Question

If the specific conductance of the 1M \[{H_2}S{O_4}\] solution is \[26 \times {10^{ - 2}}S\,c{m^{ - 1}}\], then the equivalent conductivity would be:
A. \[1.3 \times {10^2}Sc{m^2}e{q^{ - 1}}\]
B. \[1.6 \times {10^2}Sc{m^2}\]
C. \[13\,Sc{m^2}mo{l^{ - 1}}\]
D. \[1.3 \times {10^3}Sc{m^2}mo{l^{ - 1}}\]

Answer 1

Hint: Equivalent conductance of an electrolyte is defined as the conductance of a volume of solution containing one equivalent weight of dissolved substance when placed between two parallel electrodes 1 cm apart, and large enough to contain between them all of the solutions.

Formula used:
Normality \[ = n \times \] molarity, \[\Lambda = \dfrac{{1000 \times \kappa }}{N}\]

Complete step by step answer:
To solve this question, you should know about resistance, conductance. The opposite force to the current is called resistance. Conductance is a reciprocal of resistance to the system. It is also a ratio of the current to the potential difference. The unit of conductance is siemens or \[oh{m^{ - 1}}\].
According to ohm’s law of resistance, resistance is directly proportional to the length and inversely proportional to the cross-sectional area. which can be written as, \[R\,\alpha \,\dfrac{l}{A}\]. Where length is I and cross-sectional area is A. It can also be written with a proportional constant \[\rho \] which is called resistivity. Therefore, the formula of resistance is \[R = \rho \dfrac{l}{A}\] .
 Conductance is a reciprocal of resistance of the system, \[G = \dfrac{1}{R}\] where G is conductance. Therefore, the formula of conductance can be written as, \[G = \dfrac{1}{R} = \dfrac{1}{\rho } \times \dfrac{A}{l}\]. Where \[\dfrac{1}{\rho }\] can be written with a new term specific conductance or conductivity. Therefore, the formula of conductance is,
  \[
  G = \kappa \times \dfrac{A}{l} \\
  or,G \times \dfrac{l}{A} = \kappa \\
 \]
 The normality of1M \[{H_2}S{O_4}\] solution is,
Normality \[ = n \times \] molarity \[2 \times 1 = 2\]
 Now the equivalent conductivity is,
 \[
  \Lambda = \dfrac{{1000 \times \kappa }}{N} \\
\Rightarrow \Lambda = \dfrac{{1000 \times 26 \times {{10}^{ - 2}}S\,c{m^{ - 1}}}}{{2\dfrac{{equiv}}{{c{m^3}}}}} \\
\Rightarrow \Lambda = \dfrac{{1000 \times 26 \times {{10}^{ - 2}}S\,c{m^{ - 1}}}}{{2\dfrac{{equiv}}{{c{m^3}}}}} \\
\Rightarrow \Lambda = 1.3 \times {10^2}S\,c{m^2}e{q^{ - 1}} \\
 \]

So, the correct option is A.

Note: With the dilution of the electrolyte the equivalent conductance and molar conductance value increases. This is because on dilution the degree of dissociation of electrolyte increases as well as the number of ions increases. Remember the definitions of resistance, conductance. Remember the formula of conductance and resistance. The ratio of length to the cross-sectional area is called cell constant. Remember the formula of molar conductivity \[{\Lambda _m} = \dfrac{{1000 \times \kappa }}{c}\].