Solveeit Logo

Question

Question: The resistance of 0.01 N \(NaC\)l solution is 200 \(\Omega \) at \({25^o}C\). Cell constant of the c...

The resistance of 0.01 N NaCNaCl solution is 200 Ω\Omega at 25oC{25^o}C. Cell constant of the conductivity cell is unity. Calculate the equivalent conductivity of the solution.

Explanation

Solution

The formula to find the equivalent conductivity (Λm{\Lambda _m}) of a solution is: Λm=κ×1000C{\Lambda _m} = \dfrac{{\kappa \times 1000}}{C}. To find the equivalent conductivity, you must also know the formula for determining the conductivity (κ\kappa ) of a solution. We are given that cell constant is unity and the quantity lA\dfrac{l}{A} is known as cell constant.

Complete step by step solution:
Given that,
Concentration of NaClNaCl solution = 0.01 N
The resistance of NaClNaCl solution = 200 Ω\Omega at 25oC{25^o}C
Cell constant is unity. The quantity lA\dfrac{l}{A} is called cell constant, where ll is the separation length of electrodes and AA is the cross-sectional area.
Thus, lA=1\dfrac{l}{A} = 1
The equivalent conductivity of the solution can be determined by the equation:
Λm=κ×1000C{\Lambda _m} = \dfrac{{\kappa \times 1000}}{C}
Here, Λm{\Lambda _m} is the equivalent conductivity, κ\kappa is the conductivity and CC is the concentration of the solution.
We are given concentration of the solution but, to find the equivalent conductivity, we also need to find the conductivity of the solution.
Conductivity of the solution is given by the equation:
κ=Cell constantR\kappa = \dfrac{{{\text{Cell constant}}}}{{\text{R}}}
Here, R is the resistance in ohm.
Therefore, substituting the given value of resistance i.e., 200 Ω\Omega and cell constant i.e., 1, we get:
κ=1200 S cm1\kappa = \dfrac{1}{{200}}{\text{ }}S{\text{ }}c{m^{ - 1}}
Now, substituting the given value of concentration of NaCl solution i.e., 0.01 N and conductivity (determined above), we get equivalent conductivity of NaCl solution as:
Λm=κ×1000C=1×1000200×0.01{\Lambda _m} = \dfrac{{\kappa \times 1000}}{C} = \dfrac{{1 \times 1000}}{{200 \times 0.01}}
Λm=500 Scm2mol1\therefore {\Lambda _m} = 500{\text{ }}Sc{m^2}mo{l^{ - 1}}.

Thus, the equivalent conductivity of NaCl solution is 500 Scm2mol1Sc{m^2}mo{l^{ - 1}} and hence this is the required answer.

Note: Always take care of the units while finding the equivalent conductivity of the solution. In the equation, Λm=κ×1000C{\Lambda _m} = \dfrac{{\kappa \times 1000}}{C}, if κ\kappa is expressed in S cm1S{\text{ }}c{m^{ - 1}} and the concentration, CC in mol cm3mol{\text{ }}c{m^{ - 3}}, then the units of Λm{\Lambda _m} are in Scm2mol1Sc{m^2}mo{l^{ - 1}}. Cell constant which is equal to lA\dfrac{l}{A} is usually denoted by the symbol, G{G^*}.