ElectrochemistryHard
Question
A resistance of 50W is registered when two electrodes are suspended into a beaker containing a dilute solution of a strong electrolyte such that exactly half of the them are submerged into solution as shown in figure. If the solution is diluted by adding pure water (negligible conductivity) so as to just completely submerge the electrodes, the new resistance offered by the solution would be
Options
A.50 Ω
B.100 Ω
C.25 Ω
D.200 Ω
Solution
R = 
The k is halved while the A is doubled. Hence R remains 50 Ω.
The k is halved while the A is doubled. Hence R remains 50 Ω.Create a free account to view solution
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