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At 25^(@)C, the resistance of 0.01 N NaC...

At `25^(@)C`, the resistance of `0.01 N NaCl` solution is `200 ohm` . If cell constant of the conductivity cell is unity, then the equivalent conductance of the solution is:

Text Solution

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The correct Answer is:
`500 ohm^(-1)`
Step I. Calculation of specific conductance
Specific conductance (k)`=("cell constant")/("R")=((1" cm"^(-1)))/((200 ohm))=0.005" ohm"^(-1) cm^(-1)`.
Step II. Calculation of equivalent conductance of solution
`Lambda_(E)=(1000xxk)/(N)=((1000 cm^(3))xx(0.005" ohm"^(-1)cm^(-1)))/((0.01"g eq"))=500" ohm"^(-1)cm^(2)("g eq")^(-1)`.
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