Each of the two platinum electrodes having area $64m{m}^2$ of a conductivity cell are separated by $8mm$. The resistance of the cell containing $7.5\times {10}^{-3}M$ $KCl$ solution at $298K$ is $1250ohm$. Calculate :
(i) Cell constant
(ii) Conductivity and
(iii) Molar conductivity

Q. The electrodes in a conductivity cell have area $1.2\times {10}^{-4}{m}^2$ and they are fixed $3\times {10}^{-3}m$ apart. A solution containing $200g$ equivalent of the electrolyte per $m^3$ of the solution has a resistance of $60ohm$ at $298K$. Calculate the equivalent conductivity of the solution.

Q. The measured resistance of a conductance cell containing $7.5\times {10}^{-3}M$ solution of $KCl$ at ${25}^{o}C$ was $1005ohms$. Calculate molar conductance $\left(oh{m}^{-1}c{m}^{2}mo{l}^{-1}\right)$ of the solution.
(Cell constant$=1.25c{m}^{-1}$)

Q. The resistance of a conductivity cell containing $0.0001MKCI$ solution at $298K\text{is}1500\Omega$. Calculate the cell constant when the conductivity of $0.0001MKCI$ solution at $298K\text{is}0.146\times {10}^{-3}Sc{m}^{-1}$:

Q. In a conductivity cell the two platinum electrodes, each of area $10{\text{cm}}^2,$ are fixed $1.5\text{cm}$ apart. The cell contains $0.05\text{N}$ solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of $50\Omega$, find specific conductance of the salt solution.

Q. In a conductivity cell the two platinum electrodes, each of area 10 sq. cm. are fixed 1.5 cm apart. The cell contained 0.05 N solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of 50 ohms, the equivalent conductance of the salt solution will be: