Electrochemistry
If $$0.01 M$$ solution of an electrolyte has a resistance of $$40$$ ohms in a cell having a cell constant $$0.4$$ $$ cm^{-1} $$ then its molar conductance would be:
A $$0.1$$ M solution of a monobasic acid has a specific resistance of 'r' ohm-cm. Its molar conductivity is:
The specific conductance is inversely proportional to the specific resistance $$ \left( k=\cfrac { 1 }{ r }\right) $$. The relationship between the molar conductivity and the specific conductance is $$ { \Lambda }_{ m }=\cfrac { \kappa \times 1000 }{ M } $$.
Substitute $$ k=\cfrac { 1 }{ r } $$ and $$ M=0.1 M $$ in the above equation,
$$ { \Lambda }_{ m }=\cfrac { \cfrac { 1 }{ r } \times 1000 }{ 0.1 } =\cfrac { { 10 }^{ 4 } }{ r }$$
Hence, option $$C$$ is correct.
If $$0.01 M$$ solution of an electrolyte has a resistance of $$40$$ ohms in a cell having a cell constant $$0.4$$ $$ cm^{-1} $$ then its molar conductance would be:
Resistance of $$0.2\ M$$ solution of an electrolyte is $$ 50\ \Omega $$. The specific conductance of the solution is $$ 1.4\ S$$ $$ m^{-1}$$. The resistance of $$ 0.5\ M$$ solution of the same electrolyte is $$ 280\ \Omega $$. The molar conductivity of $$ 0.5\ M$$ solution of the electrolyte in $$ S$$ $$ m^{2}$$ $$ mol^{-1}$$ is:
The molar conductivity of a 0.5 mol/dm$$^3$$ solution of $$AgNO_3$$ with electrolytic conductivity of $$5.76 \times 10^{-3} S cm^{-1}$$ at 298 K is :
Calculate $$\Delta ^{\infty}_{HOAc}$$ $$( in\, S\, cm^2\, mol^{-1})$$ using appropriate molar conductances of the electrolytes listed in the given table at infinite dilution in $$H_2O$$ at $$25^0$$ C. Electrolyte KCl KNO$$_3$$ HCl NaOAc NaCl $$\Delta^{\infty} (S\, cm^2 \, mol^{-1}$$ 149.9 145.0 426.2 91.0 126.5
The equivalent conductivity of a solution containing 2.54g of $$CuSO_{4}$$ per litre is $$91.0\ \Omega^{-1}cm^{2}eq^{-1}$$. Its conductivity would be :
The resistance of $$\dfrac{1}{10}$$M solution is $$2.5\times 10^{3}$$ ohm. What is the molar conductivity of solution? (cell constant$$=1.25$$ $$cm^{-1}$$)
Conductivity of $$0.00241\ M$$ acetic acid is $$7.896 \times 10^{-5} S cm^{-1}$$. Calculate its molar conductivity. If $$\wedge^{0}_{m}$$ for acetic acid is $$390.5\ S\ cm^{2} mol^{-1}$$, what is its dissociation constant?
Limiting molar conductivity of NaBr is:
Express the relation among cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity?
The conductivity of $$0.20\ M$$ solution of $$KCl$$ at $$298\ K$$ is $$0.025\ S\ cm^{-1}$$. Calculate its molar conductivity.