
A 5.0 m mol d \[{m^{ - 3}}\] aqueous solution of KCl has a conductance of 0.55 mS when measured in a cell of cell constant 1.3 \[c{m^{ - 1}}\]. What is the molar conductivity of this solution ?(Round off to the nearest integer).
Answer
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Hint: Molar conductivity is the conductance property of a solution containing one mole of an electrolyte. It is a function of the ionic strength of a solution or the concentration of the particular ionic salt. In other words, molar conductivity can be defined as the conducting power of all the ions that are formed by dissolving a unit mole of an electrolyte in a solution. Molar conductivity is a property of an electrolyte solution that is used to determine the efficiency with which an electrolyte conducts electricity in a solution. Hence, it is not a constant.
Complete Step by Step Solution:
\[{\lambda _M} = \dfrac{K}{C}\] is the equation used to denote the molar conductivity of a solution. Here, “K” stand for the specific conductivity and “c” stands for the concentration in mole per litre.
\[{G_{KCl}}\] = 0.55 mS = 55 \[ \times {10^{ - 5}}\]s
Cell constant = \[\dfrac{I}{A} = 1.3c{m^{ - 1}}\]
\[{\lambda _M}\] = ??
\[K = G(\dfrac{I}{A}) = 55 \times {10^{ - 5 \times 1.3}}S\]\[c{m^{ - 1}}\]
\[{\lambda _M} = \dfrac{{K \times 1000}}{{Molarity}} = \dfrac{{55 \times 1.3 \times {{10}^{ - 5}} \times 1000}}{{5 \times {{10}^{ - 3}}}}\]
\[{\lambda _M} = 11 \times 1.3 \times 10 = 11 \times 13 = 143\]S \[c{m^2}\] \[mo{l^{ - 1}}\]
\[{\lambda _M} = \dfrac{{143 \times 1000 \times {{10}^{ - 3}}S}}{{{{({{10}^{ - 2}}m)}^{ - 2}}}}mo{l^{ - 1}}\]
\[{\lambda _M} = 143 \times 1000 \times {10^{ - 4}}(ms){m^2}mo{l^{ - 1}}\]
\[{\lambda _M} = 14.3(ms){m^2}mo{l^{ - 1}}\]
Therefore, after rounding off to the nearest integer we get the answer as \[{\lambda _M} = 14(ms){m^2}mo{l^{ - 1}}\].
Note: The conducting power of all the ions created by dissolving one mole of an electrolyte in the solution is the general definition of molar conductivity. With a fall in concentration or an increase in dilution, the molar conductivity of both weak and strong electrolytes increases. We know that molar conductivity is the conductivity offered by a unit mole of ions. Even after dilution, considering the same unit mole of ions are present in the solution, the increased dilution results in the dissociation of more electrolytes into ions, and effectively the number of active ions in the solution is increased. Therefore, these active ions impart more conductivity.
Complete Step by Step Solution:
\[{\lambda _M} = \dfrac{K}{C}\] is the equation used to denote the molar conductivity of a solution. Here, “K” stand for the specific conductivity and “c” stands for the concentration in mole per litre.
\[{G_{KCl}}\] = 0.55 mS = 55 \[ \times {10^{ - 5}}\]s
Cell constant = \[\dfrac{I}{A} = 1.3c{m^{ - 1}}\]
\[{\lambda _M}\] = ??
\[K = G(\dfrac{I}{A}) = 55 \times {10^{ - 5 \times 1.3}}S\]\[c{m^{ - 1}}\]
\[{\lambda _M} = \dfrac{{K \times 1000}}{{Molarity}} = \dfrac{{55 \times 1.3 \times {{10}^{ - 5}} \times 1000}}{{5 \times {{10}^{ - 3}}}}\]
\[{\lambda _M} = 11 \times 1.3 \times 10 = 11 \times 13 = 143\]S \[c{m^2}\] \[mo{l^{ - 1}}\]
\[{\lambda _M} = \dfrac{{143 \times 1000 \times {{10}^{ - 3}}S}}{{{{({{10}^{ - 2}}m)}^{ - 2}}}}mo{l^{ - 1}}\]
\[{\lambda _M} = 143 \times 1000 \times {10^{ - 4}}(ms){m^2}mo{l^{ - 1}}\]
\[{\lambda _M} = 14.3(ms){m^2}mo{l^{ - 1}}\]
Therefore, after rounding off to the nearest integer we get the answer as \[{\lambda _M} = 14(ms){m^2}mo{l^{ - 1}}\].
Note: The conducting power of all the ions created by dissolving one mole of an electrolyte in the solution is the general definition of molar conductivity. With a fall in concentration or an increase in dilution, the molar conductivity of both weak and strong electrolytes increases. We know that molar conductivity is the conductivity offered by a unit mole of ions. Even after dilution, considering the same unit mole of ions are present in the solution, the increased dilution results in the dissociation of more electrolytes into ions, and effectively the number of active ions in the solution is increased. Therefore, these active ions impart more conductivity.
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