The resistance of 0.1 N solution of formic acid is 200 ohm and cell constant is 2.0cm^(-2) . The equivalent conductivity (in Scm^(2)eg^(-1) ) of 0.1 N formic acid is 1.0 xx 10^(x) , x is
The resistance of 0.05M KCl solution at 25^@C is 100 ohm in a cell whose cell constant is 0.3765cm^(-1) . The specific conductivity of KCl solution is
What is cell constant of a conductivity cell ?
The resistance of 0.1N KCl solution is found 702Omega awhen measured in a conductivity cell. The specific conductance of 0.1N KCl is 0.14807 Omega^(-1)m^(-1) . Calculate the cell constant.
The specific conductance of 0.1N KCl solution at M 23^@C is 0.012 ohm^(-1)cm^(-1) . The resistance of cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be
0.05M NaOH solution offered a resistance of 31.69 Omega in a conductivity cell at 298K. If the cell constant of the conductivity cell is 0.367cm^(-1) the molar conductivity of the NaOH solution is
The resistnce of 0.1 N solution of a salt is found to be 2.5xx10^(3) ohms. The equivalent conductance of the solution is ( cell constant =1.15cm^(-1) )
The resistance of 1N solution of acetic acid is 250 ohm when measured in a cell of cell constant 1.15cm^(-1) . The equivalent conductance ( in ohm^(-1)cm^(2)equi^(-1) ) of 1N acetic acid is
The resistance of 0.01N solution of an electrolyte AB at 328K is 100 ohm. The specific conductance of solution is ( cell constant =1 cm^(-1) )
A student has 100mL of 0.1 M KCl solution. To make it 0.2 M, he has to
