Solveeit Logo
Login

Question

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

Resistance of 0.2M0.2M solution of an electrolyte is 50Ω50\Omega . The specific conductance of the solution is 1.4Sm−11.4S{{m}^{-1}}. The resistance of 0.5M0.5M solution of the same electrolyte is 280Ω280\Omega . The molar conductivity of 0.5M0.5M solution of the electrolyte is Sm2mol−1S{{m}^{2}}mo{{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

The conductance of any material is that property of material due to which it allows the flow of ions through itself and as a result they conduct electricity. It is generally defined as the reciprocal of resistance of that material. SI unit of conductance is Siemens represented by the symbol S.

Complete answer:
We know that resistance is given by the formula:
R=1κ×lAR=\dfrac{1}{\kappa }\times \dfrac{l}{A}
Where RR is resistance which is given as 50Ω50\Omega , κ\kappa is specific conductance which is 1.4Sm−11.4S{{m}^{-1}} given, ll is length which is generally taken as 1 and AA is area of cross section, Put the values in the equation
50=11.4×1A50=\dfrac{1}{1.4}\times \dfrac{1}{A}
1A=70m−1\dfrac{1}{A}=70{{m}^{-1}}
Here for resistance given 280Ω280\Omega the specific conductance will be calculated by the same formula
R=1κ×lAR=\dfrac{1}{\kappa }\times \dfrac{l}{A}
Here, R=280ΩR=280\Omega , l=1l=1 and 1A=70m−1\dfrac{1}{A}=70{{m}^{-1}}, so κ\kappa will be
280=1κ×70280=\dfrac{1}{\kappa }\times 70, κ=14Sm−1\kappa =\dfrac{1}{4}S{{m}^{-1}}
Now we have to calculate the molar conductivity of the solution and In general manner molar conductivity can also be defined as the ratio of specific conductivity and the concentration of the electrolyte. It is represented by the termΛm{{\Lambda }_{m}}. It can be calculated by:
Λm=κ×1000M×(10−2)3{{\Lambda }_{m}}=\dfrac{\kappa \times 1000}{M}\times {{({{10}^{-2}})}^{3}}
Where M is molarity i.e. 0.5M and κ=14Sm−1\kappa =\dfrac{1}{4}S{{m}^{-1}}, by putting all the values in equation
Λm=1×10004×0.5×10−6{{\Lambda }_{m}}=\dfrac{1\times 1000}{4\times 0.5}\times {{10}^{-6}}
Λm=500×10−6{{\Lambda }_{m}}=500\times {{10}^{-6}}
i.e. Λm=5×10−4{{\Lambda }_{m}}=5\times {{10}^{-4}}

Hence we can say that option C is the correct answer.

Note:
The molar conductivity of any solution can also be defined as conductance at any given concentration of the volume of solution containing one mole of electrolyte kept between two electrodes with the unit area of cross-section and distance of unit length.