Q101P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate each of the following quantities: (a) Volume of 2.050 M copper(II) nitrate that must be diluted with water to prepare 750.0 mL of a 0.8543 M solution (b) Volume of 1.63 M calcium chloride that must be diluted with water to prepare 350. mL of a 2.86102 M chloride ion solution (c) Final volume of a 0.0700 M solution prepared by diluting 18.0 mL of 0.155 M lithium carbonate with water.

Step-by-Step Solution

Answer

(a) The volume of 2.050 M copper(II) nitrate that must be diluted with water to prepare 750.0 mL of a 0.8543 M solution is a 313 mL Cu(NO₃)₂ solution.

(b) The volume of 1.63 M calcium chloride that must be diluted with water to prepare 350 mL of a 2.86102 M chloride ion solution is a 3.07 mL CaCl₂ solution.

(c) The volume of a 0.0700 M solution prepared by diluting 18.0 mL of 0.155 M lithium carbonate with water is a 39.9 mL Li₂CO₃ solution.

a. Step 1: Finding the volume of 2.050 M copper (II) nitrate that must be diluted with water to prepare 750.0 mL of a 0.8543 M solution

On substituting and solving,

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

b. Step 2: Finding the volume of 1.63 M calcium chloride that must be diluted with water to prepare 350 mL of a 2.86102 M chloride ion solution

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.

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

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

$=3.07 mL CaC{l}_2 solution$

c. Step 3: Finding the volume of a 0.0700 M solution prepared by diluting 18.0 mL of 0.155 M lithium carbonate with water

On and solving,

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