Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
Question
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Chapter 16, Problem 16B.5P

(a)

Interpretation Introduction

Interpretation:

The Kohlrausch’s law has to be verified for the molar conductivity.  The limiting molar conductivity Λmo  and coefficient k have to be calculated.

Concept introduction:

Conductivity is a property of a material that depends upon the charge carriers in a sample.  Molar conductivity is the conductivity per unit molar concentration.  According to Kohlrausch’s law, the molar conducivities of the dilute solutions containing strong electrolytes depend upon the square root of the concentration.  The conductivity of a solution when the concentration tends to zero is known as the limiting molar conductivity.  Limiting molar conductivity is the sum of contributions from individual ions.

(a)

Expert Solution
Check Mark

Answer to Problem 16B.5P

The graph between molar conductivity and c12 gives a straight line which proves that that molar conductivity follows the Kohlrausch’s law.

The value of limiting molar conductivity (Λmo) is 0.125Sdm3cm1mol1_.

The value of Kohlrausch’s constant (k) is 0.066Sdm3cm1mol1(moldm3)12_.

Explanation of Solution

The equation for the Kohlrausch’s law is shown below.

    Λm=Λmokc12        (1)

Where,

  • Λm is the molar conductivity.
  • Λmo is the limiting molar conductivity.
  • k is the Kohlrausch’s constant.
  • c is the molar concentration.

The molar conductivity is calculated by the formula shown below.

    Λm=κc        (2)

Where,

  • Λm is the molar conductivity.
  • κ is the conductivity.
  • c is the molar concentration.

The conductivity of a solution is defined as shown below.

  κ=CR        (3)

Where,

  • C is cell constant.
  • R is the resistance.

Substitute κ=CR in equation (2).

    Λm=CRc        (4)

The given data is shown below.

c/(moldm3)R/Ω
0.000503314
0.00101669
0.0050342.1
0.010174.1
0.02089.08
0.05037.14

The value of cell constant is given as 0.2063cm1.

The molar conductivity at each concentration is calculated using equation (4) as shown below.

c/(moldm3)R/ΩΛm=CRc(Sdm3cm1mol1)
0.0005033140.1245
0.001016690.1236
0.0050342.10.1206
0.010174.10.1184
0.02089.080.1158
0.05037.140.1111

The value of c12 is calculated as shown below.

c/(moldm3)R/ΩΛm=CRc(Sdm3cm1mol1)c12
0.0005033140.12450.0224
0.001016690.12360.0316
0.0050342.10.12060.0707
0.010174.10.11840.1000
0.02089.080.11580.1414
0.05037.140.11110.2236

Now plot a graph between Λm and c12 as shown below.

Atkins' Physical Chemistry, Chapter 16, Problem 16B.5P

Figure 1

The equation obtained from the graph is y=0.066x+0.125.  In the graph, y-axis represents Λm and x-axis represents c12.  Therefore, the equation can be written as Λm=0.1250.066c12.

Compare the equation Λm=0.1250.066c12 with equation (1) to obtain the value of Λmo and k.

Therefore, the value of Λmo is 0.125Sdm3cm1mol1_ and the value of Kohlrausch’s constant (k) is 0.066Sdm3cm1mol1(moldm3)12_

(b)

Interpretation Introduction

Interpretation:

The value of molar conductivity, conductivity and resistance of the 0.010moldm3NaI solution has to be calculated.

Concept introduction:

As mentioned in concept introduction in part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 16B.5P

The value of molar conductivity of 0.010moldm3NaI solution is 12.03mSm2mol1_.

The value of conductivity of 0.010moldm3NaI solution is 120.3mSm1_.

The value of resistance of the 0.010moldm3NaI solution is 171.49Ω_.

Explanation of Solution

The value of molar conductivity of the 0.010moldm3NaI solution is calculated by the formula shown below.

    Λm=Λmokc12        (1)

Where,

  • Λm is the molar conductivity.
  • Λmo is the limiting molar conductivity.
  • k is the Kohlrausch’s constant.
  • c is the molar concentration.

The molar conductivity of NaI solution is calculated as shown below.

    Λmo=λ(Na+)+λ(I)        (2)

Where,

  • λ(Na+) is the molar conductivity of sodium ion.
  • λ(I) is the molar conductivity of iodide ion.

The value of λ(Na+) and λ(I) is 5.01mSm2mol1 and 7.68mSm2mol1 respectively.

Substitute the value of λ(Na+) and λ(I) in equation (2).

    Λmo=λ(Na+)+λ(I)=5.01mSm2mol1+7.68mSm2mol1=12.69mSm2mol1

The value of Kohlrausch’s constant (k) and c is 0.066Sdm3cm1mol1(moldm3)12 and 0.010moldm3 respectively.

Substitute the value of Kohlrausch’s constant (k) and c in equation (1).

    Λm=Λmokc12=12.69mSm2mol10.066Sdm3cm1mol1(moldm3)12×(0.010moldm3)12=12.69mSm2mol10.0066S(103mS1S)dm3(103m31dm3)cm1(1cm102m)mol1=12.69mSm2mol10.66mSm2mol1=12.03mSm2mol1_

Therefore, the value of molar conductivity of 0.010moldm3NaI solution is 12.03mSm2mol1_.

The conductivity is calculated by the formula shown below.

    κ=Λmc        (3)

Where,

  • Λm is the molar conductivity.
  • κ is the conductivity.
  • c is the molar concentration.

The value of c is 0.010moldm3.

Substitute the value of c and Λm in equation (3).

    κ=Λmc=12.03mSm2mol1×0.010moldm3(1dm3103m3)=120.3mSm1_

Therefore, the value of conductivity of 0.010moldm3NaI solution is 120.3mSm1_.

The resistance of a solution is calculated as shown below.

  R=Cκ        (4)

Where,

  • C is cell constant.
  • R is the resistance.
  • κ is the conductivity.

The value of cell constant (C) is given as 0.2063cm1.

Substitute the value of C and κ in equation (4)

    R=Cκ=0.2063cm1(1cm102m)120.3mS(103S1mS)m1=20630120.3S1(1Ω1S1)=171.49Ω_

Therefore, the value of resistance of the 0.010moldm3NaI solution is 171.49Ω_.

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Chapter 16 Solutions

Atkins' Physical Chemistry

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