1.

At a particular temperature the ratio of equivalent conductance to specific conductance of a 0.01 (N) NaCl solution is

Answer»

`10^( 5)cm^(3)`
`10^(3) cm^(3)`
`10 cm^(3)`
`10^(5)cm`

Solution :`lambda_(eq) = K XX (1000)/(N)`
`(lambda_(eq))/(K) = (ohm^(-1) cm^(2) ge q^(-1))/(ohm^(-1) cm^(-1))= cm^(3) g eq^(-1)`
`(lambda_eq)/(K) = (1000)/(N) = (1000)/(10^(-2)) = 10^(5) cm^(3) g eq^(-1)`


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