Chapter 4, Problem 4.60QP

(a)

Interpretation Introduction

Interpretation:

Mass of solute needed to make $2.50\times {10}^{2}mL$ of $0.100M$ of solution has to be identified.

Concept introduction:

 $Molarityofthesolution=\frac{Molesofsolute}{Volumeinliter}$

 The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

 $\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

 The mass of compound is given by the multiplication of mole of the compound and molar mass of the compound in gram.

 $\text{Mass}\text{=}\text{Mole×}\text{Molar}\text{mass}$

Explanation of Solution

Given,

    The concentration of solution is $\text{0}\text{.100M}$

    The volume of solution is $2.50\times {10}^{2}mL$

The number of mole of cesium iodide can be calculated by using

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

$\begin{array}{l}\text{Number }\text{of Moles}\text{=}\text{0}\text{.100M}\text{×}\text{2}\text{.50}\times {\text{10}}^2\times {10}^{-3}\text{L}\\ \text{=}\text{0}\text{.025mol}\end{array}$

The mass of cesium iodide can be calculated by using,

From the standard data molar mass of $\text{CsI=}259.8\text{g/mol}$

$\begin{array}{l}\text{=}\text{0}\text{.025mol×}259.8\text{g/mol}\\ \text{=}\text{6}\text{.495}\text{g=6}\text{.50g}\end{array}$

The $\text{6}\text{.50g}$ of cesium iodide is used to  $2.50\times {10}^{2}mL$ of a $0.100M$ of solution.

(b)

Interpretation Introduction

Interpretation:

Mass of solute needed to make $2.50\times {10}^{2}mL$ of $0.100M$ of solution has to be identified.

Concept introduction:

 $Molarityofthesolution=\frac{Molesofsolute}{Volumeinliter}$

 The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

 $\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

 The mass of compound is given by the multiplication of mole of the compound and molar mass of the compound in gram.

 $\text{Mass}\text{=}\text{Mole×}\text{Molar}\text{mass}$

Explanation of Solution

Given,

    The concentration of solution is $\text{0}\text{.100M}$

    The volume of solution is $2.50\times {10}^{2}mL$

The number of mole of $H_{2}S{O}_4$ can be calculated by using

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

$\begin{array}{l}\text{Number }\text{of Moles}\text{=}\text{0}\text{.100M}\text{×}\text{2}\text{.50}\times {\text{10}}^2\times {10}^{-3}\text{L}\\ \text{=}\text{0}\text{.025mol}\end{array}$

The mass of $H_{2}S{O}_4$ can be calculated by using,

From the standard data molar mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=}\text{98}\text{.086g/mol}$

$\begin{array}{l}\text{=}\text{0}\text{.025mol×}\text{98}\text{.086g/mol}\\ \text{=}\text{2}\text{.45g}\end{array}$

The $\text{2}\text{.45g}$ of $H_{2}S{O}_4$ is used to  $2.50\times {10}^{2}mL$ of a $0.100M$ of solution.

(c)

Interpretation Introduction

Interpretation:

Mass of solute needed to make $2.50\times {10}^{2}mL$ of a $0.100M$ of solution has to be identified.

Concept introduction:

 $Molarityofthesolution=\frac{Molesofsolute}{Volumeinliter}$

 The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

 $\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

 The mass of compound is given by the multiplication of mole of the compound and molar mass of the compound in gram.

 $\text{Mass}\text{=}\text{Mole×}\text{Molar}\text{mass}$

Explanation of Solution

Given,

    The concentration of solution is $\text{0}\text{.100M}$

    The volume of solution is $2.50\times {10}^{2}mL$

The number of mole can be calculated by using

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

$\begin{array}{l}\text{Number }\text{of Moles}\text{=}\text{0}\text{.100M}\text{×}\text{2}\text{.50}\times {\text{10}}^2\times {10}^{-3}\text{L}\\ \text{=}\text{0}\text{.025mol}\end{array}$

The mass of $N{a}_{2}C{O}_3$ can be calculated by using,

From the standard data molar mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{=}\text{105}\text{.99g/mol}$

$\begin{array}{l}\text{=}\text{0}\text{.025mol×}\text{105}\text{.99g/mol}\\ \text{=2}\text{.649g}\end{array}$

The $\text{2}\text{.649g}$ of $N{a}_{2}C{O}_3$ is used to  $2.50\times {10}^{2}mL$ of a $0.100M$ of solution.

(d)

Interpretation Introduction

Interpretation:

Mass of solute needed to make $2.50\times {10}^{2}mL$ of a $0.100M$ of solution has to be identified.

Concept introduction:

 $Molarityofthesolution=\frac{Molesofsolute}{Volumeinliter}$

 The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

 $\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

 The mass of compound is given by the multiplication of mole of the compound and molar mass of the compound in gram.

 $\text{Mass}\text{=}\text{Mole×}\text{Molar}\text{mass}$

Explanation of Solution

Given,

    The concentration of solution is $\text{0}\text{.100M}$

    The volume of solution is $2.50\times {10}^{2}mL$

The number of mole can be calculated by using

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

$\begin{array}{l}\text{Number }\text{of Moles}\text{=}\text{0}\text{.100M}\text{×}\text{2}\text{.50}\times {\text{10}}^2\times {10}^{-3}\text{L}\\ \text{=}\text{0}\text{.025mol}\end{array}$

The mass of $K_{2}C{r}_2{O}_7$ can be calculated by using,

From the standard data molar mass of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{=}\text{294}\text{.2g/mol}$

$\begin{array}{l}\text{=}\text{0}\text{.025mol×}294.2\text{g/mol}\\ \text{=}\text{7}\text{.355g}\end{array}$

The $\text{7}\text{.355g}$ of $K_{2}C{r}_2{O}_7$ is used to  $2.50\times {10}^{2}mL$ of a $0.100M$ of solution.

(e)

Interpretation Introduction

Interpretation:

Mass of solute needed to make $2.50\times {10}^{2}mL$ of a $0.100M$ of solution has to be identified.

Concept introduction:

 $Molarityofthesolution=\frac{Molesofsolute}{Volumeinliter}$

 The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

 $\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

 The mass of compound is given by the multiplication of mole of the compound and molar mass of the compound in gram.

 $\text{Mass}\text{=}\text{Mole×}\text{Molar}\text{mass}$

Explanation of Solution

Given,

    The concentration of solution is $\text{0}\text{.100M}$

    The volume of solution is $2.50\times {10}^{2}mL$

The number of mole can be calculated by using

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

$\begin{array}{l}\text{Number }\text{of Moles}\text{=}\text{0}\text{.100M}\text{×}\text{2}\text{.50}\times {\text{10}}^2\times {10}^{-3}\text{L}\\ \text{=}\text{0}\text{.025mol}\end{array}$

The mass of $KMn{O}_4$ can be calculated by using,

From the standard data molar mass of ${\text{KMnO}}_{\text{4}}\text{=}\text{158}\text{.04g/mol}$

$\begin{array}{l}\text{=}\text{0}\text{.025mol×}\text{158}\text{.04g/mol}\\ \text{=}\text{3}\text{.95g}\end{array}$

The $\text{3}\text{.95g}$ of $KMn{O}_4$ is used to  $2.50\times {10}^{2}mL$ of a $0.100M$ of solution.