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Electrochemistry question

2021 · 1 Sep · Shift 2 · Q22
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Electrochemistry question

2021 · 1 Sep · Shift 2 · Q22

JEE MainChemistryElectrochemistryNumerical+4 / −1
If the conductivity of mercury at 0 ∘^\circ∘ C is 1.07 ×\times× 106 S m −-− 1 and the resistance of a cell containing mercury is 0.243 Ω\OmegaΩ, then the cell constant of the cell is x ×\times× 104 m −-− 1. The value of x is ‾\underline{\hspace{2cm}}​. (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 26

  1. Use the relation between conductivity, resistance, and cell constant

For a conductivity cell,

R=ρlAR = \rho \frac{l}{A}R=ρAl​

Since conductivity κ=1ρ\kappa = \dfrac{1}{\rho}κ=ρ1​, we get

lA=κR\frac{l}{A} = \kappa RAl​=κR

But cell constant is

Cell constant=lA\text{Cell constant} = \frac{l}{A}Cell constant=Al​

So,

Cell constant=κR\text{Cell constant} = \kappa RCell constant=κR
  1. Substitute the given values

Given:

κ=1.07×106 S m−1\kappa = 1.07 \times 10^6\ \text{S m}^{-1}κ=1.07×106 S m−1 R=0.243 ΩR = 0.243\ \OmegaR=0.243 Ω

Hence,

Cell constant=(1.07×106)(0.243)\text{Cell constant} = (1.07 \times 10^6)(0.243)Cell constant=(1.07×106)(0.243)
  1. Calculate

First multiply 1.071.071.07 and 0.2430.2430.243:

1.07×0.243=0.260011.07 \times 0.243 = 0.260011.07×0.243=0.26001

Therefore,

Cell constant=0.26001×106\text{Cell constant} = 0.26001 \times 10^6Cell constant=0.26001×106 =2.6001×105 m−1= 2.6001 \times 10^5\ \text{m}^{-1}=2.6001×105 m−1
  1. Match with the required form

We need

Cell constant=x×104 m−1\text{Cell constant} = x \times 10^4\ \text{m}^{-1}Cell constant=x×104 m−1

Now,

2.6001×105=26.001×1042.6001 \times 10^5 = 26.001 \times 10^42.6001×105=26.001×104

So,

x≈26x \approx 26x≈26
  1. Nearest integer
26\boxed{26}26​
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