Answer:
The final molarity of nitrate anion in the solution is 0.258 M.
Explanation:
![molarity=\frac{moles}{volume(L)}](https://tex.z-dn.net/?f=molarity%3D%5Cfrac%7Bmoles%7D%7Bvolume%28L%29%7D)
Mass of barium nitrate 1.73 g
Moles of barium nitrate =![\frac{1.73 g}{261.34 g/mol}=0.00662 mol](https://tex.z-dn.net/?f=%5Cfrac%7B1.73%20g%7D%7B261.34%20g%2Fmol%7D%3D0.00662%20mol)
![Ba(NO_3)_2(aq)\rightarrow Ba^{2+}(aq)+NO_3^{-}(aq)](https://tex.z-dn.net/?f=Ba%28NO_3%29_2%28aq%29%5Crightarrow%20Ba%5E%7B2%2B%7D%28aq%29%2BNO_3%5E%7B-%7D%28aq%29)
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
![n=0.100 L\times 0.063 M=0.0063 mol](https://tex.z-dn.net/?f=n%3D0.100%20L%5Ctimes%200.063%20M%3D0.0063%20mol)
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](https://tex.z-dn.net/?f=%5Cfrac%7B0.02584%20mol%7D%7B0.100%20L%7D%3D0.2584%20M%5Capprox%200.258%20M)
The final molarity of nitrate anion in the solution is 0.258 M.