The equivalent conductivity of a solution containing 2.54 g of ${\rm{CuS}}{{\rm{O}}_4}{\kern 1pt} {\kern 1pt} {\rm{per L is}}{\kern 1pt} {\kern 1pt} {\kern 1pt} 91.0{{\rm{\Omega }}^{ - 1}}{\rm{c}}{{\rm{m}}^2}{\rm{e}}{{\rm{q}}^{ - 1}}$. Its conductivity would be

$2.9 \times {10^{ - 3}}{{\rm{\Omega }}^{ - 1}}{\rm{c}}{{\rm{m}}^2}$

$1.8 \times {10^{ - 2}}{{\rm{\Omega }}^{ - 1}}{\rm{c}}{{\rm{m}}^2}$

$2.4 \times {10^{ - 4}}{{\rm{\Omega }}^{ - 1}}{\rm{c}}{{\rm{m}}^2}$

$3.6 \times {10^{ - 3}}{{\rm{\Omega }}^{ - 1}}{\rm{c}}{{\rm{m}}^2}$

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

1 Easy

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Application +4 | -1

The equivalent conductivity of a solution containing 2.54 g of

\mathrm{CuS}{\mathrm{O}_4}{\kern 1pt} {\kern 1pt} \mathrm{per L is}{\kern 1pt} {\kern 1pt} {\kern 1pt} 91.0{\mathrm{\Omega }^{ - 1}}\mathrm{c}{\mathrm{m}^2}\mathrm{e}{\mathrm{q}^{ - 1}}

. Its conductivity would be

View answer and explanation

Correct option: A

2.9 \times {10^{ - 3}}{\mathrm{\Omega }^{ - 1}}\mathrm{c}{\mathrm{m}^2}

Option A (

2.9 \times {10^{ - 3}}{\mathrm{\Omega }^{ - 1}}\mathrm{c}{\mathrm{m}^2}

) is correct because conductivity is calculated by dividing equivalent conductivity by 1000 and multiplying by the concentration in normality.