Respuesta :
Answer:
The final molarity of nitrate anion in the solution is 0.258 M.
Explanation:
[tex]molarity=\frac{moles}{volume(L)}[/tex]
Mass of barium nitrate 1.73 g
Moles of barium nitrate =[tex]\frac{1.73 g}{261.34 g/mol}=0.00662 mol[/tex]
[tex]Ba(NO_3)_2(aq)\rightarrow Ba^{2+}(aq)+NO_3^{-}(aq)[/tex]
1 mole of barium nitrate gives 2 moles of nitrate ions,then 0.00662 mol of barium nitrate will give :
[tex]2\times 0.00662 mol =0.01324 mol[/tex] 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
[tex]n=0.100 L\times 0.063 M=0.0063 mol[/tex]
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.:
[tex]\frac{0.02584 mol}{0.100 L}=0.2584 M\approx 0.258 M[/tex]
The final molarity of nitrate anion in the solution is 0.258 M.
