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Calculate conductance of 1 M AgNO(3) sol...

Calculate conductance of 1 M `AgNO_(3)` solution at 298 K if the inter electrode distance is 5 cm and the area of each electrode is `2 cm^(2)`. The equivalent conductance of the solution `L_(E)=94.3" S "cm^(2)equiv^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
`37.74xx10^(-3)S`
Step I. Calculation of specific conductance (k)
`1"M " AgNO_(3)=1"N " AgNO_(3) , Lambda_(E)=(1000xxk)/(N)`
or `(94.3" S "cm^(2)" equiv"^(-1))=((1000 cm^(3))xxk)/((1" equiv".))`
`k=((94.3s cm^(2)" equiv"^(-1))xx1("equiv"))/((1000 cm^(3)))=0.09435" S cm"^(-1)`
Step II. Calculation of conductance (C )
`k="Conductance"xx"cell constant"`
Conductance (C )`=(k)/("cell constant")=((0.09435" S cm"^(-1)))/((5//2 cm^(-1)))=0.03774" S "=37.74xx10^(-3)" S "`.
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