1.

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)

Answer»


Solution :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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