Chapter 15, Problem 57A

(a)

Interpretation Introduction

Interpretation:

The new molarity when $\text{150}\text{mL}$ of water is added to $\text{125}\text{mL}$ of $\text{0}\text{.200}\text{M}\text{HBr}$ has to be calculated.

Concept Introduction: Molarity: Molarity is defined as the number of moles of solute in one liter of solution. Molarity is the preferred concentration unit for stoichiometry calculations. The formula is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Concentration of the solution is calculated using the formula,

$M_1{V}_1=M_2{V}_2$

Answer to Problem 57A

The new molarity when $\text{150}\text{mL}$ of water is added to $\text{125}\text{mL}$ of $\text{0}\text{.200}\text{M}\text{HBr}$ is $\text{0}\text{.0909}\text{M}\text{.}$

Explanation of Solution

Given,

Molarity of hydroboric acid solution $\left(M_{\text{1}}\right)$ is $\text{0}\text{.200}\text{M}\text{.}$

Volume of hydroboric acid solution $\left(V_{\text{1}}\right)$ is $125\text{mL}\text{.}$

Volume of water $\left(V_{\text{2}}\right)$ is $\text{275}\text{mL}\left(\text{150}\text{.0+125}\text{mL}\right)\text{.}$

Molarity is defined as the number of moles of solute in one liter of solution. The formula for calculating molarity is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Concentration of the solution is calculated using the formula,

$M_1{V}_1=M_2{V}_2$

Here,

$M_1\text{=}$ Initial molarity of the solution

$V_1=$ Initial volume of the solution

$M_2=$ Final molarity of the solution

$V_2=$ Final volume of the solution

The new molarity of the solution is calculated as,

New molarity= $\frac{M_1{V}_1}{V_2}$

New molarity= $\frac{\left(\text{0}\text{.200}\text{M}\right)\left(\text{225}\text{.0}\text{mL}\right)}{\text{275}\text{mL}}$

New molarity= $\text{0}\text{.0909}\text{M}$

The new molarity when $\text{150}\text{mL}$ of water is added to $\text{125}\text{mL}$ of $\text{0}\text{.200}\text{M}\text{HBr}$ is $\text{0}\text{.0909}\text{M}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The new molarity when $\text{150}\text{mL}$ of water is added to $\text{155}\text{mL}$ of $\text{0}\text{.250}\text{M}\text{Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ has to be calculated.

Concept Introduction: Molarity: Molarity is defined as the number of moles of solute in one liter of solution. Molarity is the preferred concentration unit for stoichiometry calculations. The formula is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Concentration of the solution is calculated using the formula,

$M_1{V}_1=M_2{V}_2$

Answer to Problem 57A

The new molarity when $\text{150}\text{mL}$ of water is added to $\text{155}\text{mL}$ of $\text{0}\text{.250}\text{M}\text{Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ is $\text{0}\text{.127}\text{M}\text{.}$

Explanation of Solution

Given,

Molarity of $\text{Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$

$\left(M_{\text{1}}\right)$ is $\text{0}\text{.250}\text{M}\text{.}$

Volume of $\text{Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$

$\left(V_{\text{1}}\right)$ is $155\text{mL}\text{.}$

Volume of water $\left(V_{\text{2}}\right)$ is $\text{305}\text{mL}\left(\text{155+150 mL}\right)\text{.}$

Molarity is defined as the number of moles of solute in one liter of solution. The formula for calculating molarity is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Concentration of the solution is calculated using the formula,

$M_1{V}_1=M_2{V}_2$

Here,

$M_1\text{=}$ Initial molarity of the solution

$V_1=$ Initial volume of the solution

$M_2=$ Final molarity of the solution

$V_2=$ Final volume of the solution

The new molarity of the solution is calculated as,

New molarity= $\frac{M_1{V}_1}{V_2}$

New molarity= $\frac{\left(\text{0}\text{.250}\text{M}\right)\left(\text{155}\text{.0}\text{mL}\right)}{\text{305}\text{mL}}$

New molarity= $\text{0}\text{.1270 M}$

The new molarity when $\text{150}\text{mL}$ of water is added to $\text{155}\text{mL}$ of $\text{0}\text{.250}\text{M}\text{Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ is $\text{0}\text{.127}\text{M}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The new molarity when $\text{150}\text{mL}$ of water is added to $\text{0}\text{.500}\text{L}$ of $\text{0}\text{.250}\text{M}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ has to be calculated.

