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Electrochemistry question

2006 · Shift 0 · Q33
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Electrochemistry question

2006 · Shift 0 · Q33

JEE MainChemistryElectrochemistryMCQ+4 / −1
Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 Ω\OmegaΩ. The conductivity of this solution is 1.29 S m–1. Resistance of the same cell when filled with 0.2 M of the same solution is 520 Ω\OmegaΩ, The molar conductivity of 0.02 M solution of the electrolyte will be
  1. A
    124 ×\times× 10–4 S m2 mol–1
  2. B
    1240 ×\times× 10–4 S m2 mol–1
  3. C
    1.24 ×\times× 10–4 S m2 mol–1
  4. D
    12.4 ×\times× 10–4 S m2 mol–1
View written solutionFree

Correct answer: A: 124 \TIMES 10^{-4}\,\TEXT{S M}^2\TEXT{ MOL}^{-1}

  1. Use the first solution to find the cell constant

For a conductivity cell,

κ=G(lA)\kappa = G\left(\frac{l}{A}\right)κ=G(Al​)

where κ\kappaκ is conductivity, G=1RG=\frac{1}{R}G=R1​ is conductance, and lA\frac{l}{A}Al​ is the cell constant.

Given for 0.1 M0.1\,\text{M}0.1M solution:

  • R1=100 ΩR_1 = 100\,\OmegaR1​=100Ω
  • κ1=1.29 S m−1\kappa_1 = 1.29\,\text{S m}^{-1}κ1​=1.29S m−1

So,

G1=1100=0.01 SG_1 = \frac{1}{100} = 0.01\,\text{S}G1​=1001​=0.01S

Hence,

cell constant=κ1G1=1.290.01=129 m−1\text{cell constant} = \frac{\kappa_1}{G_1} = \frac{1.29}{0.01} = 129\,\text{m}^{-1}cell constant=G1​κ1​​=0.011.29​=129m−1
  1. Find conductivity of the second solution

For the 0.02 M0.02\,\text{M}0.02M solution (the options clearly correspond to this; the statement appears to have a typo where it says 0.2 M0.2\,\text{M}0.2M), resistance is

R2=520 ΩR_2 = 520\,\OmegaR2​=520Ω

Thus,

G2=1520 SG_2 = \frac{1}{520}\,\text{S}G2​=5201​S

Now,

κ2=G2×cell constant=1520×129\kappa_2 = G_2 \times \text{cell constant} = \frac{1}{520}\times 129κ2​=G2​×cell constant=5201​×129 κ2≈0.248 S m−1\kappa_2 \approx 0.248\,\text{S m}^{-1}κ2​≈0.248S m−1
  1. Calculate molar conductivity

Molar conductivity is

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

with CCC in mol m−3\text{mol m}^{-3}mol m−3.

Given concentration:

0.02 M=0.02 mol L−1=20 mol m−30.02\,\text{M} = 0.02\,\text{mol L}^{-1} = 20\,\text{mol m}^{-3}0.02M=0.02mol L−1=20mol m−3

Therefore,

Λm=0.24820=0.0124 S m2 mol−1\Lambda_m = \frac{0.248}{20} = 0.0124\,\text{S m}^2\text{ mol}^{-1}Λm​=200.248​=0.0124S m2 mol−1

Now write in the form given in options:

0.0124=124×10−40.0124 = 124\times 10^{-4}0.0124=124×10−4

So,

Λm=124×10−4 S m2 mol−1\Lambda_m = 124\times 10^{-4}\,\text{S m}^2\text{ mol}^{-1}Λm​=124×10−4S m2 mol−1
  1. Match with the options

This corresponds to Option A.

  1. Comparison with stored answer

Stored correct answer is D: 12.4×10−412.4\times 10^{-4}12.4×10−4.

But,

12.4×10−4=0.0012412.4\times 10^{-4} = 0.0012412.4×10−4=0.00124

which is 10 times smaller than the correct value.

Hence the stored answer appears to be incorrect. The derived answer is A.

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