The resistance of $1 N$ solution of acetic acid is $250 o h m .$ when measured in a cell of cell constant $1.15$ $c m^{- 1}$ . The equivalent conductance of $1 N$ acetic acid (in $o h m^{- 1}$ $c m^{2}$ $e q v^{- 1}$ ) is:

Join BYJU'S Learning Program

The specific conductance of a solution is 0.3568 (ohm)−1 cm−1When placed in a cell the conductance is 0.0268ohm−1. The cell constant is:

The correct option is B 13.31cm−1The specific conductance (κ) is the product of conductance (C) and cell constant (la). Hencela=κC=0.3568ohm−1cm−10.0268ohm−1=13.31cm−1Hence, option B is correct.

The specific conductance of a solution is $0.3568$ $(o h m)^{- 1}$ $c m^{- 1}$ When placed in a cell the conductance is $0.0268 o h m$ $^{- 1}$ .
The cell constant is:

The correct option is B $13.31 c m$ $^{- 1}$
The specific conductance $(κ)$ is the product of conductance $(C)$ and cell constant $(\frac{l}{a})$ .

Hence
$\frac{l}{a} = \frac{κ}{C} = \frac{0.3568 {o h m}^{- 1} {c m}^{- 1}}{0.0268 {o h m}^{- 1}} = 13.31 {c m}^{- 1}$

Hence, option B is correct.