You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.11

No worries! We‘ve got your back. Try BYJU‘S free classes today!

110

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $C$

Suggest Corrections

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:

Join BYJU'S Learning Program

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 $C$