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The correct order of equivalent conductance at infinite dilution of \(\mathrm{LiCl}, \mathrm{NaCl}\) and \(\mathrm{KCl}\) is (a) \(\mathrm{LiCl}>\mathrm{NaCl}>\mathrm{KCl}\) (b) \(\mathrm{KCl}>\mathrm{NaCl}>\mathrm{LiCl}\) (c) \(\mathrm{NaCl}>\mathrm{KCl}>\mathrm{LiCl}\) (d) \(\mathrm{LiCl}>\mathrm{KCl}>\mathrm{NaCl}\)

Short Answer

Expert verified
(b) \(\mathrm{KCl} > \mathrm{NaCl} > \mathrm{LiCl}\).

Step by step solution

01

Understanding Equivalent Conductance

Equivalent conductance at infinite dilution ( abla_r^0) depends on the size and mobility of ions in the solution. Smaller ions usually move faster, leading to higher conductance.
02

Analyzing Ion Sizes

In the case of LiCl, NaCl, and KCl, the cations are Li鈦, Na鈦, and K鈦. Among these, Li鈦 is the smallest ion, followed by Na鈦, and the largest is K鈦.
03

Relating Ion Size to Conductance

Ion mobility generally increases with the size of the ion's solvation shell, where the smallest ion has the largest solvation shell and thus the highest conductance at infinite dilution.
04

Conclusion on Conductance Order

Based on the concept that larger solvated ions have higher equivalent conductance, the order of equivalent conductance is typically abla_r^0 (KCl) > abla_r^0 (NaCl) > abla_r^0 (LiCl).

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Key Concepts

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

Infinite Dilution's Role in Conductance
Infinite dilution is a critical concept when discussing equivalent conductance. At infinite dilution, a solution contains a very low concentration of solute, so low that the solute's ions are far apart and do not interact with each other. This allows each ion to move freely, minimizing interference and maximizing conductance.

When ions can move without hindrance from each other, the measured conductance reflects the intrinsic properties of the individual ions in the solution. Therefore, measuring conductance at this point helps understand the ion's behavior when they are essentially isolated, giving a clearer view of their true conductance potential.

Studying conductance under the conditions of infinite dilution allows us to focus on how factors like ion size and solvation affect mobility without the complications of ion-ion interactions.
Understanding Ion Mobility
Ion mobility is how quickly an ion moves through a solution when influenced by an electric field. Mobility can significantly impact the conductance observed at infinite dilution.

Several factors affect ion mobility:
  • Ion Size: Smaller ions can move more quickly due to less resistance. However, they often carry more water molecules around them, forming a solvation shell which impacts their effective size.
  • Charge: Higher charged ions can be more strongly influenced by the electric field, increasing mobility.
In the real world, the smallest cation is often the most mobile due to smaller physical size and larger solvation shells, which allow them to move through the solution more efficiently than larger ions.
Impact of Ion Solvation
Ion solvation involves the surrounding of an ion by solvent molecules. It significantly affects properties like mobility and conductance.

  • Solvation Shell Size: Small ions have larger solvation shells because they have higher charge density. A larger shell makes them bigger in the solution, impacting how they move.
  • Hydration: In aqueous solutions, water molecules orient themselves around ions, forming hydration shells. This interaction stabilizes ions but also alters their effective size and mobility.
Understanding solvation helps us see that even though an ion might be small, the 'bigger picture' involves how their charge interacts with surrounding molecules, vastly influencing conductance.
Cations in Solution and Their Behavior
Cations are positively charged ions and their interactions in solutions can be quite complex. Their size, solvation, and mobility all determine how they influence conductance.

When looking at specific cations like Li鈦, Na鈦, and 碍鈦:
  • 尝颈鈦: Small size but large hydration shell makes it less mobile than expected.
  • 狈补鈦: Medium size and medium solvation effect leads to moderate mobility.
  • 碍鈦: Larger size but less extensive solvation facilitates higher mobility.
Cations like these help us understand the intricate dance between size and solvation. It's not just about the physical size of the ion, but the whole solvation complex that determines their path through a solution.

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Most popular questions from this chapter

The reversible reduction potential of pure water is \(-0.413 \mathrm{~V}\) under \(1.00 \mathrm{~atm} \mathrm{H}_{2}\) pressure. If the reduction is considered to be \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}\), calculate \(\mathrm{pH}\) of pure water. (a) 6 (b) 7 (c) 3 (d) 5

One faraday of electricity is passed separately through one litre of one molar aqueous solutions of (i) \(\mathrm{AgNO}_{3}\) (ii) \(\mathrm{SnCl}_{4}\) and (iii) \(\mathrm{CuSO}_{4}\). The number of moles of \(\mathrm{Ag}, \mathrm{Sn}\), and \(\mathrm{Cu}\) deposited at cathode are respectively (a) \(1.0,0.25,0.5\) (b) \(1.0,0.5,0.25\) (c) \(0.5,1.0,0.5\) (d) \(0.25,0.25,0.5\)

\(2 \mathrm{MnO}_{4}^{-}+\mathrm{aH}^{+}+\mathrm{bH}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{Mn}^{+2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{dO}_{2}\) What are the values of \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) and \(\mathrm{d} ?\) (a) \(4,5,4,5\) (b) \(6,5,8,5\) (c) \(3,2,4,8\) (d) \(2,5,4,6\)

Aluminium oxide may be electrolysed at \(1000^{\circ} \mathrm{C}\) to furnish aluminium metal (atomic mass \(=27\) amu; 1 faraday \(=965000\) coulombs). The cathode reaction is \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}\) To prepare \(5.12 \mathrm{~kg}\) of aluminium metal by this method would require \([2008]\) (a) \(5.49 \times 10^{7} \mathrm{C}\) of electricity (b) \(1.83 \times 10^{7} \mathrm{C}\) of electricity (c) \(5.49 \times 10^{4}\) C of electricity (d) \(5.49 \times 10^{10} \mathrm{C}\) of electricity

Match the following Column-I \(\quad\) Column-II (a) Charge on one mole of (p) 1 Faraday electron. (b) \(108 \mathrm{~g}\) of silver deposited (q) 96500 coulomb at electrode. $$ \begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \text { (c) 22.4 L of hydrogen at } & \text { (r) 2 Faraday } \\ \text { STP collected. } & \\ \text { (d) } 8 \mathrm{~g} \text { of oxygen collected. } & \text { (t) } 5.6 \mathrm{~L} \text { at STP } \\ \hline \end{array} $$

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