Chapter 9, Problem 9.68QP

(a)

Interpretation Introduction

Interpretation:

The masses of given solutes require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ for the respective given solutions 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.

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

Answer to Problem 9.68QP

The mass of solutes require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solution is $\text{6}\text{.49}\text{gCsI}$

Explanation of Solution

Given,

Record the given data,

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

The volumes for solution are $\text{250}\text{mL}$

Determine the mole for given $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solutions

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

Determine the mass of $\text{CsI}$ required to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$$\text{CsI}$ solution.

From the standard data the molar mass of $\text{CsI}$ is $\text{259}\text{.81 g/mol}$

$\begin{array}{l}\text{=}\text{0}\text{.0250}\text{mol}\text{CsI}\text{×}\frac{\text{259}\text{.81 g CsI}}{\text{1mol}\text{CsI}}\\ \text{=}6.49\text{g}\text{CsI}\end{array}$

• The calculated mole and molar mass of the solute is substituted in the above equation to give the required mass of solute to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solution.
• Therefore, the require mass of $\text{CsI}$ to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$$\text{CsI}$ solution is $\text{6}\text{.49}\text{g}$.

(b)

Interpretation Introduction

Interpretation:

The masses of given solutes require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ for the respective given solutions 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.

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

Answer to Problem 9.68QP

The mass of solute require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solution is $\text{2}{\text{.452 g H}}_{\text{2}}{\text{SO}}_{\text{4}}$.

Explanation of Solution

Given,

Record the given data,

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

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

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

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

Determine the mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solutions.

From the standard data the molar mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{98}\text{.079 g/mol}$

$\begin{array}{l}\text{=}\text{0}\text{.0250}\text{mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{×}\frac{\text{98}{\text{.079 g H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{1mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\\ \text{=}2.452{\text{gH}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

• The calculated mole and molar mass of the solute is substituted 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 ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ to prepare the $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solution is $\text{2}\text{.452}\text{g}$.

(c)

Interpretation Introduction

Interpretation:

The masses of given solutes require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ for the respective given solutions 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.

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

Answer to Problem 9.68QP

The mass of solutes require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solution is $\text{2}{\text{.65 g Na}}_2{\text{CO}}_3$.

Explanation of Solution

Given,

Record the given data,

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

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

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

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

Determine the mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution.

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

$\begin{array}{l}\text{=}\text{0}\text{.0250}\text{mol}{\text{Na}}_2{\text{CO}}_3\text{×}\frac{\text{105}{\text{.99 g Na}}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{1mol}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}}\\ \text{=}2.65\text{g}{\text{Na}}_2{\text{CO}}_3\end{array}$

• The calculated mole and molar mass of the solute is substituted in the above equation to give the required mass of solute to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solution.
• The require mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution is $\text{2}\text{.65}\text{g}$.

(d)

Interpretation Introduction

Interpretation:

The masses of given solutes require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ for the respective given solutions 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.

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

Answer to Problem 9.68QP

The mass of solutes require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solution is $\text{7}{\text{.355 g K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$

Explanation of Solution

Given,

Record the given data,

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

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

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

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

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

From the standard data the molar mass of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is $\text{294}\text{.185 g}$

$\begin{array}{l}\text{=}\text{0}\text{.0250}\text{mol}{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{×}\frac{\text{294}{\text{.185 gK}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}{\text{1mol}{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}\\ \text{=}\text{7}\text{.355}{\text{g K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\end{array}$

• The calculated mole and molar mass of the solute is substituted 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 ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ solution is $\text{7}\text{.355g}$.

(e)

Interpretation Introduction

Interpretation:

The masses of given solutes require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ for the respective given solutions 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.

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

Answer to Problem 9.68QP

The mass of solute require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ solution is $\text{3}{\text{.951 g KMnO}}_4$

Explanation of Solution

Given,

Record the given data,

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

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

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

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

Determine the mass of ${\text{KMnO}}_4$ require to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$ ${\text{KMnO}}_4$ given solutions.

From the standard data the molar mass of ${\text{KMnO}}_4$ is $\text{158}\text{.034 g}$

$\begin{array}{l}\text{=}\text{0}\text{.0250}\text{mol}{\text{KMnO}}_4\text{×}\frac{\text{158}{\text{.034 gKMnO}}_4}{\text{1mol}{\text{KMnO}}_4}\\ \text{=}\text{3}\text{.951}{\text{g KMnO}}_4\end{array}$

• The calculated mole and molar mass of the solute is substituted 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 ${\text{KMnO}}_4$ to prepare $\text{250}\text{mL}$ of $\text{0}\text{.100}\text{M}$${\text{KMnO}}_4$ solution is $\text{3}\text{.951g}$.