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Sagot :
Certainly! Let's approach the problem step-by-step.
Step 1: Calculate the molar conductivity
The molar conductivity [tex]\(\Lambda_m\)[/tex] is given by the formula:
[tex]\[ \Lambda_m = \frac{\kappa}{c} \times 1000 \][/tex]
where:
- [tex]\(\kappa\)[/tex] is the conductivity of the solution ([tex]\(2.48 \times 10^{-2} S \, cm^{-1}\)[/tex])
- [tex]\(c\)[/tex] is the concentration (0.2 M)
- 1000 is the factor to convert from [tex]\(S \, cm^{-1}\)[/tex] to [tex]\(S \, cm^2 \, mol^{-1}\)[/tex]
Plug in the given values:
[tex]\[ \Lambda_m = \frac{2.48 \times 10^{-2}}{0.2} \times 1000 \][/tex]
Simplify the expression:
[tex]\[ \Lambda_m = \frac{2.48 \times 10^{-2}}{0.2} \times 1000 = 0.124 \times 1000 = 124.0 \, S \, cm^2 \, mol^{-1} \][/tex]
So, the molar conductivity of the solution is [tex]\(124.0 \, S \, cm^2 \, mol^{-1}\)[/tex].
Step 2: Calculate the theoretical molar conductivity at infinite dilution
The theoretical molar conductivity at infinite dilution [tex]\(\Lambda_m^0\)[/tex] can be calculated by summing the individual ionic conductivities at infinite dilution for [tex]\(K^+\)[/tex] and [tex]\(Cl^-\)[/tex]:
[tex]\[ \Lambda_m^0 = \lambda_{ K ^{+}}^{ o } + \lambda_{ Cl ^{- }}^{ o } \][/tex]
Given:
[tex]\[ \lambda_{ K ^{+}}^o = 73.5 \, S \, cm^2 \, mol^{-1} \][/tex]
[tex]\[ \lambda_{ Cl ^{- }}^o = 76.5 \, S \, cm^2 \, mol^{-1} \][/tex]
Add these two values:
[tex]\[ \Lambda_m^0 = 73.5 + 76.5 = 150.0 \, S \, cm^2 \, mol^{-1} \][/tex]
So, the theoretical molar conductivity at infinite dilution is [tex]\(150.0 \, S \, cm^2 \, mol^{-1}\)[/tex].
Step 3: Calculate the degree of dissociation ([tex]\(\alpha\)[/tex])
The degree of dissociation [tex]\(\alpha\)[/tex] can be calculated using the formula:
[tex]\[ \alpha = \frac{\Lambda_m}{\Lambda_m^0} \][/tex]
Plug in the values for [tex]\(\Lambda_m\)[/tex] and [tex]\(\Lambda_m^0\)[/tex]:
[tex]\[ \alpha = \frac{124.0}{150.0} \][/tex]
Simplify the fraction:
[tex]\[ \alpha = 0.8267 \][/tex]
So, the degree of dissociation is approximately [tex]\(0.827\)[/tex] (or [tex]\(82.7\%\)[/tex]).
Summary:
1. Molar conductivity ([tex]\(\Lambda_m\)[/tex]): [tex]\(124.0 \, S \, cm^2 \, mol^{-1}\)[/tex]
2. Theoretical molar conductivity at infinite dilution ([tex]\(\Lambda_m^0\)[/tex]): [tex]\(150.0 \, S \, cm^2 \, mol^{-1}\)[/tex]
3. Degree of dissociation ([tex]\(\alpha\)[/tex]): [tex]\(0.827\)[/tex] or [tex]\(82.7\%\)[/tex]
These are the calculated values based on the given information.
Step 1: Calculate the molar conductivity
The molar conductivity [tex]\(\Lambda_m\)[/tex] is given by the formula:
[tex]\[ \Lambda_m = \frac{\kappa}{c} \times 1000 \][/tex]
where:
- [tex]\(\kappa\)[/tex] is the conductivity of the solution ([tex]\(2.48 \times 10^{-2} S \, cm^{-1}\)[/tex])
- [tex]\(c\)[/tex] is the concentration (0.2 M)
- 1000 is the factor to convert from [tex]\(S \, cm^{-1}\)[/tex] to [tex]\(S \, cm^2 \, mol^{-1}\)[/tex]
Plug in the given values:
[tex]\[ \Lambda_m = \frac{2.48 \times 10^{-2}}{0.2} \times 1000 \][/tex]
Simplify the expression:
[tex]\[ \Lambda_m = \frac{2.48 \times 10^{-2}}{0.2} \times 1000 = 0.124 \times 1000 = 124.0 \, S \, cm^2 \, mol^{-1} \][/tex]
So, the molar conductivity of the solution is [tex]\(124.0 \, S \, cm^2 \, mol^{-1}\)[/tex].
Step 2: Calculate the theoretical molar conductivity at infinite dilution
The theoretical molar conductivity at infinite dilution [tex]\(\Lambda_m^0\)[/tex] can be calculated by summing the individual ionic conductivities at infinite dilution for [tex]\(K^+\)[/tex] and [tex]\(Cl^-\)[/tex]:
[tex]\[ \Lambda_m^0 = \lambda_{ K ^{+}}^{ o } + \lambda_{ Cl ^{- }}^{ o } \][/tex]
Given:
[tex]\[ \lambda_{ K ^{+}}^o = 73.5 \, S \, cm^2 \, mol^{-1} \][/tex]
[tex]\[ \lambda_{ Cl ^{- }}^o = 76.5 \, S \, cm^2 \, mol^{-1} \][/tex]
Add these two values:
[tex]\[ \Lambda_m^0 = 73.5 + 76.5 = 150.0 \, S \, cm^2 \, mol^{-1} \][/tex]
So, the theoretical molar conductivity at infinite dilution is [tex]\(150.0 \, S \, cm^2 \, mol^{-1}\)[/tex].
Step 3: Calculate the degree of dissociation ([tex]\(\alpha\)[/tex])
The degree of dissociation [tex]\(\alpha\)[/tex] can be calculated using the formula:
[tex]\[ \alpha = \frac{\Lambda_m}{\Lambda_m^0} \][/tex]
Plug in the values for [tex]\(\Lambda_m\)[/tex] and [tex]\(\Lambda_m^0\)[/tex]:
[tex]\[ \alpha = \frac{124.0}{150.0} \][/tex]
Simplify the fraction:
[tex]\[ \alpha = 0.8267 \][/tex]
So, the degree of dissociation is approximately [tex]\(0.827\)[/tex] (or [tex]\(82.7\%\)[/tex]).
Summary:
1. Molar conductivity ([tex]\(\Lambda_m\)[/tex]): [tex]\(124.0 \, S \, cm^2 \, mol^{-1}\)[/tex]
2. Theoretical molar conductivity at infinite dilution ([tex]\(\Lambda_m^0\)[/tex]): [tex]\(150.0 \, S \, cm^2 \, mol^{-1}\)[/tex]
3. Degree of dissociation ([tex]\(\alpha\)[/tex]): [tex]\(0.827\)[/tex] or [tex]\(82.7\%\)[/tex]
These are the calculated values based on the given information.
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