Calculate λ m ∘ at 298 K of NH 4 OH at infinite dilution, given λ ∘ OH - = 174 S cm 2 mol - 1 , λ ∘ Cl - = 66 S cm 2 mol - 1 and λ ∘ NH 4 Cl = 130 S cm 2 mol - 1 .

Given, λ ∘ OH - = 174 S cm 2 mol - 1 , λ ∘ Cl - = 66 S cm 2 mol - 1 and λ ∘ NH 4 Cl = 130 S cm 2 mol - 1 . We know from Kohlrausch law that, λ ∘ NH 4 OH = λ ∘ NH 4 + + λ ∘ OH - = λ ∘ NH 4 Cl - λ ∘ Cl - + λ ∘ OH - Therefore, λ ∘ NH 4 OH = 130 - 66 + 174 = 238 S cm 2 mol - 1 .

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How is the molar conductivity of a weak electrolyte at infinite dilution determined?

Important Questions on Electrochemistry

EASY

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

Λ_{m}^{0}

values for

NaCl, HCl

and

{CH}_{3} COONa

are

126.4 S {cm}^{2} {mol}^{- 1}, 425.9 S {cm}^{2} {mol}^{- 1}

and

91.0 S {cm}^{2} {mol}^{- 1}

respectively.

Λ_{m}^{0}

value for

{CH}_{3} COOH

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

λ ° m

for

{NH}_{4} Cl, NaOH

and

NaCl

are

130, 248

and

126 . 5

{ohm}^{- 1} {em}^{2} {mo}^{- 1}

respectively. The

λ ° m

{NH}_{4} OH

will be,

EASY

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

Which one is correct for acetic acid?

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Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

The molar conductivity of $0.007 M$ acetic acid is $20 S {cm}^{2} {mol}^{- 1} .$ What is the dissociation constant of acetic acid? Choose the correct option.

$λ_{H^{+}}^{\circ} = 350 S {cm}^{2} {mol}^{- 1} λ_{{CH}_{3} {COO}^{-}}^{\circ} = 50 S {cm}^{2} {mol}^{- 1}$

EASY

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

The specific conductance

(K)

0 .02 M

aqueous acetic acid solution at

298 K

1 .65 \times 10^{- 4} S {cm}^{- 1} .

The degree of dissociation of acetic acid is

[Given: equivalent conductance at infinite dilution of

H^{+} = 349 .1 S {cm}^{2} {mol}^{- 1}

and

{CH}_{3} {COO}^{-} = 40 .9 S {cm}^{2} {mol}^{- 1}

]

EASY

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

The molar conductance of $NaCl, HCl$ and ${CH}_{3} COONa$ at infinite dilution are $126.45, 426.16$ and $91.0 {Scm}^{2} {mol}^{- 1}$ respectively. The molar conductance of ${CH}_{3} COOH$ at infinite dilution is. Choose the right option for your answer.

HARD

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

State and explain kohlrausch's law of independent migration of ions.

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

Calculate

{λ_{m}}^{\circ}

298 K

{NH}_{4} OH

at infinite dilution, given

λ^{\circ} ({OH}^{-}) = 174 S {cm}^{2} {mol}^{- 1}

λ^{\circ} ({Cl}^{-}) = 66 S {cm}^{2} {mol}^{- 1}

and

λ^{\circ} ({NH}_{4} Cl) = 130 S {cm}^{2} {mol}^{- 1}

EASY

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

State Kohlrausch law.

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

Why can limiting molar conductivity of

{CH}_{3} COOH

not be determined experimentally ?

EASY

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

Define the following terms: Kohlrausch’s Law.

EASY

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

If the molar conductivities

(in S {cm}^{2} {mol}^{- 1})

NaCl, KCl and NaOH

at infinite dilution are 126 and 150 and 250 respectively, the molar conductivity of

KOH (in S {cm}^{2} {mol}^{- 1})

is -

EASY

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

State Kohlrausch’s law. Calculate the molar conductivity of an aqueous solution at infinite dilution for

{BaCl}_{2}

, when ionic conductance of

{Ba}^{2 +}

and

{Cl}^{-}

ions are

127.30 S {cm}^{2} {mol}^{- 1}

and

76.34 S {cm}^{2} {mol}^{- 1}

respectively.

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

The conductivity of a weak acid

HA

of concentration

0.001 mol L^{- 1}

2.0 \times 10^{- 5} S {cm}^{- 1}

. If

Λ_{m}^{0} (HA) = 190 S {cm}^{2} {mol}^{- 1},

the ionization constant

(K_{a})

HA

is equal to

____

\times 10^{- 6}

(Round off to the Nearest Integer)

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

The conductivity of

0.00241 M

acetic acid is

7.896 \times 10^{- 5} S {cm}^{- 1}

. Calculate its molar conductivity and its degree of dissociation. (Given

{Λ^{\circ}}_{m}

for acetic acid is

390.5 S {cm}^{2} {mol}^{- 1}

)

EASY

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

The limiting molar conductivity and molar conductivity of acetic acid are $390.5 S {cm}^{2} . {mol}^{- 1}$ and $48.15 S {cm}^{2} . {mol}^{- 1}$ respectively. Calculate the degree acid are of dissociation of the weak acid?

EASY

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

State Kohlrausch law.

MEDIUM

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

0.002 M

solution of a weak acid has an equivalent conductance

(Λ) 60 {ohm}^{- 1} {cm}^{2}

{eq}^{- 1}

. What will be the

pH

? (Given :

Λ ° = 400 {ohm}^{- 1} {cm}^{2} {eq}^{- 1}

]

HARD

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

The equation that is incorrect is:

HARD

Chemistry>Physical Chemistry>Electrochemistry>Conductance in Electrolytic Solutions

\land_{m}^{o}

for

NaCl, HCl

and

NaA

are

126.4, 425.9

and

100 .5 S {cm}^{2} {mol}^{- 1}

respectively. If the conductivity of

0 .001 M HA

5 \times 10^{- 5} S {cm}^{- 1}

, degree of dissociation of

HA

How is the molar conductivity of a weak electrolyte at infinite dilution determined?

Important Questions on Electrochemistry

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