
The molar conductivity of acetic acid at infinite dilution is $387\;S\;c{m^2}mo{l^{ - 1}}$ At the same temperature, $0.001$M solution of acetic acid, molar conductivity is $55\;S\;c{m^2}mo{l^{ - 1}}$ ,$\alpha $ is the degree of dissociation of 0.1 N acetic acids What is the value of $600\alpha $?
Assume $1 - \alpha = 1$ for $0.1$ N acid
Answer
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Hint: Molar conductivity is the conductance property of a solution containing one mole of the electrolyte or it is a function of the ionic strength of a solution or the concentration of salt. It is therefore not a constant.
Degree of dissociation $\alpha $
= molar conductance at the particular concentration molar conductance at infinite dilution
Complete step-by-step answer:We have been given the molar conductivity at infinite dilution.
And the molar conductivity of acetic acid.
So for $0.001M\;C{H_3}COOH$ ,$\alpha = \dfrac{{{\Delta _m}}}{{{\Delta _\infty }}}$
$\therefore \alpha = \dfrac{{55}}{{387}} = 0.142$or $14.2\% $
${K_\alpha } = \dfrac{{c{\alpha ^2}}}{{1 - \alpha }}$
${K_\alpha } = \dfrac{{0.001 \times {{(0.142)}^2}}}{{1 - 0.142}} = 2.35 \times {10^{ - 5}}$
For $0.1N$ acid,${K_\alpha } = c{\alpha ^2}$
$2.35 \times {10^{ - 5}} = 0.1 \times {\alpha ^2}$
$\therefore \alpha = 0.015\;or\;1.5\% $
Therefore
$600\alpha = 600 \times 0.015 = 9$
Additional information: When the solution is dilute the number of ions available per unit volume reduces, resulting in the reduction of the conductivity. ... The molar conductivity of both weak and strong electrolytes increases with a decrease in concentration or dilution. Molar conductance is dependent on the concentration of the electrolyte. When the solution is dilute the number of ions available per unit volume reduces, resulting in reduction of the conductivity.
Hence the correct answer is 9.
Note: the 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-centimetre apart. The molar conductivity of both weak and strong electrolytes increases with a decrease in concentration or dilution.
Degree of dissociation $\alpha $
= molar conductance at the particular concentration molar conductance at infinite dilution
Complete step-by-step answer:We have been given the molar conductivity at infinite dilution.
And the molar conductivity of acetic acid.
So for $0.001M\;C{H_3}COOH$ ,$\alpha = \dfrac{{{\Delta _m}}}{{{\Delta _\infty }}}$
$\therefore \alpha = \dfrac{{55}}{{387}} = 0.142$or $14.2\% $
${K_\alpha } = \dfrac{{c{\alpha ^2}}}{{1 - \alpha }}$
${K_\alpha } = \dfrac{{0.001 \times {{(0.142)}^2}}}{{1 - 0.142}} = 2.35 \times {10^{ - 5}}$
For $0.1N$ acid,${K_\alpha } = c{\alpha ^2}$
$2.35 \times {10^{ - 5}} = 0.1 \times {\alpha ^2}$
$\therefore \alpha = 0.015\;or\;1.5\% $
Therefore
$600\alpha = 600 \times 0.015 = 9$
Additional information: When the solution is dilute the number of ions available per unit volume reduces, resulting in the reduction of the conductivity. ... The molar conductivity of both weak and strong electrolytes increases with a decrease in concentration or dilution. Molar conductance is dependent on the concentration of the electrolyte. When the solution is dilute the number of ions available per unit volume reduces, resulting in reduction of the conductivity.
Hence the correct answer is 9.
Note: the 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-centimetre apart. The molar conductivity of both weak and strong electrolytes increases with a decrease in concentration or dilution.
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