Calculating grams:

If you wanted to make 0.800 Liters of a 0.531 M solution of Nickel (II) chloride, NiCl2, how many grams of NiCl2 would you need?
M = 0.531 M
n = 0.4248 mol
V = 0.800 L

n = M x V 0.4248 = 0.531 x 0.800

Now that you know the number of moles, you will need the molar mass of NiCl2 to calculate the grams of NiCl2.

Ni = 58.6934 = 59
Cl = 35.453 x 2 = 35 x 2 = 70
NiCl2 = 129 g/mol

Now use the unit canceling method to determine the grams of NiCl2 required:

.4248 mol NiCl2 x 129 g NiCl2 / 1 mol NiCl2 = 54.799 g NiCl2

Is this correct??

The answer to your question is 55.224 g of NiCl₂

Explanation:

I will solve it and then I compare the results.

Data

Volume = 0.8 l

Molarity = 0.531

substance = NiCl₂

Process

1.- Calculate the molar mass of NiCl₂

NiCl₂ = 59 + (35.5 x 2)

         = 130 g

2.- Calculate the number of moles

Molarity = moles / volume

-Solve for moles

  moles = Molarity x volume

-Substitution

  moles = 0.531 x 0.8

-Result

  moles = 0.4248

3.- Convert the moles to mass

                   130 g 1 mol of NiCl₂

                      x     0.4248 moles

                           x = (0.4248 x 130) / 1

-Simplifying

                         x = 55.224 / 1

-Result

                        x = 55.224 g of NiCl₂

4.- We need 55.224 g of NiCl₂ to prepare 0.8l of a solution 0.531 M.

You have similar results, just our molar mass is different.

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