Prepare 1.00L 0.50M NiCl2 from NiCl2*6H2O

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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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To prepare 1.00 L of 0.50M NiCl2 solution, we will need to start with NiCl2*6H2O salt. The first step is to determine the molar mass of NiCl2*6H2O, which is 237.69 g/mol. Next, we need to calculate the amount of NiCl2*6H2O needed to make 1.00 L of 0.50M solution.

1.00 L of 0.50M solution means we need 0.50 moles of NiCl2 in 1.00 L of solution. Since the molar mass of NiCl2*6H2O is 237.69 g/mol, we will need (0.50 mol) * (237.69 g/mol) = 118.845 g of NiCl2*6H2O.

To prepare the solution, we will need to dissolve 118.845 g of NiCl2*6H2O in enough water to make 1.00 L of solution. We can use a volumetric flask to accurately measure 1.00 L of water and then add the NiCl2*6H2O salt to it. Make sure to stir the solution until all of the salt is dissolved.

Finally, we can use a graduated cylinder or a pipette to accurately measure 1.00 L of the prepared solution. This solution will have a concentration of 0.50M NiCl2, meaning that for every liter of solution, there will be 0.50 moles of NiCl2 present.

In summary, to prepare 1.00 L of 0.50M NiCl2 solution from NiCl2*6H2O salt, we will need to dissolve 118.845 g of the salt in 1.00 L of water. This will result in a solution with a concentration of 0.50M NiCl2.