Assuming the volumes are additive, what is the [Cl - ] ina solution obtained by
ID: 690578 • Letter: A
Question
Assuming the volumes are additive, what is the [Cl-] ina solution obtained by mixing 225 mL of .625M KCl and 615 mL of0.385 M MgCl2?I don't understand if i need to write achemical equation and then find? Why do they give you the volumesof each and how do they work as a conversion factor? I don'tunderstand the steps to solve.
I don't understand if i need to write achemical equation and then find? Why do they give you the volumesof each and how do they work as a conversion factor? I don'tunderstand the steps to solve.
Explanation / Answer
Assuming the volumes are additive, what is the[Cl-] in a solution obtained by mixing 225 mL of .625MKCl and 615 mL of 0.385 M MgCl2 No . of moles of KCl , n = Molarity * Volume in L = 0.625 M * 0.225 L = 0.140625 moles 1 mole of KCl contains 1 mole K + & 1 mole of Cl- 0.140625 mole of KCl contains 0.140625 mole K + & 0.140625 mole of Cl - No . of moles of MgCl2 , n ' = Molarity * Volume in L = 0.385 * 0.615 L = 0.236775 mol 1 mole of MgCl2 contains 1 mole Mg 2 + & 2mole of Cl - 0.236775 mole ofMgCl2 contains 0.236775 mole Mg 2 + & 2* 0.236775 mole of Cl - = 0.47355 moles of Cl- Total no . of moles of Cl - is N = 0.47355 + 0.140625 = 0.614175 moles Total Volume , V = 0.225 L + 0.615 L = 0.84 L So the [Cl-] in a solution is = N / V = 0.7312 moles / L 0.236775 mole ofMgCl2 contains 0.236775 mole Mg 2 + & 2* 0.236775 mole of Cl - = 0.47355 moles of Cl- Total no . of moles of Cl - is N = 0.47355 + 0.140625 = 0.614175 moles Total Volume , V = 0.225 L + 0.615 L = 0.84 L So the [Cl-] in a solution is = N / V = 0.7312 moles / LRelated Questions
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