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Standard VII

Chemistry

First Order Reaction

The molar con...

Question

The molar conductance of $N H_{4} O H$ at 0.01M concentration is 11.3 $o h m^{- 1} c m^{2} m o l^{- 1}$ . The degree of dissociation
of $N H_{4} O H$ is
(molar conductance at infinite dilution is 271.1)

4.2 %

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1.3 %

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12.6 %

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41 %

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Solution

The correct option is A $4.2 %$
$α = \frac{11.3}{271.1}$ = 0.042 or 4.2 $%$

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Similar questions

Q. The molar conductance of

N H_{4} O H

at 0.01M concentration is 11.3

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

. The degree of dissociation
of

N H_{4} O H

is
(molar conductance at infinite dilution is 271.1)

Q. The molar conductance of

N H_{4} O H

at 0.01M concentration is 11.3

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

. The degree of dissociation
of

N H_{4} O H

is
(molar conductance at infinite dilution is 271.1)

The molar conductance of $N H_{4} O H$ at 0.01 M concentration is 11.3 $o h m^{- 1} c m^{2} m o l^{- 1}$ . The degree of dissociation of $N H_{4} O H$ is: (Molar conductance at infinite dilution = 271.1 $o h m^{- 1} c m^{2} m o l^{- 1}$

Q. The molar conductance of

N H_{4} O H

at 0.01M concentration is 11.3

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

. The degree of dissociation
of

N H_{4} O H

is
(molar conductance at infinite dilution is 271.1)

Q. At

25^{\circ} C

molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is

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

and at infinite dilution its molar conductance is

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

. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is

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The molar conductance of NH4OH at 0.01M concentration is 11.3 ohm−1 cm2 mol−1. The degree of dissociation of NH4OH is (molar conductance at infinite dilution is 271.1)

The correct option is A 4.2%α=11.3271.1 = 0.042 or 4.2 %

The molar conductance of $N H_{4} O H$ at 0.01M concentration is 11.3 $o h m^{- 1} c m^{2} m o l^{- 1}$ . The degree of dissociation
of $N H_{4} O H$ is
(molar conductance at infinite dilution is 271.1)

The correct option is A $4.2 %$
$α = \frac{11.3}{271.1}$ = 0.042 or 4.2 $%$