Prepare 1.00L 0.50M NiCl2 from NiCl2*6H2O

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To prepare 1.00 L of a 0.50 M solution of NiCl2 from NiCl2 * 6H2O, the first step is to calculate the number of moles required. For a 0.50 M solution, 0.50 moles of NiCl2 are needed in 1.00 L. The formula weight of NiCl2 * 6H2O must be determined to accurately weigh the hydrated compound. The molar mass of NiCl2 * 6H2O includes the contributions from nickel, chlorine, and the six water molecules. Once the formula weight is established, the appropriate mass of NiCl2 * 6H2O can be weighed to provide the necessary moles of NiCl2 for the solution preparation. The moles of NiCl2 from the hydrated salt will be equivalent to the moles of anhydrous NiCl2 required for the solution.
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How would you prepare 1.00 L of 0.50M solution sof NiCl2 from the salt NiCl2 * 6H2O
 
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The initial target is the number of MOLES of NiCl2. The moles you use of dry hydrated NiCl2 and the moles of the anhydrous NiCl2 would be identical. In the task you will actually weigh the hydrated compound. First determine formula weight of the sexa-hydrate.
 

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