Chapter 9, Problem 9.68QP

(a)

Interpretation Introduction

Interpretation:

The masses of given solutes require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solutions are should be determined.

Concept introduction:

Mole:

Mole of the solute in the solution is given by the multiplication of concentration and volume in liter.

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

Mass:

The mass of solute present in the solution is given by multiplication of mole and molar mass of the solute.

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

Explanation of Solution

Record the given data,

The molarities of solutions are $\text{0}\text{.100}\text{M}$

The volumes of solution are $\text{250}\text{L}$

The required molarities and volumes of solutions are recorded as shown above.

The mole of given $\text{250}\text{mL}$of $\text{0}\text{.100}\text{M}$ solutions

$\begin{array}{l}\text{Mole}\text{=}\text{0}\text{.100}\text{M×}\text{0}\text{.250}\text{L}\\ \text{=}\text{0}\text{.0250}\text{mole}\end{array}$

The multiplication of given molarity and volume to give mole of the each solutions

The mass of $\text{CsBr}$require to prepare $\text{250}\text{mL}$of $\text{0}\text{.100}\text{M}$ $\text{CsBr}$solutions.

From the standard data the molar mass of $\text{CsBr}$ is $\text{212}\text{.81 g}$

$\begin{array}{l}\text{=}\text{0}\text{.0250}\text{mol}\text{CsBr}\text{×}\frac{\text{212}\text{.81 g CsBr}}{\text{1mol}\text{CsBr}}\\ \text{=}\text{5}\text{.32}\text{g}\text{CsBr}\end{array}$

• The calculated mole and molar mass of the solute is plugged in the above equation to give the require mass of solute to prepare the $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solution.
• The require mass of $\text{CsBr}$ to prepare the $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ of $\text{CsBr}$ solution is $\text{5}\text{.32}\text{g}$.

(b)

Interpretation Introduction

Interpretation:

The masses of given solutes require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solutions are should be determined.

Concept introduction:

Mole:

Mole of the solute in the solution is given by the multiplication of concentration and volume in liter.

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

Mass:

The mass of solute present in the solution is given by multiplication of mole and molar mass of the solute.

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

Explanation of Solution

Record the given data,

The molarities of solutions are $\text{0}\text{.100}\text{M}$

The volumes of solution are $\text{250}\text{L}$

The required molarities and volumes of solutions are recorded as shown above.

The mole of given $\text{250}\text{mL}$of $\text{0}\text{.100}\text{M}$ solutions

$\begin{array}{l}\text{Mole}\text{=}\text{0}\text{.100}\text{M×}\text{0}\text{.250}\text{L}\\ \text{=}\text{0}\text{.0250}\text{mole}\end{array}$

The multiplication of given molarity and volume to give mole of the each solutions

The mass of ${\text{CaSO}}_{\text{4}}$require to prepare $\text{250}\text{mL}$of $\text{0}\text{.100}\text{M}$ ${\text{CaSO}}_{\text{4}}$solutions.

From the standard data the molar mass of ${\text{CaSO}}_{\text{4}}$ is $\text{136}\text{.14 g}$

$\begin{array}{l}\text{=}\text{0}\text{.0250}\text{mol}{\text{CaSO}}_{\text{4}}\text{×}\frac{\text{136}{\text{.14 g CaSO}}_{\text{4}}}{\text{1mol}{\text{CaSO}}_{\text{4}}}\\ \text{=}3.40\text{g}{\text{CaSO}}_{\text{4}}\end{array}$

The calculated mole and molar mass of the solute is plugged in the above equation to give the require mass of solute to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solution.

The require mass of ${\text{CaSO}}_{\text{4}}$ to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ ${\text{CaSO}}_{\text{4}}$ solution is $\text{5}\text{.32}\text{g}$.

(c)

Interpretation Introduction

Interpretation:

The masses of given solutes require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solutions are should be determined.

Concept introduction:

Mole

Mole of the solute in the solution is given by the multiplication of concentration and volume in liter.

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

Mass:

The mass of solute present in the solution is given by multiplication of mole and molar mass of the solute.

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

Explanation of Solution

Record the given data,

The molarities of solutions are $\text{0}\text{.100}\text{M}$

The volumes of solution are $\text{250}\text{L}$

The required molarities and volumes of solutions are recorded as shown above.

The mole of given $\text{250}\text{mL}$of $\text{0}\text{.100}\text{M}$ solutions

$\begin{array}{l}\text{Mole}\text{=}\text{0}\text{.100}\text{M×}\text{0}\text{.250}\text{L}\\ \text{=}\text{0}\text{.0250}\text{mole}\end{array}$

The multiplication of given molarity and volume to give mole of the each solutions

The mass of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ require to prepare $\text{250}\text{mL}$of $\text{0}\text{.100}\text{M}$ ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$solutions.

From the standard data the molar mass of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $\text{163}\text{.94 g}$

$\begin{array}{l}\text{=}\text{0}\text{.0250}\text{mol}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\text{×}\frac{\text{163}{\text{.94 g Na}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{1mol}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}}\\ \text{=}4.10\text{g}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\end{array}$

• The calculated mole and molar mass of the solute is plugged in the above equation to give the require mass of solute to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solution.
• The require mass of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ to prepare the $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $\text{4}\text{.10}\text{g}$.

(d)

Interpretation Introduction

Interpretation:

The masses of given solutes require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solutions are should be determined.

Concept introduction:

Mole:

Mole of the solute in the solution is given by the multiplication of concentration and volume in liter.

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

Mass:

The mass of solute present in the solution is given by multiplication of mole and molar mass of the solute.

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

Explanation of Solution

Record the given data,

The molarities of solutions are $\text{0}\text{.100}\text{M}$

The volumes of solution are $\text{250}\text{L}$

The required molarities and volumes of solutions are recorded as shown above.

The mole of given $\text{250}\text{mL}$of $\text{0}\text{.100}\text{M}$ solutions

$\begin{array}{l}\text{Mole}\text{=}\text{0}\text{.100}\text{M×}\text{0}\text{.250}\text{L}\\ \text{=}\text{0}\text{.0250}\text{mole}\end{array}$

The multiplication of given molarity and volume to give mole of the each solutions

The mass of ${\text{Li}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ require to prepare $\text{250}\text{mL}$of $\text{0}\text{.100}\text{M}$ ${\text{Li}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$given solutions.

From the standard data the molar mass of ${\text{Li}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is $\text{229}\text{.87 g}$

$\begin{array}{l}\text{=}\text{0}\text{.0250}\text{mol}{\text{Li}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{×}\frac{\text{229}{\text{.87 g Li}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}{\text{1mol}{\text{Li}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}\\ \text{=}\text{5}\text{.75}\text{g}{\text{Li}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\end{array}$

• The calculated mole and molar mass of the solute is plugged in the above equation to give the require mass of solute to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solution.
• The require mass of ${\text{Li}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ ${\text{Li}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ solution is $\text{5}\text{.75}\text{g}$.