Question

The resistance of 1 N solution of acetic acid is 250 ohm,
when measured in a cell of cell constant 1.15 cm⁻¹. The
equivalent conductance (in ohm⁻¹ cm² equiv⁻¹) of 1 N acetic
acid will be
(a) 4.6 (b) 9.2
(c) 18.4 (d) 0.023

Answer 1

Explanation:

The

equivalent conductance (in ohm⁻¹ cm² equiv⁻¹) of 1 N acetic

acid will be

option (a) 4.6 answer

Answer 2

The  equivalent conductance (in ohm⁻¹ cm² equiv⁻¹) of 1 N acetic  acid will be (a) 4.6 ohm⁻¹ cm² equiv⁻¹

Given:

Resistance of solution = 250 ohm

Cell constant = 1.15 cm⁻¹

To find:

Equivalent conductance = ?

Solution:

The equivalent conductance is given by the formula:

$\Lambda_{e q}=\frac{k \times 1000}{N}$

Where,

k = Reciprocal of resistivity

N = Molarity

Now, the reciprocal of resistivity is given by the formula:

$\mathrm{k}=\frac{1}{R} \times G$

Where,

R = Resistance

G = Cell constant

On substituting the values, we get,

$k=\frac{1}{250} \times 1.15$

$\therefore k=0.0046 \ S C m^{-1}$

Now, on substituting the values in equivalent conductance formula, we get,

$\Lambda_{e q}=\frac{0.0046 \times 1000}{1}$

$\therefore \Lambda_{e q}=4.6 \ \mathrm{ohm}^{-1} \mathrm{cm}^{2} \mathrm{eq}^{-1}$