The resistance of a 0.02 mol dm-3 solution of acetic acid in a cell (cell constant= 0.2063cm-1 ) was found to be 888 ohm. What is the degree of ionization of the acid at this concentration?
Answers
Answer:
the degree of ionization is 0.98
Explanation:
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Answer:
An atom or molecule becomes an ion by the process of ionization, which produces an electrically charged particle.
Transferring electrons from one atom or molecule to another typically starts the process.
Explanation:
The formula below can be used to determine how much acetic acid is ionized at this concentration:
α = (1/R) * (1/C) * (1/K)
the cell constant K, the concentration C, and the resistance R.
As a result, = (1/888)(1/0.02 mol dm-3)(1/0.2063 cm-1) α = 0.0019
The acetic acid is therefore 0.0019 times more ionized at this concentration.
The degree of ionization of a different atom or molecule is the fraction of its atoms or molecules that have been ionized, either through the loss of electrons (in the case of an atom) or the dissociation of molecules into ions (in the case of a molecule).
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https://brainly.in/question/2718187
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