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Electronic Configuration

The conductiv...

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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 resul†an t solution decreases to 0.0055 S cm–1. The equivalent conductivity in S cm2 eq–1 of NaCl solution is

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Q. The conductivity of

0.1 N

N a O H

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

Q. The conductivity of

0.1 N N a O H

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}

N a C l

Q. The conductivity oh

0.1

N NaOH solution is

0.022

^{- 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

^{- 1}

. The equivalent conductivity of NaCl solution is S cm

^{2}

^{- 1}

Q. Resistance of a

0.1

K C l

solution in a conductance cell is

300

ohm and specific conductance of

0.1

K C l

1.29 \times 10^{- 2}

o h m^{- 1}

c m^{- 1}

. The resistance of

0.1

N a C l

solution in the same cell is

380

ohm. Calculate the equivalent conductance of the

0.1

N a C l

The resistance of 0.1 N solution of a salt is found to be $2.5 \times 10^{3}$ . The equivalent conductance of the solution is: (cell constant=1.15 $c m^{- 1}$ )

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