Question

0.01 solution of NaCl offered a resistance of 200 Ohm in a conductivity cell at 298 Kelvin if the cell constant of the cell is unity calculate molar conductivity of the solution.

Answer 1

hyy mate ✌️

κ =

R

cellconstant

κ = 200^1

∧ m =κ M1000

∧ m = 0.01∗200^1000

^ m = 500 Scm 2 mol −1

hope it will help you ❣️❣️

Answer 2

Answer:

• Molar Conductivity of the solution is 500 S cm² mol⁻¹

Given:

1. Molarity (M) = 0.01 M
2. Resistance (R) = 200 Ω
3. Cell constant = 1 cm⁻¹

Explanation:

$\rule{300}{1.5}$

We Know from the formula,

⇒ Cell constant = κ × R

Here,

• κ Denotes conductivity.
• R Denotes resistance.

⇒ 1 = κ × 200

⇒ κ = 1/200

⇒ κ = 1/200 S cm⁻¹

∴ Conductivity (κ) of the solution is 1/200 S cm⁻¹.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the formula we know,

$\bigstar\;\underline{\boxed{\sf \Lambda_{m}=\dfrac{\kappa \times 1000}{M}}}$

Here,

• κ Denotes conductivity.
• M Denotes Molarity.

Substituting the values,

$\longrightarrow\sf \Lambda_{m}=\dfrac{\frac{1}{200} \times 1000}{0.01}\\\\\\\longrightarrow\sf \Lambda_{m}=\dfrac{1\times 1000}{200\times 0.01}\\\\\\\longrightarrow\sf \Lambda_{m}=\dfrac{1000}{2}\\\\\\\longrightarrow\sf \Lambda_{m}=500\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf \Lambda_{m}=500\;Scm^{2}mol^{-1}}}}}$

∴ Molar Conductivity of the solution is 500 S cm² mol⁻¹.

$\rule{300}{1.5}$