Question

The electrical resistance of a column of 0.05 M KOH solution of diameter 1 cm and length 45.5 cm is 4.55 ×1000 ohm . Calculate its molar conductivity

Answer 1

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Answer 2

Answer: $255Scm^2mol^{-1}$

Explanation:

Specific conductance is the conductance for unit volume of a solution.

$\kappa=C\times \frac{l}{a}$

where,

$\kappa$ =Specific conductance

C = Conductance =$\frac{1}{Resistance}=\frac{1}{4.55\times 10^{3}ohm}=0.22\times 10^{-3}S$

$\frac{l}{a}$= cell constant = $\frac{l}{\pi r^2}cm^{-1}=\frac{45.5cm}{3.14\times (0.5)^2cm^2}=57.96cm^{-1}$      ($r=\frac{d}{2}$)

$\kappa=0.22\times 10^{-3}S\times 57.96cm^{-1}=12.75\times 10^{-3}Scm^{-1}$

Molar conductance is defined as the conductance of all the ions produced by ionization of 1 g mole of an electrolyte when present in V mL of solution.

$\mu=\frac{\kappa\times 1000}{C}$

where,  

$\mu$= molar conductance

$\kappa$ =Specific conductance

c = concentration of the solution in mole per litre

$\mu=\frac{12.75\times 10^{-3}Scm^{-1}\times 1000}{0.05M}=255Scm^2mol^{-1}$