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Calculate the amount of chloride ions (Cl-) present in a 0.500 g sample of a sodium chloride (NaCl) solution using coulometry. The electric current used in the process was 2.26 A, and the time taken to completely react the sample was 120 s. The molecular weight of NaCl is 58.44 g/mol, and 1 mol of NaCl contains 1 mol of Cl-.

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To calculate the amount of chloride ions  Cl-  present in the 0.500 g sample of sodium chloride  NaCl  solution using coulometry, we need to first determine the number of moles of electrons transferred during the reaction. Coulometry is based on Faraday's laws of electrolysis, which states that the amount of substance consumed or produced at an electrode during electrolysis is directly proportional to the amount of electric charge passed through the solution. The relationship between the amount of substance and the charge is given by:Amount of substance  in moles  =  Charge in coulombs  /  Faraday's constant  n where n is the number of electrons transferred in the reaction, and Faraday's constant  F  is approximately 96485 C/mol.First, we need to calculate the total charge passed through the solution:Charge  Q  = Current  I   Time  t Q = 2.26 A  120 sQ = 271.2 CIn the reaction, NaCl is dissociated into Na+ and Cl- ions. The reaction involves the transfer of one electron per chloride ion  Cl- :NaCl  Na+ + Cl-Now, we can calculate the number of moles of Cl- ions using Faraday's law:Amount of Cl-  in moles  =  Charge in coulombs  /  Faraday's constant  n Amount of Cl- = 271.2 C /  96485 C/mol  1 Amount of Cl-  0.00281 molNow that we have the number of moles of Cl-, we can calculate the mass of Cl- in the sample:Mass of Cl- =  Amount of Cl- in moles    Molecular weight of NaCl Mass of Cl- = 0.00281 mol  58.44 g/molMass of Cl-  0.164 gSo, there are approximately 0.164 g of chloride ions  Cl-  present in the 0.500 g sample of sodium chloride  NaCl  solution.

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