
The specific conductance of a solution is \[{\rm{0}}{\rm{.2oh}}{{\rm{m}}^{{\rm{ - 1}}}}{\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\] and the conductance is \[{\rm{0}}{\rm{.04oh}}{{\rm{m}}^{{\rm{ - 1}}}}\]. The cell constant would be:
A. \[{\rm{1c}}{{\rm{m}}^{{\rm{ - 1}}}}\]
B. \[{\rm{0c}}{{\rm{m}}^{{\rm{ - 1}}}}\]
C. \[{\rm{5c}}{{\rm{m}}^{{\rm{ - 1}}}}\]
D. \[{\rm{0}}{\rm{.2c}}{{\rm{m}}^{{\rm{ - 1}}}}\]
Answer
162k+ views
Hint: Conductance is the extent to which current flows through materials. It is the reciprocal of resistance. It is denoted by G. Conductivity or specific conductance is defined as the conductance of a solution of 1cm length and having 1sq.cm as the cross-section area.
Formula Used:
\[\kappa {\rm{ = G}}\left( {\frac{{\rm{l}}}{{\rm{a}}}} \right)\]
where \[\kappa \]=specific conductance or conductivity
G=conductance
\[\left( {\frac{{\rm{l}}}{{\rm{a}}}} \right)\]=cell constant
Complete Step by Step Solution:
We know that electrolytic solutions conduct electricity due to the movement of ions.
This flow of electricity is measured by conductance. Resistance is exhibited by an electrolyte solution due to factors like interionic attractions, the viscosity of the solvent, etc. Conductance, denoted by G is the extent to which current flows through materials. It is the reciprocal of resistance.
We know that,
\[\kappa {\rm{ = G}}\left( {\frac{{\rm{l}}}{{\rm{a}}}} \right)\]
where \[\kappa \]=specific conductance or conductivity
G=conductance
The ratio of length to cross-section i.e., \[\left( {\frac{{\rm{l}}}{{\rm{a}}}} \right)\] represents the cell constant.
If l=1cm and a=1sq.cm, then \[\kappa \]=G.
Conductivity or specific conductance is defined as the conductance of a solution of 1cm length and having 1sq.cm as the cross-section area.
We are given that ,
G=\[{\rm{0}}{\rm{.04oh}}{{\rm{m}}^{{\rm{ - 1}}}}\]
κ=\[{\rm{0}}{\rm{.2oh}}{{\rm{m}}^{{\rm{ - 1}}}}{\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\]
We have to find out the cell constant i.e., la
So, \[\kappa {\rm{ = G}}\left( {\frac{{\rm{l}}}{{\rm{a}}}} \right)\]
\[ \Rightarrow \frac{\kappa }{{\rm{G}}}{\rm{ = }}\left( {\frac{{\rm{l}}}{{\rm{a}}}} \right)\]
\[ \Rightarrow \left( {\frac{{\rm{l}}}{{\rm{a}}}} \right){\rm{ = }}\frac{{{\rm{0}}{\rm{.2oh}}{{\rm{m}}^{{\rm{ - 1}}}}{\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}}}{{{\rm{0}}{\rm{.04oh}}{{\rm{m}}^{{\rm{ - 1}}}}}}\]
So, cell constant=\[{\rm{5c}}{{\rm{m}}^{{\rm{ - 1}}}}\]
So, option C is correct.
Additional Information: Conductivity or specific conductance can be alternatively defined as the conductance of a one-centimetre cube of solution of the electrolyte.
Note: Flow of electricity through solutions of electrolytes is due to migration of ions when a potential difference is applied between two electrodes. While attempting the question, one must mention the unit of conductance and conductivity in the calculation steps. The unit of cell constant in the conclusion statement must be mentioned.
Formula Used:
\[\kappa {\rm{ = G}}\left( {\frac{{\rm{l}}}{{\rm{a}}}} \right)\]
where \[\kappa \]=specific conductance or conductivity
G=conductance
\[\left( {\frac{{\rm{l}}}{{\rm{a}}}} \right)\]=cell constant
Complete Step by Step Solution:
We know that electrolytic solutions conduct electricity due to the movement of ions.
This flow of electricity is measured by conductance. Resistance is exhibited by an electrolyte solution due to factors like interionic attractions, the viscosity of the solvent, etc. Conductance, denoted by G is the extent to which current flows through materials. It is the reciprocal of resistance.
We know that,
\[\kappa {\rm{ = G}}\left( {\frac{{\rm{l}}}{{\rm{a}}}} \right)\]
where \[\kappa \]=specific conductance or conductivity
G=conductance
The ratio of length to cross-section i.e., \[\left( {\frac{{\rm{l}}}{{\rm{a}}}} \right)\] represents the cell constant.
If l=1cm and a=1sq.cm, then \[\kappa \]=G.
Conductivity or specific conductance is defined as the conductance of a solution of 1cm length and having 1sq.cm as the cross-section area.
We are given that ,
G=\[{\rm{0}}{\rm{.04oh}}{{\rm{m}}^{{\rm{ - 1}}}}\]
κ=\[{\rm{0}}{\rm{.2oh}}{{\rm{m}}^{{\rm{ - 1}}}}{\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\]
We have to find out the cell constant i.e., la
So, \[\kappa {\rm{ = G}}\left( {\frac{{\rm{l}}}{{\rm{a}}}} \right)\]
\[ \Rightarrow \frac{\kappa }{{\rm{G}}}{\rm{ = }}\left( {\frac{{\rm{l}}}{{\rm{a}}}} \right)\]
\[ \Rightarrow \left( {\frac{{\rm{l}}}{{\rm{a}}}} \right){\rm{ = }}\frac{{{\rm{0}}{\rm{.2oh}}{{\rm{m}}^{{\rm{ - 1}}}}{\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}}}{{{\rm{0}}{\rm{.04oh}}{{\rm{m}}^{{\rm{ - 1}}}}}}\]
So, cell constant=\[{\rm{5c}}{{\rm{m}}^{{\rm{ - 1}}}}\]
So, option C is correct.
Additional Information: Conductivity or specific conductance can be alternatively defined as the conductance of a one-centimetre cube of solution of the electrolyte.
Note: Flow of electricity through solutions of electrolytes is due to migration of ions when a potential difference is applied between two electrodes. While attempting the question, one must mention the unit of conductance and conductivity in the calculation steps. The unit of cell constant in the conclusion statement must be mentioned.
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