Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Electrochemistry question

2021 · 16 Mar · Shift 2 · Q16
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Electrochemistry
  5. /2021 · 16 Mar · Shift 2 · Q16

Electrochemistry question

2021 · 16 Mar · Shift 2 · Q16

JEE MainChemistryElectrochemistryNumerical+4 / −1
A 5.0 m mol dm −-− 3 aqueous solution of KClKClKCl has a conductance of 0.55 mS when measured in a cell of cell constant 1.3 cm −-− 1. The molar conductivity of this solution is ‾\underline{\hspace{2cm}}​ mSm2 mol −-− 1. (Round off to the Nearest Integer).
Numerical answer
View written solutionFree

Correct answer: 14

  1. Given data
  • Concentration of KClKClKCl solution: 5.0 mmol dm−35.0\ \text{mmol dm}^{-3}5.0 mmol dm−3
  • Conductance: G=0.55 mSG = 0.55\ \text{mS}G=0.55 mS
  • Cell constant: lA=1.3 cm−1\dfrac{l}{A} = 1.3\ \text{cm}^{-1}Al​=1.3 cm−1

We need to find the molar conductivity Λm\Lambda_mΛm​ in mS m2 mol−1\text{mS m}^2\text{ mol}^{-1}mS m2 mol−1.

  1. Find conductivity

Conductivity κ\kappaκ is related to conductance by:

κ=G×cell constant\kappa = G \times \text{cell constant}κ=G×cell constant

Substitute the values:

κ=0.55 mS×1.3 cm−1=0.715 mS cm−1\kappa = 0.55\ \text{mS} \times 1.3\ \text{cm}^{-1} = 0.715\ \text{mS cm}^{-1}κ=0.55 mS×1.3 cm−1=0.715 mS cm−1
  1. Convert concentration into mol m−3^{-3}−3

Given:

5.0 mmol dm−3=5.0×10−3 mol dm−35.0\ \text{mmol dm}^{-3} = 5.0 \times 10^{-3}\ \text{mol dm}^{-3}5.0 mmol dm−3=5.0×10−3 mol dm−3

Now use:

1 dm3=10−3 m31\ \text{dm}^3 = 10^{-3}\ \text{m}^31 dm3=10−3 m3

So,

5.0×10−3 mol dm−3=5.0 mol m−35.0 \times 10^{-3}\ \text{mol dm}^{-3} = 5.0\ \text{mol m}^{-3}5.0×10−3 mol dm−3=5.0 mol m−3
  1. Convert conductivity into SI-compatible form

We have:

0.715 mS cm−10.715\ \text{mS cm}^{-1}0.715 mS cm−1

Since:

1 mS cm−1=100 mS m−11\ \text{mS cm}^{-1} = 100\ \text{mS m}^{-1}1 mS cm−1=100 mS m−1

therefore,

κ=0.715×100=71.5 mS m−1\kappa = 0.715 \times 100 = 71.5\ \text{mS m}^{-1}κ=0.715×100=71.5 mS m−1
  1. Calculate molar conductivity

Formula:

Λm=κc\Lambda_m = \frac{\kappa}{c}Λm​=cκ​

where

  • κ=71.5 mS m−1\kappa = 71.5\ \text{mS m}^{-1}κ=71.5 mS m−1
  • c=5.0 mol m−3c = 5.0\ \text{mol m}^{-3}c=5.0 mol m−3

Thus,

Λm=71.55.0=14.3 mS m2 mol−1\Lambda_m = \frac{71.5}{5.0} = 14.3\ \text{mS m}^2\text{ mol}^{-1}Λm​=5.071.5​=14.3 mS m2 mol−1

Rounded to nearest integer:

14\boxed{14}14​
  1. Comparison with stored correct answer

Stored correct answer = 141414

Our derived answer = 141414

So, the answer agrees with the stored correct answer.

PreviousNext

More from Electrochemistry

  • A KCl solution of conductivity 0.14 S m − 1 shows a resistance of 4.19 Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03 Ω. The conductivity of the HCl solution is ​×…2021 · Numerical
  • For the reaction 2Fe3+(aq) + 2I −(aq) → 2Fe2+(aq) + I2​(s) the magnitude of the standard molar Gibbs free energy change, Δ rG mo​=−​ kJ (Round off to the Nearest Integer). [EFe2+/Fe(s)o​=−0.440V;EI2​/2I−o​=0.539V;​EFe3+/Fe(s)o​=−0.036VF=96500C​]…2021 · Numerical
  • The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are 280, 860 and 426 S cm2 mol − 1 respectively. The molar conductivity at infinite dilution of barium sulphate is ​…2021 · Numerical
  • Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation. 6OH− + Cl−→ ClO3−​ + 3H2​O + 6e− A current of xA has to be passed for 10h to produce 10.0g of potassium chlorate. The…2021 · Numerical
  • Assume a cell with the following reaction Cu(s)​+2Ag+(1×10−3M)→Cu2+(0.250M)+2Ag(s)​ EcellΘ​=2.97 V Ecell for the above reaction is ​ V. (Nearest integer)…2021 · Numerical
  • The electrode potential of M2+/M of 3d-series elements shows positive value for :2021 · MCQ
  • The magnitude of the change in oxidising power of the MnO4−​/Mn2+ couple is x × 10 − 4 V, if the H+ concentration is decreased from 1M to 10 − 4 M at 25 ∘ C. (Assume concentration of MnO4−​ and Mn2+…2021 · Numerical
  • Copper reduces NO 3−​ into NO and NO2​ depending upon the concentration of HNO3​ in solution. (Assuming fixed [Cu2+] and PNO = PNO2), the HNO3​ concentration at which the thermodynamic tendency for reduction of NO 3−​ into NO…2021 · Numerical