Advertisements
Advertisements
प्रश्न
The conductivity of 0.001 mol L-1 solution of CH3COOH is 3.905× 10-5 S cm-1. Calculate its molar conductivity and degree of dissociation (α) Given λ°(H+)= 349.6 S cm2 mol-1 and λ°(CH3COO)= 40.9S cm2mol-1.
Advertisements
उत्तर
Given:
Conductivity (k) = 3.905 x 10-5 S cm-1
Concentration ofelectrolyte (c) = 0.001 mol L-1
`^^_m=k/cxx1000=39.05S cm^2mol^(-1)`
`^^_m^@=lambda_(CH_3COO^-)^@+lambda_(H+)^@`
= 40.9 + 349.6
= 390.5 S cm2 mol-1
Degree of dissociation = `39.05/390.5=0.1`
APPEARS IN
संबंधित प्रश्न
Define “Molar conductivity”.
State Kohlrausch’s law of independent migration of ions.
Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Λm of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer.
When acidulated water (dil.H2SO4 solution) is electrolysed, will the pH of the solution be affected? Justify your answer.
Assertion: Λm for weak electrolytes shows a sharp increase when the electrolytic solution is diluted.
Reason: For weak electrolytes degree of dissociation increases with dilution of solution.
Consider figure and answer the question to given below.
How will the concentration of Zn2+ ions and Ag+ ions be affected after the cell becomes ‘dead’?
Given below are two statements:
Statements I: The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH3COOH (weak electrolyte).
Statement II: Molar conductivity decreases with decrease in concentration of electrolyte.
In the light of the above statements, choose the most appropriate answer from the options given below:
The variation of molar conductivity with concentration of an electrolyte (X) m aqueous solution is shown in the given figure.
The electrolyte X is ______.
The unit of molar conductivity is ______.
Discuss the variation of conductivity and molar conductivity with concentration.
