If - -c for NH-4OH is 11-5 ohm-1 cm-2 mol-1- its degree of dissociation would be -given that -NH-4- - 73-4 and -OH- - 197-6 ohm-1cm-2 mol-1 - -0-04240-02120-08480-004

The correct option is A 0.0424α=Λ _c/Λ^∘=115/73.4+197.6=115/271.0=0.0424 ...

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Standard X

Chemistry

Electrolytic Dissociation

If Λ _c for N...

Question

If $Λ_{c}$ for $N H_{4} O H$ is 11.5 $o h m^{- 1} c m^{2} m o l^{- 1}$ , its degree of dissociation would be
(given that $λ^{\circ} N H_{4}^{+}$ = 73.4 and $λ^{\circ} O H^{-}$ = 197.6 $o h m^{- 1} c m^{2} m o l^{- 1}$ )

0.0848

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0.0424

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0.0212

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0.004

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Solution

The correct option is B 0.0424
$α = \frac{Λ_{c}}{Λ^{\circ}} = \frac{115}{73.4 + 197.6} = \frac{115}{271.0} = 0.0424$

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Similar questions

Q. If

Λ_{c}

for

N H_{4} O H

is 11.5

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

, its degree of dissociation would be
(given that

λ^{\circ} N H_{4}^{+}

= 73.4 and

λ^{\circ} O H^{-}

= 197.6

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

)

Q. If

Λ_{c}

for

N H_{4} O H

is 11.5

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

, its degree of dissociation would be
(given that

λ^{\circ} N H_{4}^{+}

= 73.4 and

λ^{\circ} O H^{-}

= 197.6

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

)

If $\land_{c} o f N H_{4} O H i s 11.5 Ω^{- 1} c m^{2} m o l^{- 1}$ , its degree of dissociation would be:

(Given $λ_{N H_{4}^{+}}^{\infty} = 73.4 Ω^{- 1} c m^{2} m o l^{- 1}$ and $λ_{O H^{-}}^{\infty} = 197.6 Ω^{- 1} c m^{2} m o l^{- 1}$ ) :

If A_c of NH₄OH is 11.5 W^-1cm² mol^-1 andA_cat infinite dilution of NH₄OH is 0.4876 W^-1cm² mol^-1, its degree of dissociation is:

Specific conductance of 0.1 M HA is $3.75 \times 10^{- 4} o h m^{- 1} c m^{- 1} .$ If $λ^{\infty} (H A) = 250 o h m^{- 1} c m^{2} m o l^{- 1},$ the dissociation constant $K_{a}$ of HA is :

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If Λc for NH4OH is 11.5 ohm−1cm2mol−1, its degree of dissociation would be (given that λ∘NH+4 = 73.4 and λ∘OH− = 197.6 ohm−1cm2mol−1 )

The correct option is B 0.0424α=ΛcΛ∘=11573.4+197.6=115271.0=0.0424

If $Λ_{c}$ for $N H_{4} O H$ is 11.5 $o h m^{- 1} c m^{2} m o l^{- 1}$ , its degree of dissociation would be
(given that $λ^{\circ} N H_{4}^{+}$ = 73.4 and $λ^{\circ} O H^{-}$ = 197.6 $o h m^{- 1} c m^{2} m o l^{- 1}$ )

The correct option is B 0.0424
$α = \frac{Λ_{c}}{Λ^{\circ}} = \frac{115}{73.4 + 197.6} = \frac{115}{271.0} = 0.0424$