/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 Specific conductance of \(0.01 \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Specific conductance of \(0.01 \mathrm{~N}\) solution of an electrolyte is \(0.00419 \mathrm{mho} \mathrm{cm}^{-1}\). The equivalent conductance of this solution will be (a) \(4.19 \mathrm{mho} \mathrm{cm}^{2}\) (b) \(419 \mathrm{mh} \mathrm{cm}^{2}\) (c) \(0.0419 \mathrm{mho} \mathrm{cm}^{2}\) (d) \(0.209 \mathrm{mho} \mathrm{cm}^{2}\)

Short Answer

Expert verified
The equivalent conductance of this solution is (b) 419 mho cm².

Step by step solution

01

Identify the Given Values

We are given that the specific conductance (also known as conductivity), \( k \), is \( 0.00419 \text{ mho cm}^{-1} \) and the concentration of the solution is \( 0.01 \text{ N} \).
02

Use the Formula for Equivalent Conductance

The formula for equivalent conductance, \( \Lambda_{eq} \), is \( \Lambda_{eq} = \frac{k \times 1000}{c} \), where \( c \) is the concentration in normality (equivalents per liter).
03

Substitute the Given Values into the Formula

Substitute \( k = 0.00419 \) and \( c = 0.01 \) into the formula: \[ \Lambda_{eq} = \frac{0.00419 \times 1000}{0.01} \].
04

Calculate Equivalent Conductance

Perform the calculation: \[ \Lambda_{eq} = \frac{0.00419 \times 1000}{0.01} = 419 \text{ mho cm}^2 / \text{equiv} \].
05

Choose the Correct Option

Comparing the calculated value with the options provided, the equivalent conductance is \( 419 \text{ mh cm}^2 \). This matches with option (b).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Conductance
Specific conductance, often just called conductivity, is a measure of how well a solution can conduct electricity. It gives us an idea of the number of ions available to carry electric charge. The more ions present, the higher the conductivity.Conductivity is expressed in units of \(\text{mho cm}^{-1}\) or \(\text{Siemens cm}^{-1}\). For example, if a solution has specific conductance of \(0.00419 \ \text{mho cm}^{-1}\), it tells us the solution’s ability to conduct electricity over a centimeter.In practice, specific conductance is vital for understanding the behavior of electrolyte solutions in various chemical processes and applications, from batteries to biochemistry.
Electrolyte Solution
An electrolyte solution is a liquid that contains ions and can conduct electricity. This concept is at the heart of many chemical and biological processes. Here are a few key points to understand about electrolyte solutions: - **Composition**: - Typically involves water as a solvent and dissolved ions like sodium, potassium, chloride and others. - Commonly formed by dissolving salts, acids, or bases in water. - **Functionality**: - In electrolyte solutions, the dissolved ions allow the solution to conduct electricity. - **Examples**: - Saline solution (common in medical use). - Seawater – naturally available electrolytic solution. Since ions carry electric current, the presence of electrolyte is crucial for any process that involves the flow of electric current through a liquid.
Conductivity Formula
The conductivity formula is a fundamental formula in electrochemistry that connects specific conductance, concentration, and equivalent conductance.The formula for equivalent conductance is given by:\[ \Lambda_{eq} = \frac{k \times 1000}{c} \]where:- \(\Lambda_{eq}\) is the equivalent conductance in \(\text{mho cm}^2/\text{equiv}\).- \(k\) is the specific conductance in \(\text{mho cm}^{-1}\).- \(c\) is the concentration in normality (N).This relation helps in transforming the specific conductance to a per unit measure for the ions, making it easier to compare different solutions regardless of their concentration. This is incredibly useful for figuring out how efficient an electrolyte solution is under different conditions.
Normality
Normality is a term used to describe the concentration of a solution in terms of equivalents per liter. An equivalent is the amount of substance that reacts with or replaces a standard quantity, like a hydrogen ion in an acid-base reaction.A few points to consider for a clearer understanding:- **Calculation**: - Normality (N) is calculated as \(\text{moles of solute} \times \text{valency factor}\).- **Uses**: - It is useful in redox reactions and acid-base equilibrium calculations. - Not as commonly used as molarity but when dealing with reactions involving ions or charge, it’s very helpful. - Allows direct use in measuring volumes for reactions due to one-to-one correspondence with reacting equivalents.Normality helps ensure precise and accurate measurements in titrations and chemical analysis where the reactivity of a solute is influenced by the amount of charge.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When an electric current is passed through acidulated water, \(112 \mathrm{~mL}\) of hydrogen gas at NTP collects at the cathode in 965 seconds. The current passed, in ampere is (a) \(0.1\) (b) \(0.5\) (c) \(1.0\) (d) \(2.0\)

A solution containing one mole per litre of each \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} ; \mathrm{AgNO}_{3} ; \mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2} ;\) is being electrolyzed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are \(\mathrm{Ag} / \mathrm{Ag}^{+}=+0.80,2 \mathrm{Hg} / \mathrm{Hg}_{2}^{+4}=+0.79\) \(\mathrm{Cu} / \mathrm{Cu}^{++}=+0.34, \mathrm{Mg} / \mathrm{Mg}^{++}=-2.37\) With increasing voltage, the sequence of deposition of metals on the cathode will be (a) \(\mathrm{Cu}, \mathrm{Hg}, \mathrm{Ag}\) (b) \(\mathrm{Ag}, \mathrm{Hg}, \mathrm{Cu}, \mathrm{Mg}\) (c) \(\mathrm{Mg}, \mathrm{Cu}, \mathrm{Hg}, \mathrm{Ag}\) (d) \(\mathrm{Ag}, \mathrm{Hg}, \mathrm{Cu}\)

Two faraday of electricity is passed through a solution of \(\mathrm{CuSO}_{4}\). The mass of copper deposited at the cathode is (at. mass of \(\mathrm{Cu}=63.5 \mathrm{amu}\) ) (a) \(0 \mathrm{~g}\) (b) \(63.5 \mathrm{~g}\) (c) \(2 \mathrm{~g}\) (d) \(127 \mathrm{~g}\)

In the electrolytic cell, flow of electrons is from (a) cathode to anode in solution (b) cathode to anode through external supply (c) cathode to anode through internal supply (d) anode to cathode through internal supply

A current of \(15 \mathrm{amp}\) is employed to plate Nickel in a \(\mathrm{NiSO}_{4}\) bath. Both \(\mathrm{Ni}\) and \(\mathrm{H}_{2}\) are formed at the cathode. If \(9.9 \mathrm{~g}\) of \(\mathrm{Ni}\) are deposited with the simultaneous liberation of \(2.51\) litres of \(\mathrm{H}_{2}\) measured at STP, what is the current efficiency for the deposition of Ni? (Atomic weight of \(\mathrm{Ni}=58.7\) ) (a) \(60 \%\) (b) \(70 \%\) (c) \(80 \%\) (d) \(56 \%\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.