Discussion Forum



1)

The equivaent conductivity of a solution containing 2.54 g of $CuSO_{4}$ per L is 91.0 $W^{-1} cm^{2} eq^{-1}$ . Its conductivity would be 


A) $2.9 \times 10^{-3} \Omega^{-1}cm^{-1}$

B) $1.8 \times 10^{-2} \Omega^{-1}cm^{-1}$

C) $2.4 \times 10^{-4} \Omega^{-1}cm^{-1}$

D) $3.6 \times 10^{-3} \Omega^{-1}cm^{-1}$

Answer:

Option A

Explanation:

We know that,

  $k=\wedge_{eq}.C$

Given,  $\wedge eq=91.0 \Omega^{-1}cm^{2} eq^{-1}$

$k=(91 \Omega^{-1} cm^{2} eq^{-1})$

                                           $\left(\frac{2.54}{159/2\times1000}eq.cm^{-3}\right)$

 $=2.9 \times 10^{-3}\Omega^{-1}cm^{-1}$