0.01 solution of NaCl offered a resistance of 200 Ohm in a conductivit

1.	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» Answer: Molar CONDUCTIVITY of the SOLUTION is 500 S cm² mol⁻¹ Given: Molarity (M) = 0.01 M Resistance (R) = 200 Ω 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. $\rule{300}{1.5}$

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»

Answer:

Molar CONDUCTIVITY of the SOLUTION is 500 S cm² mol⁻¹

Given:

Molarity (M) = 0.01 M
Resistance (R) = 200 Ω
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.

$\rule{300}{1.5}$

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.

$\rule{300}{1.5}$