The resistance of N/10 solution is found to be 2.5× 103 ohm. The equivalent conductance of the solution is (cell constant = 1.25 cm-1 )

Question

The resistance of N/10 solution is found to be 2.5× 103 ohm. The equivalent conductance of the solution is (cell constant = 1.25 cm-1 ) (A)  2.5 ohm-1cm2equiv-1  (B)  5.0 ohm-1cm2equiv-1  (C)  2.5 ohm-1cm2equiv-1  (D)  5.0 ohm-1cm2equiv-1

Tardigrade Admin · Accepted Answer

Resistance of N/10 solution =2.5× 103Ω kappa =(1/ textresistance)× textcell textconstant =(1/2.5× 103)× 1.25 =(1.25× 10-3/2.5)=5× 10-4 textoh textm-1 textc textm-1 Equivalent conductance =( kappa × 1000/M) =(5× 10-4× 1000/1/10) =5 textoh textm-1 textc textm2 textequ texti-1

Q. The resistance of N/10 solution is found to be $2.5 \times 10^{3}$ ohm. The equivalent conductance of the solution is (cell constant = 1.25 $c m^{- 1}$ )

$2.5 o h m^{- 1} c m^{2} e q u i v^{- 1}$

$5.0 o h m^{- 1} c m^{2} e q u i v^{- 1}$

$2.5 o h m^{- 1} c m^{2} e q u i v^{- 1}$

$5.0 o h m^{- 1} c m^{2} e q u i v^{- 1}$

Q. The resistance of N/10 solution is found to be 2.5×103 ohm. The equivalent conductance of the solution is (cell constant = 1.25 cm−1 )

2.5ohm−1cm2equiv−1

5.0ohm−1cm2equiv−1

2.5ohm−1cm2equiv−1

5.0ohm−1cm2equiv−1

Resistance of N/10 solution =2.5×103Ω κ=resistance1​×cellconstant =2.5×1031​×1.25 =2.51.25×10−3​=5×10−4ohm−1cm−1 Equivalent conductance =Mκ×1000​ =1/105×10−4×1000​ =5ohm−1cm2equi−1

Q. The resistance of N/10 solution is found to be $2.5 \times 10^{3}$ ohm. The equivalent conductance of the solution is (cell constant = 1.25 $c m^{- 1}$ )

$2.5 o h m^{- 1} c m^{2} e q u i v^{- 1}$

$5.0 o h m^{- 1} c m^{2} e q u i v^{- 1}$

$2.5 o h m^{- 1} c m^{2} e q u i v^{- 1}$

$5.0 o h m^{- 1} c m^{2} e q u i v^{- 1}$