In the laboratory, a student adds 21.7 g of manganese(II) chloride to a 125 mL volumetric flask and adds water to the mark on the neck of the flask. Calculate the concentration (in mol/L) of manganese(II) chloride, the manganese(II) ion and the chloride ion in the solution. [MnCl,] %3D [Mn2+] [CF] M M M
In the laboratory, a student adds 21.7 g of manganese(II) chloride to a 125 mL volumetric flask and adds water to the mark on the neck of the flask. Calculate the concentration (in mol/L) of manganese(II) chloride, the manganese(II) ion and the chloride ion in the solution. [MnCl,] %3D [Mn2+] [CF] M M M
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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![In the laboratory, a student adds 21.7 g of manganese(II) chloride to a 125 mL volumetric
flask and adds water to the mark on the neck of the flask. Calculate the concentration (in
mol/L) of manganese(II) chloride, the manganese(II) ion and the chloride ion in the
solution.
[MnCl,] =
%3D
[Mn²*] =
[CF]
MMM
||](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a11bc88-2929-4922-a603-cf674ecc8c9c%2F27cee320-0056-4f91-8a2f-a54b36c6acf4%2F29t0bmj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:In the laboratory, a student adds 21.7 g of manganese(II) chloride to a 125 mL volumetric
flask and adds water to the mark on the neck of the flask. Calculate the concentration (in
mol/L) of manganese(II) chloride, the manganese(II) ion and the chloride ion in the
solution.
[MnCl,] =
%3D
[Mn²*] =
[CF]
MMM
||
Expert Solution
Step 1
Given : Mass of MnCl2 i.e Manganese (II) chloride = 21.7 g
And volume of solution = 125 mL = 0.125 L (since 1 L = 1000 mL)
Molar mass of MnCl2 = Atomic mass of Mn + Atomic mass of Cl X 2 = 54.9 + 35.5 X 2 = 125.9 g/mol
Since mass = moles X molar mass
=> 21.7 = moles of MnCl2 X 125.9
=> moles of MnCl2 = 0.17236 mol approx.
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