Concept Introduction: Molarity: Molarity is defined as the number of moles of solute in one liter of solution. Molarity is the preferred concentration unit for stoichiometry calculations. The formula is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Concentration of the solution is calculated using the formula,

$M_1{V}_1=M_2{V}_2$

Answer to Problem 57A

The new molarity when $\text{150}\text{mL}$ of water is added to $\text{0}\text{.500}\text{L}$ of $\text{0}\text{.250}\text{M}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $\text{0}\text{.1923}\text{M}\text{.}$

Explanation of Solution

Given,

Molarity of phosphoric acid solution $\left(M_{\text{1}}\right)$ is $\text{0}\text{.250}\text{M}\text{.}$

Volume of phosphoric acid solution $\left(V_{\text{1}}\right)$ is $\text{0}\text{.500}\text{L}\left(\text{500}\text{mL}\right)\text{.}$

Volume of water $\left(V_{\text{2}}\right)$ is $\text{650}\text{mL}\left(\text{150+500}\text{mL}\right)\text{.}$

Molarity is defined as the number of moles of solute in one liter of solution. The formula for calculating molarity is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Concentration of the solution is calculated using the formula,

$M_1{V}_1=M_2{V}_2$

Here,

$M_1\text{=}$ Initial molarity of the solution

$V_1=$ Initial volume of the solution

$M_2=$ Final molarity of the solution

$V_2=$ Final volume of the solution

The new molarity of the solution is calculated as,

New molarity= $\frac{M_1{V}_1}{V_2}$

New molarity= $\frac{\left(\text{0}\text{.250}\text{M}\right)\left(\text{500}\text{mL}\right)}{\text{650}\text{mL}}$

New molarity= $\text{0}\text{.1923}\text{M}$

The new molarity when $\text{150}\text{mL}$ of water is added to $\text{0}\text{.500}\text{L}$ of $\text{0}\text{.250}\text{M}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $\text{0}\text{.1923}\text{M}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The new molarity when $\text{150}\text{mL}$ of water is added to $\text{15}\text{mL}$ of $\text{18}\text{.0}\text{M}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ has to be calculated.

Concept Introduction: Molarity: Molarity is defined as the number of moles of solute in one liter of solution. Molarity is the preferred concentration unit for stoichiometry calculations. The formula is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Concentration of the solution is calculated using the formula,

$M_1{V}_1=M_2{V}_2$

Answer to Problem 57A

The new molarity when $\text{150}\text{mL}$ of water is added to $\text{15}\text{mL}$ of $\text{18}\text{.0}\text{M}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{1}\text{.6}\text{M}\text{.}$

Explanation of Solution

Given,

Molarity of sulfuric acid solution $\left(M_{\text{1}}\right)$ is $\text{18}\text{.0}\text{M}\text{.}$

Volume of sulfuric acid solution $\left(V_{\text{1}}\right)$ is $15\text{mL}\text{.}$

Volume of water $\left(V_{\text{2}}\right)$ is $\text{165}\text{mL}\left(\text{150}\text{.0+15}\text{mL}\right)\text{.}$

Molarity is defined as the number of moles of solute in one liter of solution. The formula for calculating molarity is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Concentration of the solution is calculated using the formula,

$M_1{V}_1=M_2{V}_2$

Here,

$M_1\text{=}$ Initial molarity of the solution

$V_1=$ Initial volume of the solution

$M_2=$ Final molarity of the solution

$V_2=$ Final volume of the solution

The new molarity of the solution is calculated as,

New molarity= $\frac{M_1{V}_1}{V_2}$

New molarity= $\frac{\left(\text{18}\text{M}\right)\left(\text{15}\text{mL}\right)}{165\text{mL}}$

New molarity= $\text{1}\text{.63}\text{M}$

The new molarity when $\text{150}\text{mL}$ of water is added to $\text{15}\text{mL}$ of $\text{18}\text{.0}\text{M}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{1}\text{.6}\text{M}\text{.}$