Question

The conductance of a salt solution \((A B)\) measured by two parallel electrodes of area \(100 \mathrm{~cm}^{2}\) separated by \(10 \mathrm{~cm}\) was found to be \(0.0001 \Omega^{-1}\). If volume enclosed between two electrode contain \(0.1\) mole of salt, what is the molar conductivity \(\left(\mathrm{Scm}^{2} \mathrm{~mol}^{-1}\right)\) of salt at same concentration : (a) 10 (b) \(0.1\) (c) 1 (d) none of these

Short answer

\(\Lambda_m = 10 \text{S cm}^2 \text{mol}^{-1}\), so the answer is (a) 10.

Step-by-step solution

Understand Molar Conductivity

Molar conductivity \(\Lambda_m\) of a solution is defined as the conductance of the solution \(G\) multiplied by the volume \(V\) in which one mole of electrolyte is dissolved, divided by the area \(A\) of the electrodes and the distance \(d\) between them. The formula for molar conductivity is \(\Lambda_m = \frac{G \times V}{A \times d}\), where G is given in \(\Omega^{-1}\), V in cubic centimeters \(\text{cm}^3\), A in square centimeters \(\text{cm}^2\), and d in centimeters \(\text{cm}\).

Retrieve Given Values

We have the following values given in the problem: Conductance \(G = 0.0001 \Omega^{-1}\), Electrode area \(A = 100 \text{cm}^2\), Distance between electrodes \(d = 10 \text{cm}\), Moles of salt \(n = 0.1 \text{mol}\), and we need to find the volume \(V\) of the solution enclosed between the electrodes.

Calculate Volume of Solution

Since we know only 0.1 mole of salt is dissolved in the solution, to find the volume \(V\) in which this amount of salt is dissolved, we consider the volume that contains exactly one mole of the salt. Since we are dealing with 0.1 mole, the volume will be 10 times larger than the volume that would contain 1 mole. Therefore, \(V = 10 \times 1000 \text{cm}^3 = 10000 \text{cm}^3\).

Calculate Molar Conductivity

Now using the formula for molar conductivity we plug in the values: \(\Lambda_m = \frac{G \times V}{A \times d} = \frac{0.0001 \times 10000}{100 \times 10}\). Calculating the values we get \(\Lambda_m = \frac{1}{100} = 0.01 \text{S cm}^2 \text{mol}^{-1}\). Since molar conductivity is usually given in \(\text{S} \text{cm}^2 \text{mol}^{-1}\) or \(\text{mS} \text{cm}^2 \text{mol}^{-1}\), we convert it to \(\text{S} \text{cm}^2 \text{mol}^{-1}\) by multiplying with 1000 and obtain \(\Lambda_m = 10 \text{S cm}^2 \text{mol}^{-1}\).

Key concepts

Electrolyte Conductance

Electrolyte conductance is a measure of an electrolyte's ability to conduct electricity in a solution. It is dependent on factors such as the type of ions, their concentration, and the temperature of the solution. Electrolytes are substances that dissociate into ions when dissolved in water, becoming capable of conducting electric current.

In a simple setup, when an electrolyte is added to water, it splits into positive and negative ions. These charged particles are what carry electric currents as they move towards electrodes of opposite charge when a potential difference is applied across the solution. The movement of these ions and their ability to carry charge is what we refer to as the 'electrolyte conductance'. In practice, measuring electrolyte conductance involves using a conductivity meter with electrodes to detect the flow of electricity.

Electrochemistry

Electrochemistry is the branch of physical chemistry that studies the relationship between electricity, as a measurable and quantitative phenomenon, and identifiable chemical change, with either electricity considered an outcome of a particular chemical change or vice versa. It encompasses a variety of phenomena, including electrolysis, battery operation, and corrosion.

• Electrolysis is the process in which electrical energy is used to drive a non-spontaneous chemical reaction.
• Batteries convert stored chemical energy into electrical energy.
• Corrosion involves electrochemical reactions that lead to the degradation of materials.

Understanding the principles of electrochemistry is essential for various applications such as energy storage, electroplating, and sensors.

Physical Chemistry JEE

In the context of the Joint Entrance Examination (JEE), which is a prestigious entrance exam for various engineering colleges in India, physical chemistry is a crucial subject. It delves into topics bridging physics and chemistry that explain the behavior of molecules and atoms, the basic components of matter.

For JEE aspirants, mastering concepts like molar conductivity, thermodynamics, chemical kinetics, electrochemistry, and equilibrium is essential. These topics not only require a deep understanding of theoretical concepts but also the ability to apply them to solve complex problems. The exercise provided is an example where students learn to apply the concept of molar conductivity to a given problem, a skill frequently tested in the JEE physical chemistry section.

Solution Conductivity

Solution conductivity refers to the ability of a solution to conduct an electric current. This property is critical in electrochemistry and is influenced by the ionic composition of the solution – the presence of ions, their mobility, and their concentration. Conductivity is often measured in siemens per meter (S/m) and is defined as the inverse of the electrical resistance.

Solution conductivity is crucial in various industries and technologies, including water purification, the manufacturing of electronic devices, and monitoring biochemical processes. In an educational context, understanding conductivity is useful to predict how substances will behave in different environments and to design systems that rely on the movement of ions through a medium.

Chemical Solution Calculations

Chemical solution calculations are quantitative evaluations used in chemistry to determine various properties of solutions, such as concentration, molarity, and molar conductivity. These calculations often require a clear understanding of concepts like Avogadro's number, the molar mass of substances, and the principles that govern the behavior of ions in a solution.

In the exercise provided, for example, using the correct formula for molar conductivity and ensuring proper unit conversion are essential to find the accurate molar conductivity of the salt solution. This exercise showcases the importance of not only understanding the formulas but also of accurately retrieving and applying the information given. Careful attention to units and basic arithmetics are key in performing these calculations successfully.