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Standard XII

Chemistry

Equivalent, Molar Conductivity and Cell Constant

The mobility ...

Question

The mobility of the $N H_{4}^{+}$ ions is $7.623 \times 10^{- 8} m^{2} / V s$ . Calculate:
(a) the molar conductivity of $N H_{4}^{+}$ ion;
(b) Velocity of the ion of $15$ volt are applied across the electrodes $25 c m$ apart.

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Solution

$λ$ = $Z F μ$ here $μ$ = mobility of ion
=7.623* $10^{- 8} m^{2} \frac{1}{V . s}$ 96500
$λ$ =735619.5 $10^{- 8} m^{2} \frac{1}{V . s}$
$μ$ = $v \times \frac{l}{V}$
v =157.623 $10^{- 8} m \frac{1}{.25 s e c}$
velocity=
4.5738* $10^{- 6} \frac{m}{s e c}$

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Similar questions

Q. A potential of

12.0

volts was applied to two electrodes placed

20 c m

apart. A dilute solution of ammonium chloride was placed between the electrodes when

N H_{4}^{+}

ions was found to cover a distance of

1.60 c m

in one hour. What is the mobility of

N H_{4}^{+}

ion?

Q. At

18^{\circ} C

the mobilities of

N H_{4}^{+}

and

C l O_{4}^{-}

ions are

6.6 \times 10^{- 4}

and

5.7 \times 10^{- 4} c m^{2} v o l t^{- 1} s e c^{- 1}

at infinite dilution. The equivalent conductance of ammonium chlorate solution is:

Q. At

18^{\circ} C

, the mobilities of

N H_{4}^{+}

and

C l O_{4}

^{-}

ions are

6.6 \times 10^{- 4}

and

5.7 \times 10^{- 4} c m^{2} v o l t^{- 1} s e c^{- 1}

at infinite dilution. Calculate the equivalent conductance of ammonium chlorate solution.

At $18^{o} C$ , the mobilities of $N H {_{4}}^{+}$ and $C l O {_{4}}^{-}$ ions are $6.6 \times 10^{- 4}$ and $5.7 \times 10^{- 4} c m^{2} v o l t^{- 1} s e c^{- 1}$ respectively, at infinite dilution.

The equivalent conductance of ammonium chlorate solution is :(in $m h o c m^{2}$ )

N H_{4}^{+}, C l^{-}, H^{+}

are monovalent ions.

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The mobility of the NH+4 ions is 7.623×10−8m2/Vs. Calculate:(a) the molar conductivity of NH+4 ion;(b) Velocity of the ion of 15 volt are applied across the electrodes 25 cm apart.

λ = ZFμ here μ = mobility of ion=7.623*10−8m21V.s*96500λ =735619.5*10−8m21V.sμ = v×lVv =15*7.623*10−8m1.25secvelocity=4.5738*10−6msec

The mobility of the $N H_{4}^{+}$ ions is $7.623 \times 10^{- 8} m^{2} / V s$ . Calculate:
(a) the molar conductivity of $N H_{4}^{+}$ ion;
(b) Velocity of the ion of $15$ volt are applied across the electrodes $25 c m$ apart.

$λ$ = $Z F μ$ here $μ$ = mobility of ion
=7.623* $10^{- 8} m^{2} \frac{1}{V . s}$ 96500
$λ$ =735619.5 $10^{- 8} m^{2} \frac{1}{V . s}$
$μ$ = $v \times \frac{l}{V}$
v =157.623 $10^{- 8} m \frac{1}{.25 s e c}$
velocity=
4.5738* $10^{- 6} \frac{m}{s e c}$