The conductivity of a $0.1N{\kern 1pt} {\rm{KCl}}$ solution at 230C is 0.012 ${\rm{oh}}{{\rm{m}}^{ - 1}}{\kern 1pt}{\kern 1pt}{\rm{c}}{{\rm{m}}^{ - 1}}$ .The resistance of the cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be :

0.66 ${\rm{c}}{{\rm{m}}^{ - 1}}$

0.918 ${\rm{c}}{{\rm{m}}^{ - 1}}$

1.142 ${\rm{c}}{{\rm{m}}^{ - 1}}$

1.12 ${\rm{c}}{{\rm{m}}^{ - 1}}$

JEE CHEMISTRY MCQ: ELECTRO CHEMISTRY

1 Medium

1 of 1

Application +4 | -1

The conductivity of a

0.1N{\kern 1pt} \mathrm{KCl}

solution at 230C is 0.012

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

.The resistance of the cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be :

View answer and explanation

Correct option: B

0.66

\mathrm{c}{\mathrm{m}^{ - 1}}

Option B (0.66

\mathrm{c}{\mathrm{m}^{ - 1}}

) is incorrect. The correct calculation using the conductivity and resistance yields 0.66

\mathrm{c}{\mathrm{m}^{ - 1}}

Practice All Chapter MCQs →

25^{\circ} \mathrm{C}

0.295 \mathrm{~V}

{p_1}{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} {\kern 1pt} {p_2}

\mathop {\mathrm{Pt}\left( {{\mathrm{H}_2}} \right)}\limits_{{p_1}} \left| {\mathop {{\mathrm{H}^ + }\left( {aq} \right)}\limits_{1\mathrm{ }M} } \right|\mathop {Pt\left( {{\mathrm{H}_2}} \right)}\limits_{{p_2}}

\mathrm{CuS}{\mathrm{O}_4}

{AlCl}_3

STP

({F}=96500 {C})

2 \mathrm{Ag}(\mathrm{s})+\mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{Ag}_2 \mathrm{S}(\mathrm{s})+2 \mathrm{e}^{-}

\mathrm{pH}=3

0.1 \mathrm{~M}-\mathrm{H}_2 \mathrm{S},

\mathrm{H}_2 \mathrm{S}, K_{\mathrm{a}_1}=10^{-8} ; K_{\mathrm{a}_2}=10^{-13}, K_{\mathrm{sp}}

\mathrm{Ag}_2 \mathrm{S}=4 \times 10^{-48},

E^{\circ}_{{\mathrm{Ag} | \mathrm{Ag}}^{+}}=0.80 \mathrm{~V},

\log 2=0.3,

2.303 R T / F=0.06