Question

The specific conductivity of $N/10KCl$ solution at ${20}^{\circ }C$ is $0.0212oh{m}^{-1}c{m}^{-1}$ and the resistance of the cell containing this solution at ${20}^{\circ }C$ is $55ohm$. The cell constant is:

• A. $4616c{m}^{-1}$
• B. $1.166c{m}^{-1}$
• C. $2.173c{m}^{-1}$
• D. $3.324c{m}^{-1}$

Answer 1

Answer: (B)

$1.166c{m}^{-1}$

$K=C\times \frac{l}{A}$

Where, $\frac{l}{A}=cellconstant$

$K=specificconductivity$

$R=Resistance$

$C=Conductance$

$\frac{l}{A}=K\times \frac{1}{C}=K\times R=0.0212\times 55=1.166c{m}^{-1}$.