# A solution contains Cr3+ and Mg2+. The addition of 1.00 L of 1.51 M 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 50.1 g .Find the mass of Cr3+ in the original solution.

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A solution contains Cr3+ and Mg2+. The addition of 1.00 L of 1.51 M 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 50.1 g .

Find the mass of Cr3+ in the original solution.

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Step 1

A solution contains Cr3+ and Mg2+. The addition of 1.00 L of 1.51 M 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 50.1 g.

The mass of Cr3+ in the original solution is to be determined.

Step 2

Moles of NaF in the solution = molarity of NaF×Volume = 1.51 moles.

Now suppose these 1.51 moles of NaF reacts with Cr3+ and Mg2+to form x grams of CrF3 precipitate and (50.1-x) grams of MgF2 precipitate.

Step 3

Determine the moles of NaF that reacts with the given Cr3+ in the solution.

Molar mass of CrF3 = 108.99 g/mol

Mass of CrF3 produced = x grams

As per the ba...

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