1.

The resistance of (N)/(10)solutionis foundto be2.5xx10^(3) omegatheequivalent conductance of the solutionis (cellconstant=1.25 cm^(-1))

Answer»

2.5 `omega^(-1) cm^(-2) equiv^(-1)`
5.0 `omega^(-1) cm^(2) equiv^(-01)`
2.5 `omega^(-1) cm^(-2 equiv^(-1)`
5.0 `omega^(-1) cm^(-2) equiv^(-1)`

Solution :Resistance of `(N)/(10)` solution`=2.5 xx10^(3) omega`
`K=(1)/("resistance ")xx` cell constant
`=(1)/(2.5 xx10^(-3))xx1.25`
`=(1.25 xx10^(-3))/(2.5)=5x10^(-4) omega^(-1) cm^(-1)`
EQUIVALENT conductance `=(Kxx1000)/(M)=(5XX10^(-4)xx1000)/(1//10)`
`= omega^(-1) cm^(2) equiv^(-1)`


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