1.

The resistance of 0.1 N solution of a salt is found to be 2.5xx 10^(3)ohm. The equivalent conductance of the solutionis (cell constnat =1.15 cm^(-1))

Answer»

`4.6`
`5.6`
`6.6`
`7.6`

Solution :Specific CONDUCTANCE = Conductance `xx` Cell CONSTANT
`k =(1)/(2.5xx 10^(3)) xx1.15,`
`^^_(cq) =(1.15)/(2.5xx10^(3)) xx(100)/(0.1) =4.6`


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