12 public questions tagged with this topic.
α = (Lambdam/Lambdam°) = (17.5/350) = 0.05 . Ka = (α² c/1 - α) = (0.05² × 0.05/1 - 0.05) = (0.0025 × 0.05/0.95) = 1.316 × 10⁻⁴ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
α = (Lambdam/Lambdam°) = (19/380) = 0.05 . Ka = (α² c/1 - α) = (0.05² × 0.02/1 - 0.05) = (0.0025 × 0.02/0.95) = 5.263 × 10⁻⁵ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
α = (Lambdam/Lambdam°) = (19/380) = 0.05 . Ka = (α² c/1 - α) = (0.05² × 0.02/1 - 0.05) = (0.0025 × 0.02/0.95) = 5.263 × 10⁻⁵ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
α = (Lambdam/Lambdam°) = (40/400) = 0.1 . Ka = (α² c/1 - α) = (0.1² × 0.1/1 - 0.1) = (0.01 × 0.1/0.9) = 1.111 × 10⁻³ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
α = (Lambdam/Lambdam°) = (40/400) = 0.1 . Ka = (α² c/1 - α) = (0.1² × 0.1/1 - 0.1) = (0.01 × 0.1/0.9) = 1.111 × 10⁻³ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
α = sqrt(Ka/c) = sqrt(1.78 × 10⁻⁵/0.001) = 0.133 . α = (Lambdam/Lambdam°) , 0.133 = (48.15/Lambdam°) , Lambdam° = 390.5 S cm² mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
α = sqrt(Ka/c) = sqrt(1.78 × 10⁻⁵/0.001) = 0.133 . α = (Lambdam/Lambdam°) , 0.133 = (48.15/Lambdam°) , Lambdam° = 390.5 S cm² mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
Dissociation constant, is consistent with established principles of cell signaling, receptor pharmacology and cellular regulation. Experimental measurements of binding parameters, genetic loss-of-function studies and pharmacological interventions all converge on the same interpretation. Related options address neighboring concepts but do not satisfy the precise criterion stated in the question.
Ref: NCERT Biology Class 11–12 Alberts et al Molecular Biology of the Cell Lodish et al, Molecular Cell Biology Cooper & Hausman, The Cell Abbas et al., Cellular and Molecular Immunology (for immunology sections)
The equilibrium dissociation constant Kd is the ratio of the dissociation rate constant to the association rate constant. A higher numerical value of Kd therefore indicates that dissociation is favored relative to association, resulting in weaker net binding. Affinity is consequently the reciprocal of Kd. This inverse relationship is fundamental to the quantitative interpretation of all reversible ligand–receptor interactions.
Ref: NCERT Biology Class 11–12 Alberts et al Molecular Biology of the Cell Lodish et al, Molecular Cell Biology Cooper & Hausman, The Cell Abbas et al., Cellular and Molecular Immunology (for immunology sections)
The equilibrium dissociation constant Kd is the ratio of the dissociation rate constant to the association rate constant. A higher numerical value of Kd therefore indicates that dissociation is favored relative to association, resulting in weaker net binding. Affinity is consequently the reciprocal of Kd. This inverse relationship is fundamental to the quantitative interpretation of all reversible ligand–receptor interactions.
Ref: NCERT Biology Class 11–12 Alberts et al Molecular Biology of the Cell Lodish et al, Molecular Cell Biology Cooper & Hausman, The Cell Abbas et al., Cellular and Molecular Immunology (for immunology sections)
The equilibrium dissociation constant Kd is the ratio of the dissociation rate constant to the association rate constant. A higher numerical value of Kd therefore indicates that dissociation is favored relative to association, resulting in weaker net binding. Affinity is consequently the reciprocal of Kd. This inverse relationship is fundamental to the quantitative interpretation of all reversible ligand–receptor interactions.
Ref: NCERT Biology Class 11–12 Alberts et al Molecular Biology of the Cell Lodish et al, Molecular Cell Biology Cooper & Hausman, The Cell Abbas et al., Cellular and Molecular Immunology (for immunology sections)
When half of the available receptors are occupied, fractional saturation equals 0.5. Substitution into the binding equation shows that the free ligand concentration at this point must equal the dissociation constant Kd. Measurement of the ligand concentration that produces half-maximal occupancy therefore yields Kd directly, provided nonspecific binding has been subtracted and equilibrium has been reached.
Ref: NCERT Biology Class 11–12 Alberts et al Molecular Biology of the Cell Lodish et al, Molecular Cell Biology Cooper & Hausman, The Cell Abbas et al., Cellular and Molecular Immunology (for immunology sections)
