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Standard XII

Chemistry

Kohlrausch Law

The conductiv...

Question

The conductivity of $0.1 N N a O H$ solution is $0.022 S c m^{- 1}$ . When equal volume of $0.1 N H C l$ solution is added, the conductivity of resultant solution is decreases to $0.0055 S c m^{- 1}$ . The equivalent conductivity is $S c m^{2}$ equivalent of $N a C l$ solution is:

0.0055

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0.11

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110

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None of these

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Solution

The correct option is C $110$
The concentration of $N a C l$ is $\frac{0.1}{2}$ .

As we know, $λ_{e q} = k \times \frac{1000}{N}$

By putting the values, we get-

$= 0.0055 \times \frac{1000}{\frac{0.1}{2}} = 110$ .

So, the correct option is $C$

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Similar questions

Q. The conductivity of

0.1 N

N a O H

solution is

0.022 s / c m

. When equal volume of the

0.1 N

H C l

solution is added, the conductivity of resultant solution decreases to

0.0055 s / c m

. The equivalent conductivity in s-cm/equivalent of

N a C l

solution is:

Q. The conductivity oh

0.1

N NaOH solution is

0.022

S cm

^{- 1}

. To this solution, an equal volume of

0.1

N HCl solution is added which results into a decrease of the conductivity of solution to

0.0055

S cm

^{- 1}

. The equivalent conductivity of NaCl solution is S cm

^{2}

equiv

^{- 1}

is:

Q. At a particular temperature, the ratio of equivalent conductance to specific conductance of a

0.01 N

NaCl solution is :

Q. The resistance of

0.01 N N a C l

solution at

25^{\circ} C

200 Ω

. Cell constant of conductivity cell is

1 {cm}^{- 1}

. The equivalent conductance is:

Q. The equivalent conductance of

C H_{3} C O O N a, H C l

and

N a C l

at infinite dilution are

91, 426

and

126 S c m^{3} e q^{- 1}

respectively at

25^{o} C

. The equivalent conductance of

1 M C H_{3} C O O H

solution is

19.55 S c m^{2} e q^{- 1}

. The pH of the solution is:

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The conductivity of 0.1 N NaOH solution is 0.022 S cm−1. When equal volume of 0.1 N HCl solution is added, the conductivity of resultant solution is decreases to 0.0055 S cm−1. The equivalent conductivity is S cm2 equivalent of NaCl solution is:

The correct option is C 110The concentration of NaCl is 0.12.As we know, λeq=k×1000NBy putting the values, we get-=0.0055×10000.12=110.So, the correct option is C

The conductivity of $0.1 N N a O H$ solution is $0.022 S c m^{- 1}$ . When equal volume of $0.1 N H C l$ solution is added, the conductivity of resultant solution is decreases to $0.0055 S c m^{- 1}$ . The equivalent conductivity is $S c m^{2}$ equivalent of $N a C l$ solution is:

The correct option is C $110$
The concentration of $N a C l$ is $\frac{0.1}{2}$ .

As we know, $λ_{e q} = k \times \frac{1000}{N}$

By putting the values, we get-

$= 0.0055 \times \frac{1000}{\frac{0.1}{2}} = 110$ .

So, the correct option is $C$