Q3.125CP - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate each of the following quantities: 

(a) Volume of 18.0 M sulfuric acid that must be added to water to prepare 2.00 L of a 0.429 M solution 

(b) Molarity of the solution obtained by diluting 80.6 mL of 0.225 M ammonium chloride to 0.250 L 

(c) Volume of water added to 0.130 L of 0.0372 M sodium hydroxide to obtain a 0.0100 M solution (assume the volumes are additive at these low concentrations)

(d) Mass of calcium nitrate in each milliliter of a solution prepared by diluting 64.0 mL of 0.745 M calcium nitrate to a final volume of 0.100 L

Step-by-Step Solution

Answer

Answer

1. Volume of sulfuric acid is 0.0477L

2. Molarity of the solution is 0.0725 M

3. Volume of water is 0.354L

4. Mass of calcium nitrate is 0.0782g/L

Step 1: (a) Calculating volume of sulfuric acid

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

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

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

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

Step 2: (b) Calculating the molarity of the solution

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 × 10^-3 = 0.0806 L

V₂ = 2.250L

M₂=?

 

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

Step 3: (c) Calculating volume of water

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

 

M₁ = 0.0372M

V₁ = 0.130L

M₂ = 0.0100M

V₂ =?

 

 

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

$\begin{array}{c}\text{Volume of water added ( L) = Final volume - Initial volume}\\ \text{ = 0.4836 - 0.130}\\ \text{ = 0.3536 L}\end{array}$

Step 4: (d) Calculating mass of calcium nitrate

M₁ = 0.745M

V₁ = 64.0mL × 10^-3 = 0.064L

V₂ = 0.100L

M₂ = ?

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

Converting from moles of solute to grams per milliliter 

 

 $\begin{array}{c}{\text{Mass of (CaNO}}_3{)}_2{\text{= (4.768mol Ca(NO}}_{\text{3}}{\text{)}}_2)\times \left(\frac{{\text{164.10g Ca(NO}}_{\text{3}}{\text{)}}_2}{1{\text{ molCa(NO}}_{\text{3}}{\text{)}}_2}\right)\times {\text{ 10}}^{-3}\text{mL }\\ {\text{ = 0.0782 g Ca(NO}}_{\text{3}}{\text{)}}_2\text{/ mL}\end{array}$