At 25 degree celcius lambda 0h349810-2 and lambda0 0h-
At 25 degree celcius λ( 0)H+=3.498*10^-2 and λ(0) 0H- =1.98*10^-2 Sm2mol-1.given specific conductance of water(kappa) is =5.7*10^-6 Sm-1.calculate Kw of water
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at 25 degree celcius lambda 0h349810-2 and lambda0 0h- 19810-2 sm2mol-1given specific conductance of waterkappa is
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