Question

\[{\lambda _{{\rm{ClC}}{{\rm{H}}_{\rm{2}}}{\rm{COONa}}}} = 220\,{\rm{oh}}{{\rm{m}}^{ - 1}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{gm}}\,{\rm{e}}{{\rm{q}}^{ - 1}}\]
\[{\lambda _{{\rm{NaCl}}}} = 38.2\,{\rm{oh}}{{\rm{m}}^{ - 1}}{\rm{c}}{{\rm{m}}^2}{\rm{gm}}\,{\rm{e}}{{\rm{q}}^{ - 1}}\]
\[{\lambda _{{\rm{HCl}}}} = 203\,{\rm{oh}}{{\rm{m}}^{ - 1}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{gm}}\,{\rm{e}}{{\rm{q}}^{ - 1}}\]
What is the value of \[{{\rm{\lambda }}_{{\rm{ClC}}{{\rm{H}}_{\rm{2}}}{\rm{COOH}}}}\] ?
Option 1: \[288.5\,{\rm{oh}}{{\rm{m}}^{ - 1}}{\rm{c}}{{\rm{m}}^{\rm{2}}}\,{\rm{gm}}\,\,{\rm{e}}{{\rm{q}}^{ - 1}}\]
Option 2: \[298.5\,{\rm{oh}}{{\rm{m}}^{ - 1}}{\rm{c}}{{\rm{m}}^{\rm{2}}}\,{\rm{gm}}\,\,{\rm{e}}{{\rm{q}}^{ - 1}}\]
Option 3: \[388.8\,{\rm{oh}}{{\rm{m}}^{ - 1}}{\rm{c}}{{\rm{m}}^{\rm{2}}}\,{\rm{gm}}\,\,{\rm{e}}{{\rm{q}}^{ - 1}}\]
Option 4: \[59.5\,{\rm{oh}}{{\rm{m}}^{ - 1}}{\rm{c}}{{\rm{m}}^{\rm{2}}}\,{\rm{gm}}\,\,{\rm{e}}{{\rm{q}}^{ - 1}}\]

Answer 1

Hint: Kohlrausch's law defines that, on infinite dilution of an electrolyte, the molar conductivity of the electrolyte is the summation of the individual conductance of the anion and cation constituting the electrolyte.

Complete Step by Step Answer:
The mathematical representation of The Kohlrausch's law is,
\[{\lambda ^0}_{{\rm{AB}}} = {\lambda _{{{\rm{A}}^ + }}} + {\lambda _{{{\rm{B}}^ - }}}\]

Here, the electrolyte AB's molar conductivity on infinite dilution is the summation of conductance of conductivity of cation A and anion B.
Let's write the equation for all the given electrolytes using Kohlrausch's law.

For the first electrolyte, the equation is,
\[{\lambda ^0}_{{\rm{ClC}}{{\rm{H}}_{\rm{2}}}{\rm{COONa}}} = 224\,{\rm{oh}}{{\rm{m}}^{ - 1}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{gm}}\,{\rm{e}}{{\rm{q}}^{ - 1}} = {\lambda _{{\rm{ClC}}{{\rm{H}}_{\rm{2}}}{\rm{CO}}{{\rm{O}}^ - }}} + {\lambda _{{\rm{N}}{{\rm{a}}^ + }}}\]--- (1)

The molar conductivity is the summation of conductivities of sodium acetate ions and chloride ions.
Similarly, we write the equation of sodium chloride using Kohlrausch's law.

\[{\lambda _{{\rm{NaCl}}}} = 38.2\,{\rm{oh}}{{\rm{m}}^{ - 1}}{\rm{c}}{{\rm{m}}^2}{\rm{gm}}\,{\rm{e}}{{\rm{q}}^{ - 1}} = {\lambda _{{\rm{N}}{{\rm{a}}^ + }}} + {\lambda _{{\rm{C}}{{\rm{l}}^ - }}}\]--- (2)

So, Sodium chloride's molar conductivity is the addition of conductivities of sodium ions and chloride ions. Similarly, we have to write the equation of hydrogen chloride using Kohlrausch's law.
\[{\lambda _{{\rm{HCl}}}} = 203\,{\rm{oh}}{{\rm{m}}^{ - 1}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{gm}}\,{\rm{e}}{{\rm{q}}^{ - 1}} = {\lambda _{{{\rm{H}}^ + }}} + {\lambda _{{\rm{C}}{{\rm{l}}^ - }}}\]--- (3)

Here also, the molar conductivity is the summation of conductivities of hydrogen and chloride ions.
Now, if we add the equation (1) and (3) and subtract equation (2) from it, we get the value of \[{{\rm{\lambda }}_{{\rm{ClC}}{{\rm{H}}_{\rm{2}}}{\rm{COOH}}}}\].
\[{\lambda _{{\rm{ClC}}{{\rm{H}}_{\rm{2}}}{\rm{CO}}{{\rm{O}}^ - }}} + {\lambda _{{\rm{N}}{{\rm{a}}^ + }}} + {\lambda _{{{\rm{H}}^ + }}} + {\lambda _{{\rm{C}}{{\rm{l}}^ - }}} - {\lambda _{{\rm{N}}{{\rm{a}}^ + }}} - {\lambda _{{\rm{C}}{{\rm{l}}^ - }}} = \left( {224 + 203 - 38.2} \right){\rm{oh}}{{\rm{m}}^{ - 1}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{gm}}\,{\rm{e}}{{\rm{q}}^{ - 1}}\]
\[{\lambda _{{\rm{ClC}}{{\rm{H}}_{\rm{2}}}{\rm{CO}}{{\rm{O}}^ - }}} + {\lambda _{{{\rm{H}}^ + }}} = 427 - 38.2\]
\[{\lambda _{{\rm{ClC}}{{\rm{H}}_{\rm{2}}}{\rm{COOH}}}} = 388.8\,{\rm{oh}}{{\rm{m}}^{ - 1}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{gm}}\,{\rm{e}}{{\rm{q}}^{ - 1}}\]
Therefore, the value of \[{{\rm{\lambda }}_{{\rm{ClC}}{{\rm{H}}_{\rm{2}}}{\rm{COOH}}}}\]is \[388.8\,{\rm{oh}}{{\rm{m}}^{ - 1}}{\rm{c}}{{\rm{m}}^{\rm{2}}}\,{\rm{gm}}\,\,{\rm{e}}{{\rm{q}}^{ - 1}}\].
Hence, option C is right.

Note: It is to be noted that, the power of conductance of all the produced ions, on dissolving an electrolyte of one mole in a solution is termed molar conductivity. It is represented in chemistry by the symbol of \[{ \wedge _m}\] and its SI unit is \[{\rm{S c}}{{\rm{m}}^{ - 1}}\].