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Electrochemistry question

2022 · 29 Jul · Shift 1 · Q16
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  5. /2022 · 29 Jul · Shift 1 · Q16

Electrochemistry question

2022 · 29 Jul · Shift 1 · Q16

JEE MainChemistryElectrochemistryNumerical+4 / −1
Resistance of a conductivity cell (cell constant 129 m−1129 \mathrm{~m}^{-1}129 m−1) filled with 74.5 ppm74.5 \,\mathrm{ppm}74.5ppm solution of KCl\mathrm{KCl}KCl is 100 Ω100 \,\Omega100Ω(labelled as solution 1). When the same cell is filled with KCl\mathrm{KCl}KCl solution of 149 ppm149 \,\mathrm{ppm}149ppm, the resistance is 50 Ω50 \,\Omega50Ω(labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. ∧1∧2=x×10−3\frac{\wedge_{1}}{\wedge_{2}}=x \times 10^{-3}∧2​∧1​​=x×10−3. The value of xxx is ‾\underline{\hspace{2cm}}​. (Nearest integer) Given, molar mass of KCl\mathrm{KCl}KCl is 74.5 g mol−174.5 \mathrm{~g} \mathrm{~mol}^{-1}74.5 g mol−1.
Numerical answer
View written solutionFree

Correct answer: 1000

  1. Use cell constant to find conductivity

For a conductivity cell,

κ=(cell constant)×1R\kappa = (\text{cell constant})\times \frac{1}{R}κ=(cell constant)×R1​

where κ\kappaκ is conductivity.

Given cell constant =129 m−1=129\,\text{m}^{-1}=129m−1.

  • For solution 1:
κ1=129×1100=1.29 S m−1\kappa_1 = 129\times \frac{1}{100} = 1.29\,\text{S m}^{-1}κ1​=129×1001​=1.29S m−1
  • For solution 2:
κ2=129×150=2.58 S m−1\kappa_2 = 129\times \frac{1}{50} = 2.58\,\text{S m}^{-1}κ2​=129×501​=2.58S m−1

Thus,

κ1κ2=1.292.58=12\frac{\kappa_1}{\kappa_2}=\frac{1.29}{2.58}=\frac12κ2​κ1​​=2.581.29​=21​
  1. Convert ppm to molar concentration

For dilute aqueous solutions,

1 ppm=1 mg L−11\,\text{ppm} = 1\,\text{mg L}^{-1}1ppm=1mg L−1

Solution 1

74.5 ppm=74.5 mg L−1=0.0745 g L−174.5\,\text{ppm}=74.5\,\text{mg L}^{-1}=0.0745\,\text{g L}^{-1}74.5ppm=74.5mg L−1=0.0745g L−1

Molarity:

C1=0.074574.5=10−3 mol L−1C_1=\frac{0.0745}{74.5}=10^{-3}\,\text{mol L}^{-1}C1​=74.50.0745​=10−3mol L−1

Solution 2

149 ppm=149 mg L−1=0.149 g L−1149\,\text{ppm}=149\,\text{mg L}^{-1}=0.149\,\text{g L}^{-1}149ppm=149mg L−1=0.149g L−1

Molarity:

C2=0.14974.5=2×10−3 mol L−1C_2=\frac{0.149}{74.5}=2\times 10^{-3}\,\text{mol L}^{-1}C2​=74.50.149​=2×10−3mol L−1

Hence,

C2C1=2\frac{C_2}{C_1}=2C1​C2​​=2
  1. Use relation for molar conductivity

Molar conductivity is

Λm=κC\Lambda_m = \frac{\kappa}{C}Λm​=Cκ​

(if concentration is taken in the same units for both solutions, ratio remains correct).

Therefore,

Λ1Λ2=κ1/C1κ2/C2=κ1κ2⋅C2C1\frac{\Lambda_1}{\Lambda_2} = \frac{\kappa_1/C_1}{\kappa_2/C_2} =\frac{\kappa_1}{\kappa_2}\cdot \frac{C_2}{C_1}Λ2​Λ1​​=κ2​/C2​κ1​/C1​​=κ2​κ1​​⋅C1​C2​​

Substitute values:

Λ1Λ2=12×2=1\frac{\Lambda_1}{\Lambda_2} = \frac12 \times 2 = 1Λ2​Λ1​​=21​×2=1

So,

Λ1Λ2=1=x×10−3\frac{\Lambda_1}{\Lambda_2}=1=x\times 10^{-3}Λ2​Λ1​​=1=x×10−3

Thus,

x=1000x=1000x=1000
  1. Final answer

The required integer is:

1000\boxed{1000}1000​
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