हिंदी

Assertion (A): Molar conductivity decreases with increase in concentration. Reason (R): When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.

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प्रश्न

Assertion (A): Molar conductivity decreases with increase in concentration.

Reason (R): When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.

विकल्प

  • Both (A) and (R) are true and (R) is the correct explanation of (A).

  • Both (A) and (R) are true, but (R) is not the correct explanation of (A).

  • (A) is true, but (R) is false.

  • (A) is false, but (R) is true.

MCQ
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उत्तर

Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Explanation:

Molar conductivity increases as concentration lowers. This is because the total volume, V, of a solution containing one mole of electrolyte increases. A drop in 'κ' following the dilution of a solution has been discovered to be more than countered by an increase in volume. The limiting molar conductivity of a solution is its molar conductivity at infinite dilution. In other words, when the electrolyte concentration approaches zero, the molar conductivity is said to be limiting.

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2022-2023 (March) Outside Delhi Set 3

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संबंधित प्रश्न

 

The molar conductivity of cation and anion of salt BA are 180 and 220 mhos respectively. The molar conductivity of salt BA at infinite dilution is_____________ .

(a) 90 mhos.cm2                                                                             

(b) 110 mhos.cm2.mol-1

(c) 200 mhos.cm2.mol-1                                                                 

(d) 400 mhos.cm2.mol-1


The conductivity of 0.20 mol L−1 solution of KCl is 2.48 × 10−2 S cm−1. Calculate its molar conductivity and degree of dissociation (α). Given λ0 (K+) = 73.5 S cm2 mol−1 and λ0 (C1−) = 76.5 S cm2 mol−1.


The conductivity of 0.02 M AgNO3 at 25°C is 2.428 × 10−3 Ω−1 cm−1. What is its molar conductivity?


Define the following terms :

Limiting molar conductivity


A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO4and ZnSO4 until 2.8g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y. 
(Molar mass: Fe=56g mol-1,Zn=65.3g mol-1,1F=96500C mol-1)


Assertion: Copper sulphate can be stored in zinc vessel.

Reason: Zinc is less reactive than copper.


Which of the following halogen acids is the strongest reducing agent?


Molar conductivity of substance “A” is 5.9 × 103 S/m and “B” is 1 × 10–16 S/m. Which of the two is most likely to be copper metal and why?


The following questions are case-based questions. Read the passage carefully and answer the questions that follow:

Rahul set up an experiment to find the resistance of aqueous KCl solution for different concentrations at 298 K using a conductivity cell connected to a Wheatstone bridge. He fed the Wheatstone bridge with a.c. power in the audio frequency range 550 to 5000 cycles per second. Once the resistance was calculated from the null point, he also calculated the conductivity K and molar conductivity ∧m and recorded his readings in tabular form.
S. No. Conc.
(M)
k S cm−1 ∧m S cm2 mol−1
1. 1.00 111.3 × 10−3 111.3
2. 0.10 12.9 × 10−3 129.0
3. 0.01 1.41 × 10−3 141.0

Answer the following questions:

(a) Why does conductivity decrease with dilution? (1)

(b) If `∧_"m"^0` of KCl is 150.0 S cm2 mol−1, calculate the degree of dissociation of 0.01 M KCI. (1)

(c) If Rahul had used HCl instead of KCl then would you expect the ∧m values to be more or less than those per KCl for a given concentration? Justify. (2)

OR

(c) Amit a classmate of Rahul repeated the same experiment with CH3COOH solution instead of KCl solution. Give one point that would be similar and one that would be different in his observations as compared to Rahul. (2)


The specific conductance of 2.5 × 10-4 M formic acid is 5.25 × 10-5 ohm-1 cm-1. Calculate its molar conductivity and degree of dissociation.

Given `λ°_("H"^+)` = 349.5 ohm-1 cm2 mol-1 and

`λ°_("HCOO"^-)  = 50.5 " ohm"^-1 "cm"^2  "mol"^-1`


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