Suppose 1.73g of barium nitrate is dissolved in 100.mL of a 63.0mM aqueous solution of sodium chromate.
Calculate the final molarity of nitrate anion in the solution. You can assume the volume of the solution doesn't change when the barium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.

The final molarity of nitrate anion in the solution is 0.258 M.

Explanation:

Mass of barium nitrate  1.73 g

Moles of barium nitrate =

1  mole of barium nitrate gives 2 moles of nitrate ions,then 0.00662 mol of barium nitrate will give :

of nitrate ions

moles of nitrate ion added to the solution = 0.01324 mol

Molarity of the barium nitrate solution = 63.0 mM = 0.063 M

1 mM = 0.001 M

Volume of barium nitrate solution = 100 mL = 0.100 L

1 m L = 0.001 mL

Moles of barium nitrate in the solution = n

Moles of nitrate ion in solution = 2 0.0063 × mol = 0.0126 mol

Total moles og nitrate ions = 0.01324 mol + 0.0126 mol = 0.02584 mol

Final molarity of nitrate anion in the solution.:

The final molarity of nitrate anion in the solution is 0.258 M.

is there multiple choice answers?