Question

Specific conductance of 0.1M NaCl solution is $1.0 \times {10^{ - 2}}$${\text{ oh}}{{\text{m}}^ - }{\text{c}}{{\text{m}}^ - }$ . Its molar conductance ${\text{ oh}}{{\text{m}}^ - }{\text{c}}{{\text{m}}^{ - 2}}{\text{mo}}{{\text{l}}^{ - 1}}$ is:
A. $1.0 \times {10^2}$
B. $1.0 \times {10^3}$
C. $1.0 \times {10^4}$
D. $1.0$

Answer 1

Hint:Specific conductance (or Conductivity) is the ability of the electrolytic solution to conduct electricity. The electricity produced in the electrolytic solution is due to the flow of cations and anions. Wherein, molar conductivity is defined as the amount of electricity passed per mole concentration of the solution.

Formula used: ${ \wedge _{\text{m}}} = \dfrac{{{\text{K}} \times {\text{1000}}}}{{\text{M}}}$ , Where K (kappa) is specific conductance
 M = molarity
${ \wedge _{\text{m}}}$ (Greek, lambda) = molar conductance

Complete step by step answer:
The question has given us the two quantities required to calculate the molar conductance of the solution.
K = $1.0 \times {10^{ - 2}}$${\text{ oh}}{{\text{m}}^ - }{\text{c}}{{\text{m}}^ - }$
M = 0.1 M
putting up the values in the formula gives us molar conductance as;
$ \Rightarrow { \wedge _{\text{m}}} = \dfrac{{{\text{K}} \times {\text{1000}}}}{{\text{M}}}$
$ \Rightarrow { \wedge _{\text{m}}} = \dfrac{{1.0 \times {{10}^{ - 2}} \times {\text{1000}}}}{{0.1}}$ ( now by carrying basic arithmetic operations we will solve the following)
$ \Rightarrow { \wedge _{\text{m}}} = \dfrac{{1.0 \times {{10}^{ - 2}} \times {\text{1}}{{\text{0}}^3}}}{{0.1 \times \dfrac{{10}}{{10}}}}$ ( we have divided and multiplied 10 by molarity to ease our solution, since $\dfrac{{10}}{{10}}$ = 1 )
$ \Rightarrow { \wedge _{\text{m}}} = \dfrac{{1.0 \times {{10}^{ - 2}} \times {\text{1}}{{\text{0}}^3}}}{{1.0 \times {{10}^{ - 1}}}}$
$ \Rightarrow { \wedge _{\text{m}}} = \dfrac{{1.0 \times {{10}^{ - 2}} \times {\text{1}}{{\text{0}}^3} \times {{10}^1}}}{{1.0}}$
$ \Rightarrow { \wedge _{\text{m}}} = \dfrac{{1.0 \times {{10}^{ - 2}} \times {\text{1}}{{\text{0}}^4}}}{{1.0}}$
$ \Rightarrow { \wedge _{\text{m}}} = \dfrac{{1.0 \times {{10}^2}}}{{1.0}}$
$ \Rightarrow { \wedge _{\text{m}}} = 1.0 \times {10^2}$${\text{ oh}}{{\text{m}}^ - }{\text{c}}{{\text{m}}^{ - 2}}{\text{mo}}{{\text{l}}^{ - 1}}$

Hence, the correct answer is option (A) i.e., $1.0 \times {10^2}$

Additional information:
The molarity of a solution is defined as the number of moles of solute present in per litre of the solution and the formula can be written as;
${\text{Molarity = }}\dfrac{{{\text{No}}{\text{. of moles}}}}{{{\text{volume of the solution(L)}}}}$

Note: The formula used to calculate the molar conductance varies with the units mentioned. E.g., if the value of specific conductance was given in ${\text{S}}{{\text{m}}^ - }$ and the concentration in ${\text{mol}} \times {{\text{m}}^{ - 3}}$, then the formula for molar conductance would have been ${ \wedge _{\text{m}}} = \dfrac{{\text{K}}}{{\text{c}}}$ and its units would be ${\text{S}} \times {{\text{m}}^{ - 3}} \times {\text{mo}}{{\text{l}}^{ - 1}}$. Therefore, it is important to pay attention to the units of specific conductance, before calculating the molar conductance.