The conductivity of $N/50$ solution of ${\rm{KCl}}$ in a cell at 250C is $0.002765{\kern 1pt} {\kern 1pt} {\rm{mho {\kern 1pt}{\kern 1pt} c}}{{\rm{m}}^{ - 1}}$ . If the resistance of a cell containing this solution is 400 ohm, the cell constant is :

$1.106{\kern 1pt} {\kern 1pt} {\rm{c}}{{\rm{m}}^{ - 1}}$

$1.106{\kern 1pt} {\kern 1pt} {\rm {cm}}$

$1{\kern 1pt} {\kern 1pt} {\rm{cm}}$

$1{\kern 1pt} {\kern 1pt} {\rm{c}}{{\rm{m}}^{ - 1}}$

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

1 Medium

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Application +4 | -1

The conductivity of

N/50

solution of

\mathrm{KCl}

in a cell at 250C is

0.002765{\kern 1pt} {\kern 1pt} {\rm{mho {\kern 1pt}{\kern 1pt} c}}{\mathrm{m}^{ - 1}}

. If the resistance of a cell containing this solution is 400 ohm, the cell constant is :

View answer and explanation

Correct option: B

1.106{\kern 1pt} {\kern 1pt} \mathrm{c}{\mathrm{m}^{ - 1}}

Option B (

1.106{\kern 1pt} {\kern 1pt} \mathrm{c}{\mathrm{m}^{ - 1}}

) is correct because the cell constant is calculated by multiplying conductivity and resistance.