91影视

The dilution problems are solved by using molarity and volume relationship.

${\mathit{M}}_{\mathbf{d}\mathbf{i}\mathbf{l}}\mathbf{}\mathbf{脳}\mathbf{}{\mathit{V}}_{\mathbf{d}\mathbf{i}\mathbf{l}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{M}}_{\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{c}}\mathbf{}\mathbf{脳}\mathbf{}{\mathit{V}}_{\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{c}}$

Where M and V stands for the molarity and volume of the dilute and concentrated solutions. In these types of dilution problems, the volume units should be same.

M1= 180M

M2= 0.429M

V2= 2.00L

V1=?

M1V1= M2V2

$$\begin{gathered} {\mathit{V}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{M}}_{\mathbf{2}}\mathbf{}{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{M}}_{\mathbf{1}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\left(0.429M\right)\left(2.00L\right)}{\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{M}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0477}\mathbf{}\mathit{L} \end{gathered}$$

The molarity is the number of moles of solute in each litre of solution.

M₁ = 0.225M

V₁ = 80.6 mL

Converting mL to L,

V₁ = 80.6 脳 10^-3= 0.0806 L

V₂ = 2.250L

M₂=?

$\begin{array}{c}{\text{M}}_2\text{=}\frac{{\text{M}}_1{\text{V}}_1}{{\text{V}}_2}\\ \text{=}\frac{\text{(0.225M)(0.0876L)}}{\text{0.250L}}\\ \text{=0.07254M}\end{array}$

The volume of water added is calculated by subtracting initial volume from final volume.

M₁ = 0.0372M

V₁ = 0.130L

M₂ = 0.0100M

V₂ =?

$\begin{array}{c}{\text{V}}_2\text{=}\frac{{\text{M}}_1{\text{V}}_1}{{\text{M}}_2}\\ \text{=}\frac{\text{(0.0372M)(0.130L)}}{\text{0.0100M}}\\ \text{=0.4836L}\\ \end{array}$

$\begin{array}{c}\text{Volumeofwateradded(L)=Finalvolume-Initialvolume}\\ \text{=0.4836-0.130}\\ \text{=0.3536L}\end{array}$

M₁ = 0.745M

V₁ = 64.0mL 脳 10^-3= 0.064L

V₂ = 0.100L

M₂= ?

$\begin{array}{c}{\text{M}}_2\text{=}\frac{{\text{M}}_1{\text{V}}_{\text{1}}}{{\text{V}}_2}\\ \text{=}\frac{\text{(0.745M)(0.0640L)}}{\text{(0.100L)}}\\ \text{=0.4768M}\end{array}$

Converting from moles of solute to grams per milliliter

$\begin{array}{c}{\text{Massof(CaNO}}_3{)}_2{\text{=(4.768molCa(NO}}_{\text{3}}{\text{)}}_2)脳\left(\frac{{\text{164.10gCa(NO}}_{\text{3}}{\text{)}}_2}{1{\text{molCa(NO}}_{\text{3}}{\text{)}}_2}\right)脳{\text{10}}^{鈭�3}\text{mL}\\ {\text{=0.0782gCa(NO}}_{\text{3}}{\text{)}}_2\text{/mL}\end{array}$