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Standard X

Chemistry

Electrolysis and Electrolytes

If the equiva...

Question

If the equivalent conductance of $1 M$ benzoic acid is $12.8 o h m^{- 1} c m^{2} e q^{- 1}$ and if the conductance of benzoate ion and $H^{+}$ ions are $42$ and $288.42 o h m^{- 1} c m^{2} e q^{- 1}$ respectively, its degree of dissociation is:

39

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3.9

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0.35

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0.039

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Solution

The correct option is C $3.9$ %
$α = \frac{\land_{m}}{\land_{m}^{\circ}} = \frac{12.8}{(42 + 288.42)} = 0.03987$ .
Percentage dissociation $= 0, 0387 \times 100 \approx 3.9$ %.

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Similar questions

Q. If equivalent conductance of 1M benzoic acid is

12.8 {o h m}^{- 1} {c m}^{2}

and if the conductance of benzoate ion and

H^{- 1}

ion are 42 and

288.42 {o h m}^{- 1} {c m}^{2}

respectively. Its degree of dissolution is:

Q. The equivalence conductances of sodium chloride, hydrochloric acid and sodium acetate at infinite dilution are

126.45, 426.16

and

91.0 {ohm}^{- 1} {cm}^{2} {eq}^{- 1}

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Calculate the equivalent conductance

({in ohm}^{- 1} {cm}^{2} {eq}^{- 1})

of acetic acid at infinite dilution.

Equivalent conductance at infinite dilution, $λ^{0}$ of NH $_{4}$ Cl, NaOH and NaCl are 128.0, 217.8 and 109.9 $o h m^{- 1} c m^{2} e q^{- 1}$ respectively. The equivalent conductance of 0.01 N $N H_{4} O H$ is 9.30 $o h m^{- 1} c m^{2} e q^{- 1}$ , then the degree of ionization of $N H_{4} O H$ at this temperature would be:

Q. The equivalent conductivity of

N / 10

solution of acetic acid at

25^{\circ} C

14.3 o h m^{- 1} c m^{2} e q^{- 1}

. Calculate the degree of dissociation of

C H_{3} C O O H

\land_{\infty C H_{3} C O O H}

390.71

Equivalent conductance of 1 M propanoic acid is 10 $o h m^{- 1} c m^{2} e q^{- 1}$ and that at infinite dilution 200 $o h m^{- 1} c m^{2} e q^{- 1}$ . The pH of the propanoic acid solution is:

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If the equivalent conductance of 1 M benzoic acid is 12.8 ohm−1cm2eq−1 and if the conductance of benzoate ion and H+ ions are 42 and 288.42 ohm−1cm2eq−1 respectively, its degree of dissociation is:

The correct option is C 3.9%α=∧m∧∘m=12.8(42+288.42)=0.03987.Percentage dissociation =0,0387×100≈3.9%.

If the equivalent conductance of $1 M$ benzoic acid is $12.8 o h m^{- 1} c m^{2} e q^{- 1}$ and if the conductance of benzoate ion and $H^{+}$ ions are $42$ and $288.42 o h m^{- 1} c m^{2} e q^{- 1}$ respectively, its degree of dissociation is:

The correct option is C $3.9$ %
$α = \frac{\land_{m}}{\land_{m}^{\circ}} = \frac{12.8}{(42 + 288.42)} = 0.03987$ .
Percentage dissociation $= 0, 0387 \times 100 \approx 3.9$ %.