Question

0.05 m Naoh solution offered a resistance of 31.6 ohm in a conductivity cell at 298k.if the cell constant of tge cell is 0.367cm, calculated the molar condivity of Naoh solution

Answer 1

Molar conductivity= k×1000/M
conductivity (k)= cell constant×1/resistance
=0.367×1/31.6=0.011
^M= 0.011×1000/0.05=220

Answer 2

Answer:

The molar conductivity of NaOH solution is $232.27 S cm^2/ mol$.

Explanation:

Concentration of the solution = 0.05 M

Resistance offered by the solution = $R=31.6\Omega$

Cell constant = k =$0.367 cm^{-1}$

Conductivity of the solution = $\kappa$

$\kappa =\fracx{1}{R}\times k=\frac{1}{31.6 \Omega}\times 0.367 cm^{-1}=0.01161 S cm^{-1}$

Molar conductivity of the NaOH solution  :

$\Lambda _m=\frac{\kappa \times 1000}{Molarity}=\frac{0.01161 S cm^{-1}\times 1000}{0.05M}=232.27 S cm^2/ mol$

The molar conductivity of NaOH solution is $232.27 S cm^2/ mol$.