91Ó°ÊÓ

On substituting and solving,

$\begin{array}{c}{V}_{conc}=\frac{M_{dil}×V_{dil}}{M_{conc}}\\              =\frac{0.8543 M×750 mL}{2.050 M}\\              =313 mL Cu(N{O}_3{)}_2 solution\end{array}$

Multiply the given calcium chloride’s molarity using the molar ratio between the calcium chloride and the chloride ion to find the chloride ion’s concentration.

$\begin{array}{c}{M}_{conc}\left(C{l}^{-}ions\right)=\frac{1.63 mol CaC{l}_2}{L so\ln }×\frac{2 mol C{l}^{-}ions}{1 mol CaC{l}_2}\\                                  =3.26 M C{l}^{-}ions\end{array}$

$\begin{array}{c}    V_{conc}=\frac{M_{dil}×V_{dil}}{M_{conc}}\\                                  =\frac{2.86×{10}^{-2}M×350mL}{3.26 M}\end{array}$

$                                 =3.07 mL CaC{l}_2 solution$

On and solving,

$\begin{array}{c}{V}_{dil}=\frac{M_{conc}×V_{conc}}{M_{dil}}\\          =\frac{0.155 M×18 mL}{0.07 M}\\          =39.9 mL L{i}_{3}C{O}_3 solution\end{array}$