The final solution had a volume of 1.0L and a molarity of 0.925. How many moles of Nickel (II) chloride were present in the solution? Show your work.

0.925 moles of Nickel (II) chloride

Explanation:

Now, we know that the number of moles in a solution is given by;

n=CV

n= number of moles

C= molar concentration of solution = 0.925 M

V= volume of solution= 1.0 L

So having known all the parameters except the unknown (number of moles of Nickel (II) chloride) we can now proceed to find the unknown.

n= 0.925M ×1.0L

n= 0.925 moles of Nickel (II) chloride

i just learnt this recently but i need to look through my notes again