Consider the following reaction: Cr(NO3)3 (aq) + 2NaF (aq) --> 3NaNO3 (aq) + CrF3 (s)
If 21.0 grams of NaF are needed to precipitate all of the Cr+3 ions present in 0.125L of a solution of Cr(NO3)3, what is the molarity of the Cr(NO3)3 solution?

Your answer should be to 2 decimal places.

2.01

Explanation:

First, let's convert grams to moles

(Na) 22.99 + (F) 18.998 = 41.988

Every mole of NaF is  41.988 grams

21/41.988 = 0.500143 moles of NaF

For every Cr+3, we will need 2 NaF, so Cr+3 will be half of NaF

0.500143/2 = 0.250071

molarity = moles/liters

0.250071/0.125 = 2.0057 M

i think it is true since you can test the answer and can use observations to complete it.