
Equivalent conductance of \[1M\] $C{H_3}COOH$ is \[10\;oh{m^{ - 1}}c{m^2}equi{v^{ - 1}}\;\] and that at infinite dilution is \[200\;oh{m^{ - 1}}c{m^2}equi{v^{ - 1}}\;\]. Hence % ionisation of $C{H_3}COOH$ is __________.
(A) $5\% $
(B) $2\% $
(C) $4\% $
(D) $1\% $
Answer
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Hint: Equivalent conductance is defined as the conductance by \[1\] gram equivalent of the electrolytic solution. Molar conductance is defined as the conductance by \[1\] mole of electrolytic solution. When the solvent is diluted infinite times such that volume tends to infinity and concentration tends to zero, then the conductance defined is known as conductance at infinite dilution.
Complete Step by Step Solution:
An important application of Kohlrausch law is to find the degree of dissociation or the degree of ionisation of weak electrolytes. For example, acetic acid $C{H_3}COOH$
Degree of dissociation, \[\;\alpha \]is defined as the ratio of dissociated amount to the initial amount.
As the value of\[\;\alpha \]increases, the number of ions also increases and thus, \[\;{\lambda _m}\]also increases. So, Degree of dissociation is related to conductance by the formula\[\;\alpha = \frac{{{\lambda _m}}}{{{\lambda _m}^\infty }}\] .
But the value of equivalent conductance is given in the question. Hence first convert equivalent conductance in molar conductance by the formula \[\;n = \frac{{{\lambda _m}}}{{{\lambda _{_{eq}}}}}\]where n is the n-factor of the solute.
N- factor for $C{H_3}COOH$is \[1\] as the number of replaceable hydrogen ion is one). Thus, for $C{H_3}COOH$the molar conductance is the same as the equivalent conductance.
Hence, molar conductance\[10\;oh{m^{ - 1}}c{m^2}equi{v^{ - 1}}\;\]
And molar conductance at infinite dilution\[ = 200\;oh{m^{ - 1}}c{m^2}equi{v^{ - 1}}\;\]
Therefore,\[\;\alpha = \frac{{10}}{{200}}\]
\[\;\alpha = 0.05 = 5\% \]
Thus, percentage ionisation of $C{H_3}COOH$. The correct option is a.
Note: Conductance at infinite dilution is also termed as limiting conductance or maximum conductance. This is because conductance is directly proportional to volume and inversely proportional to concentration. So at infinite volume, conductance also maximises. The degree of ionisation is independent of the concentration of the electrolyte.
Complete Step by Step Solution:
An important application of Kohlrausch law is to find the degree of dissociation or the degree of ionisation of weak electrolytes. For example, acetic acid $C{H_3}COOH$
Degree of dissociation, \[\;\alpha \]is defined as the ratio of dissociated amount to the initial amount.
As the value of\[\;\alpha \]increases, the number of ions also increases and thus, \[\;{\lambda _m}\]also increases. So, Degree of dissociation is related to conductance by the formula\[\;\alpha = \frac{{{\lambda _m}}}{{{\lambda _m}^\infty }}\] .
But the value of equivalent conductance is given in the question. Hence first convert equivalent conductance in molar conductance by the formula \[\;n = \frac{{{\lambda _m}}}{{{\lambda _{_{eq}}}}}\]where n is the n-factor of the solute.
N- factor for $C{H_3}COOH$is \[1\] as the number of replaceable hydrogen ion is one). Thus, for $C{H_3}COOH$the molar conductance is the same as the equivalent conductance.
Hence, molar conductance\[10\;oh{m^{ - 1}}c{m^2}equi{v^{ - 1}}\;\]
And molar conductance at infinite dilution\[ = 200\;oh{m^{ - 1}}c{m^2}equi{v^{ - 1}}\;\]
Therefore,\[\;\alpha = \frac{{10}}{{200}}\]
\[\;\alpha = 0.05 = 5\% \]
Thus, percentage ionisation of $C{H_3}COOH$. The correct option is a.
Note: Conductance at infinite dilution is also termed as limiting conductance or maximum conductance. This is because conductance is directly proportional to volume and inversely proportional to concentration. So at infinite volume, conductance also maximises. The degree of ionisation is independent of the concentration of the electrolyte.
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