Solveeit Logo

Question

Question: Molar conductivities (\({{\Lambda }_{m}}^{o}\)) at infinite dilution of NaCl, HCl and \(C{{H}_{3}}CO...

Molar conductivities (Λmo{{\Lambda }_{m}}^{o}) at infinite dilution of NaCl, HCl and CH3COONaC{{H}_{3}}COONa are 126.4, 425.9 and 91.0 Scm2mol1Sc{{m}^{2}}mo{{l}^{-1}} respectively. Λmo{{\Lambda }_{m}}^{o} for CH3COOHC{{H}_{3}}COOH will be:
A. 180.5 Scm2mol1Sc{{m}^{2}}mo{{l}^{-1}}
B. 290.8 Scm2mol1Sc{{m}^{2}}mo{{l}^{-1}}
C. 390.5 Scm2mol1Sc{{m}^{2}}mo{{l}^{-1}}
D. 425.5 Scm2mol1Sc{{m}^{2}}mo{{l}^{-1}}

Explanation

Solution

. Molar conductivity is defined as the conductance property of a solution which contains one mole of the electrolyte or we can say that it generally tells about the ionic strength of a solution or concentration of the salt.

Complete step by step answer:
Molar conductivity can defined as the conducting power of all the ions that are formed by dissolving a mole of electrolyte in a solution or we can say it is the property of an electrolyte solution that is mainly used in determining the efficiency of a given electrolyte in conducting electricity in a solution. Molar conductivity of an electrolytic solution is the conductance of the volume of the solution containing a unit mole of electrolyte that is placed between two electrodes of unit area cross-section or at a distance of one-centimeter apart. The unit of molar conductivity is Scm2mol1Sc{{m}^{2}}mo{{l}^{-1}}. The molar conductivity of both weak and strong electrolytes increases with a decrease in concentration or dilution. Electrolytic conductance increases with increase in dilution because conductance of electrolyte is due to the presence of ions in the solution. The greater the number of ions the greater is the conductance. As with dilution more ions are produced in solution so conductance also increases. On dilution there are weak interactions between the ions so they can move freely which increases conductance.

ΛoNaCl=126.4Scm2mol1{{\Lambda }^{o}}_{NaCl}=126.4Sc{{m}^{2}}mo{{l}^{-1}}; ΛoHCl=425.9Scm2mol1{{\Lambda }^{o}}_{HCl}=425.9Sc{{m}^{2}}mo{{l}^{-1}}and ΛoCH3COONa=91.0Scm2mol1{{\Lambda }^{o}}_{C{{H}_{3}}COONa}=91.0Sc{{m}^{2}}mo{{l}^{-1}}
ΛoCH3COOH=ΛoCH3COONa+ΛoHClΛoNaCl{{\Lambda }^{o}}C{{H}_{3}}COOH={{\Lambda }^{o}}C{{H}_{3}}COONa+{{\Lambda }^{o}}HCl-{{\Lambda }^{o}}NaCl
ΛoCH3COOH=91+425.9126.4{{\Lambda }^{o}}C{{H}_{3}}COOH=91+425.9-126.4
ΛoCH3COOH=390.5Scm2mol1{{\Lambda }^{o}}C{{H}_{3}}COOH=390.5Sc{{m}^{2}}mo{{l}^{-1}}
So, the correct answer is “Option C”.

Note: In strong electrolytes increase in concentration produces a sharp increase in conductivity. However at lower concentration weak electrolytes possess very low values of specific conductivity and the value gradually increases as the concentration is increased. This is due to the increase in the number of active ions in the solution due to concentration.