty} = 71.7 S c m^{2} m o l^{- 1}$ .