A solution contains Cr3+ ion and Mg2+ ion. The addition of 1.00 L of 1.55M NaF solution is required to cause the complete precipitation of these ions as CrF3(s) and MgF2(s). The total mass of the precipitate is 49.9g .Find the mass of Cr3+ in the original solution.

Mass of Cr3+ = (49.9 g) x (3 mol Cr3+/1 mol CrF3) x (1 mol Cr3+/1 mol Mg2+) x (24.3 g Cr3+/1 mol Cr3+) = 37.2 g Cr3+

To find the mass of Cr³⁺ in the original solution, we need to determine the moles of Cr³⁺ and then convert it to mass using the molar mass.

Given:
Volume of NaF solution added = 1.00 L
Concentration of NaF solution = 1.55 M
Mass of precipitate (CrF₃ + MgF₂) = 49.9 g

Step 1: Determine the moles of NaF added.
Moles of NaF = concentration of NaF solution × volume of NaF solution
Moles of NaF = 1.55 mol/L × 1.00 L
Moles of NaF = 1.55 mol

Step 2: Determine the moles of F⁻ ions.
Since each NaF molecule provides 1 F⁻ ion, the moles of F⁻ ions will be the same as the moles of NaF.
Moles of F⁻ ions = 1.55 mol

Step 3: Determine the moles of Cr³⁺ and Mg²⁺ ions.
Since the stoichiometry of the reaction is 1:3 (1 Cr³⁺ : 3 F⁻), the moles of Cr³⁺ will be 1/3 of the moles of F⁻ ions.
Moles of Cr³⁺ = 1/3 × 1.55 mol
Moles of Cr³⁺ = 0.5167 mol

Step 4: Determine the mass of Cr³⁺ in the original solution.
Molar mass of Cr³⁺ = 3 × 52 g/mol (Cr³⁺ has a molar mass of 52 g/mol)
Mass of Cr³⁺ = moles of Cr³⁺ × molar mass of Cr³⁺
Mass of Cr³⁺ = 0.5167 mol × 3 × 52 g/mol
Mass of Cr³⁺ = 80.4156 g

Therefore, the mass of Cr³⁺ in the original solution is approximately 80.42 g.

To find the mass of Cr3+ in the original solution, we can start by determining the moles of Cr3+ and Mg2+ ions present in the solution.

Let's assume that x moles of Cr3+ ions and y moles of Mg2+ ions were initially present in the solution.

The balanced chemical equation for the precipitation reaction is:

3Cr3+ + 6F- → CrF3(s)
3Mg2+ + 6F- → MgF2(s)

From the equation, we can see that the molar ratio between Cr3+ ions and CrF3 is 3:1. This means that for every 3 moles of Cr3+ ions, we will get 1 mole of CrF3 precipitate.

Similarly, the molar ratio between Mg2+ ions and MgF2 is also 3:1.

We know that the total volume of the solution is 1.00 L and the concentration of NaF solution is 1.55M. Therefore, the number of moles of F- ions in the solution is:

1.00 L × 1.55 mol/L = 1.55 moles of F-

Since the molar ratio between F- ions and both Cr3+ and Mg2+ ions is 6:1, we can calculate the moles of Cr3+ and Mg2+ ions as follows:

For Cr3+ ions:
(1.55 moles of F-) / 6 = 0.258 moles of Cr3+

For Mg2+ ions:
(1.55 moles of F-) / 6 = 0.258 moles of Mg2+

Now, we need to find the mass of Cr3+ in the solution. To do that, we'll use the molar mass of Cr3+, which is 51.996 grams/mol.

Mass of Cr3+ = moles of Cr3+ × molar mass of Cr3+
Mass of Cr3+ = 0.258 moles × 51.996 g/mol ≈ 13.428 grams

Therefore, the mass of Cr3+ in the original solution is approximately 13.428 grams.