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Question: The resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 ...

The resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 ohm. The conductivity of this solution is 1.29 S  m−1S\;{m^{ - 1}}. Resistance of the same cell when filled with 0.2 M of the same solution is 520 ohm. The molar conductivity of 0.02M solution of the electrolyte will be:
A. 124×10−4Sm2mol−1124 \times {10^{ - 4}}S{m^2}mo{l^{ - 1}}
B. 1240×10−4Sm2mol−11240 \times {10^{ - 4}}S{m^2}mo{l^{ - 1}}
C. 1.24×10−4Sm2mol−11.24 \times {10^{ - 4}}S{m^2}mo{l^{ - 1}}
D. 12.4×10−4Sm2mol−112.4 \times {10^{ - 4}}S{m^2}mo{l^{ - 1}}

Explanation

Solution

The conductance is defined as the inverse of resistance. The conductivity of the solution is equal to length divided by area multiplied by conductance. Volume is calculated in cm3c{m^3}which is related to the molar conductance. The molar conductance is calculated by multiplying conductivity multiplied by volume.

Complete step by step answer:
Given
Resistance for electrolyte containing 0.1M solution is 100 ohm.
Concentration of electrolyte is 0.1M.
The conductivity of the solution is 1.29 S  m−1S\;{m^{ - 1}}.
The relationship between the conductivity and resistance is shown below.
Resistance for a cell containing 0.2 M solution is 520 ohm.
K=1R(la)K = \dfrac{1}{R}\left( {\dfrac{l}{a}} \right)
Where,
K is the conductivity
R is the resistance
l is the length
a is the area.
Substitute the values in the above equation.
⇒1.29=1100(la)\Rightarrow 1.29 = \dfrac{1}{{100}}\left( {\dfrac{l}{a}} \right)
⇒(la)=129m−1\Rightarrow \left( {\dfrac{l}{a}} \right) = 129{m^{ - 1}}
For 0.2 M solution
⇒k=1520(129)Ω−1m−1\Rightarrow k = \dfrac{1}{{520}}\left( {129} \right){\Omega ^{ - 1}}{m^{ - 1}}
The relation between the molar conductance and conductivity is shown below.
The formula to calculate the molar conductance is shown below.
μ=k×V\mu = k \times V
Where,
μ\mu is molar conductance
K is the conductivity.
V is the volume.
Substitute the values in the above equation.
⇒μ=1520×129×10000.02×10−6m3\Rightarrow \mu = \dfrac{1}{{520}} \times 129 \times \dfrac{{1000}}{{0.02}} \times {10^{ - 6}}{m^3}
⇒μ=129520×10000.02×10−6m3\Rightarrow \mu = \dfrac{{129}}{{520}} \times \dfrac{{1000}}{{0.02}} \times {10^{ - 6}}{m^3}
⇒μ=124×10−4Sm2mol−1\Rightarrow \mu = 124 \times {10^{ - 4}}S{m^2}mo{l^{ - 1}}
Thus, the molar conductivity of the solution is 124×10−4Sm2mol−1124 \times {10^{ - 4}}S{m^2}mo{l^{ - 1}}.
Therefore, the correct option is A.

Note:
The resistance of the conductor varies directly to its length (l) and inversely to the cross sectional area (A). Make sure to convert the value in cm3c{m^3}. l = 1 cm and A = 1 cm2c{m^2}.