$0.05$ M NaOH solution offered a resistance of $31.6$ ohm in a conductivity cell at $298$ K. If the cell constant of the cell is $0.367 c m^{- 1}$ , calculate the molar conductivity of the NaOH solution.

Open in App

Solution

Given: Molarity of NaOH solution $= 0.05$ m
Resistance $= 31.6$ ohm
Cell constant $= 0.367 c m^{- 1}$
Now, Molar conductivity $= \frac{Specific conductivity \times 1000}{M o l a r i t y}$
$= \frac{C e l l c o n s t a n t}{R e s i s t a n c e} \times \frac{1000}{M o l a r i t y}$

$= \frac{0.367 \times 1000}{31.6 \times 0.05}$

$\land_{m} = 232.3 o h m^{- 1} c m^{2} m o l^{- 1}$ .

Suggest Corrections

Similar questions

0.5 M N a O H

31.6 o h m

N a O H

0.367 c m^{- 1}

0.05 M N a O H

36.1 Ω

0.357 c m^{- 1}

0.05 M

N H_{4} O H

50 o h m

298 K

0.50 {c m}^{- 1}

N H_{4} O H

471.4 {o h m}^{- 1} {c m}^{2} {m o l}^{- 1}

c m^{- 1}

The resistance of $0.01 M$ NACl solution is $200 o h m$ . The cell constant of conductivity cell is unity. Calculate the molar conductance.

Join BYJU'S Learning Program