Question

Suppose 1.73g of barium nitrate is dissolved in 100.mL of a 63.0mM aqueous solution of sodium chromate.

Answer 1

Answer:

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

Explanation:

$molarity=\frac{moles}{volume(L)}$

Mass of barium nitrate  1.73 g

Moles of barium nitrate =$\frac{1.73 g}{261.34 g/mol}=0.00662 mol$

$Ba(NO_3)_2(aq)\rightarrow Ba^{2+}(aq)+NO_3^{-}(aq)$

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

$2\times 0.00662 mol =0.01324 mol$ 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

$n=0.100 L\times 0.063 M=0.0063 mol$

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.:

$\frac{0.02584 mol}{0.100 L}=0.2584 M\approx 0.258 M$

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