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Question: Resistance of 0.2 M solution of an electrolyte is 50 \(\Omega\). The specific conductance of the sol...

Resistance of 0.2 M solution of an electrolyte is 50 Ω\Omega. The specific conductance of the solution is 1.4 S m−1{m^{ - 1}}. The resistance of 0.5 M solution of the same electrolyte is 280 Ω\Omega. The molar conductivity of 0.5 M solution of the electrolyte is S m2{m^2} mol−1mo{l^{ - 1}}.
A. 5×1035 \times {10^3}
B. 5×1025 \times {10^2}
C. 5×10−45 \times {10^{ - 4}}
D. 5×10−35 \times {10^{ - 3}}

Explanation

Solution

To solve this question, first calculate the cell constant value of the first solution. Then use the cell constant value to calculate the k of the second solution. At last, molar conductivity is calculated using k and M where k is specific conductance and M is molarity.

Complete step by step answer:
Given,
The resistance of 2.0 M solution is 50Ω\Omega.
The specific conductance is 1.4 S m−1{m^{ - 1}}.
The resistance of 0.5 M solution of the same electrolyte is 280 Ω\Omega.
The formula to calculate the specific conductance is shown below.
k=1R×lAk = \dfrac{1}{R} \times \dfrac{l}{A}
Where,
K is specific conductance
R is the resistance
l is the length
A is the area of cross section
For first solution
To calculate the specific conductance substitute the values in the above equation.
⇒1.4Sm−1=150×lA\Rightarrow 1.4S{m^{ - 1}} = \dfrac{1}{{50}} \times \dfrac{l}{A}
⇒lA=50×1.4m−1\Rightarrow \dfrac{l}{A} = 50 \times 1.4{m^{ - 1}}
For second reaction
⇒k=1280×50×1.4\Rightarrow k = \dfrac{1}{{280}} \times 50 \times 1.4
⇒k=14\Rightarrow k = \dfrac{1}{4}
The formula to calculate the molar conductivity is shown below.
Λm=kC{\Lambda _m} = \dfrac{k}{C}
Where
Λm{\Lambda _m} is the molar conductivity
k is the specific conductance
C is the concentration
To calculate the molar conductivity, substitute the values in the above equation.
⇒Λm=1/41000×0.5\Rightarrow {\Lambda _m} = \dfrac{{1/4}}{{1000 \times 0.5}}
⇒Λm=12000\Rightarrow {\Lambda _m} = \dfrac{1}{{2000}}
⇒Λm=5×10−4Sm2mol−1\Rightarrow {\Lambda _m} = 5 \times {10^{ - 4}}S{m^2}mo{l^{ - 1}}
Thus, the molar conductivity 0.5 M solution of the electrolyte is 5×10−45 \times {10^{ - 4}} S m2{m^2} mol−1mo{l^{ - 1}}.
Therefore, the correct option is C.

Note:
Make sure to multiply the molarity with 1000 as molarity is calculated in terms of mole per liter. Don’t get confused while calculating the values for both the solutions as only concentration is different but both are the same electrolyte.