Chapter 13, Problem 54E

Interpretation Introduction

(a)

The molar concentration of $\text{1}\text{.00 g KCl}$ in $\text{75}\text{.0 mL}$ of solution is to be stated.

Concept introduction:

The molarity is defined as the number of moles present in one liter of solution. General expression is shown below.

$\text{Molarity}=\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}}$

Measuring unit of molarity is $\text{mol}/\text{L}$ or $\text{molar}\left(\text{M}\right)$.

Answer to Problem 54E

The molar concentration of $\text{1}\text{.00 g KCl}$ in $\text{75}\text{.0 mL}$ of solution is $\text{0}\text{.178 }M$.

Explanation of Solution

The molar concentration is calculated by the formula shown below.

$M=\frac{w\left(g\right)\times 1000}{{\text{M}}_{\text{o}}\left(g/mol\right)\times \text{volume of solution }\left(\text{in mL}\right)}$ …(1)

Where,

• $w$ is the mass of $\text{KCl}$.

• ${\text{M}}_{\text{o}}$is the molar mass of $\text{KCl}$.

• $M$ is the molarity of $\text{KCl}$.

The given value of $w$, ${\text{M}}_{\text{o}}$ and volume of solution is $\text{1}\text{.00 g}$, $\text{74}\text{.55 }\text{g}/\text{mol}$ and $\text{75}\text{.0}\text{mL}$ respectively.

Substitute all the values of $w$, ${\text{M}}_{\text{o}}$ and volume of solution in equation 1 as shown below.

$\begin{array}{rll}M& =& \frac{w\left(g\right)\times 1000}{{\text{M}}_{\text{o}}\left(g/mol\right)\times \text{volume of solution }\left(\text{in mL}\right)}\\ & =& \frac{\text{1}\text{.00 g}\times 1000}{\text{74}\text{.55 }\text{g}/\text{mol}\times \text{75}\text{.0 mL}}\\ & =& \text{0}\text{.178 }M\end{array}$

Therefore, the molar concentration of $\text{1}\text{.00 g KCl}$ in $\text{75}\text{.0 mL}$ of solution is $\text{0}\text{.178 }M$.

Conclusion

The molar concentration of $\text{1}\text{.00 g KCl}$ in $\text{75}\text{.0 mL}$ of solution is calculated as $\text{0}\text{.178 }M$.

Interpretation Introduction

(b)

Interpretation:

The molar concentration of $\text{1}{\text{.00 g Na}}_2{\text{CrO}}_4$ in $\text{75}\text{.0 mL}$ of solution is to be stated.

Concept introduction:

The molarity is defined as the number of moles present in one liter of solution. General expression is shown below.

$\text{Molarity}=\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}}$

Measuring unit of molarity is $\text{mol}/\text{L}$ or $\text{molar}\left(\text{M}\right)$.

Answer to Problem 54E

The molar concentration of $\text{1}{\text{.00 g Na}}_2{\text{CrO}}_4$ in $\text{75}\text{.0 mL}$ of solution is $\text{0}\text{.0820 }M$.

Explanation of Solution

The molar concentration is calculated by the formula shown below.

$M=\frac{w\left(g\right)\times 1000}{{\text{M}}_{\text{o}}\left(g/mol\right)\times \text{volume of solution }\left(\text{in mL}\right)}$ …(1)

Where,

• $w$ is the mass of ${\text{Na}}_2{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$.

• ${\text{M}}_{\text{o}}$is the molar mass of ${\text{Na}}_2{\text{CrO}}_4$.

• $M$ is the molarity of ${\text{Na}}_2{\text{CrO}}_4$.

The given value of $w$, ${\text{M}}_{\text{o}}$ and volume of solution is $\text{1}\text{.00 g}$, $\text{161}\text{.97 }\text{g}/\text{mol}$ and $\text{75}\text{.0}\text{mL}$ respectively.

Substitute all the values of $w$, ${\text{M}}_{\text{o}}$ and volume of solution in equation 1 as shown below.

$\begin{array}{rll}M& =& \frac{w\left(g\right)\times 1000}{{\text{M}}_{\text{o}}\left(g/mol\right)\times \text{volume of solution }\left(\text{in mL}\right)}\\ & =& \frac{\text{1}\text{.00 g}\times 1000}{\text{161}\text{.97 }\text{g}/\text{mol}\times \text{75}\text{.0 mL}}\\ & =& \text{0}\text{.0820 }M\end{array}$

Therefore, the molar concentration of $\text{1}{\text{.00 g Na}}_2{\text{CrO}}_4$ in $\text{75}\text{.0 mL}$ of solution is $\text{0}\text{.0820 }M$.

Conclusion

The molar concentration of $\text{1}{\text{.00 g Na}}_2{\text{CrO}}_4$ in $\text{75}\text{.0 mL}$ of solution is calculated as $\text{0}\text{.0820 }M$.

Interpretation Introduction

(c)

Interpretation:

The molar concentration of $\text{20}{\text{.0 g MgBr}}_{\text{2}}$ in $\text{250}\text{.0 mL}$ of solution is to be stated.

Concept introduction:

The molarity is defined as the number of moles present in one liter of solution. General expression is shown below.

$\text{Molarity}=\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}}$

Measuring unit of molarity is $\text{mol}/\text{L}$ or $\text{molar}\left(\text{M}\right)$.

Answer to Problem 54E

The molar concentration of $\text{20}{\text{.0 g MgBr}}_{\text{2}}$ in $\text{250}\text{.0 mL}$ is $\text{0}\text{.434 }M$.

Explanation of Solution

The molar concentration is calculated by the formula shown below.

$M=\frac{w\left(g\right)\times 1000}{{\text{M}}_{\text{o}}\left(g/mol\right)\times \text{volume of solution }\left(\text{in mL}\right)}$ … (1)

Where,

• $w$ is the mass of ${\text{MgBr}}_{\text{2}}$.

• ${\text{M}}_{\text{o}}$is the molar mass of ${\text{MgBr}}_{\text{2}}$.

• $M$ is the molarity of ${\text{MgBr}}_{\text{2}}$.

The given value of $w$, ${\text{M}}_{\text{o}}$ and volume of solution is $\text{20}\text{.0 g}$, $\text{184}\text{.13 }\text{g}/\text{mol}$ and $\text{250}\text{.0 mL}$ respectively.

Substitute all the values of $w$, ${\text{M}}_{\text{o}}$ and volume of solution in equation 1 as shown below.

$\begin{array}{rll}M& =& \frac{w\left(g\right)\times 1000}{{\text{M}}_{\text{o}}\left(g/mol\right)\times \text{volume of solution }\left(\text{in mL}\right)}\\ & =& \frac{\text{20}\text{.0 g}\times 1000}{\text{184}\text{.13 }\text{g}/\text{mol}\times \text{250}\text{.0 mL}}\\ & =& \text{0}\text{.434 }M\end{array}$

Therefore, the molar concentration of $\text{20}{\text{.0 g MgBr}}_{\text{2}}$ in $\text{250}\text{.0 mL}$ is $\text{0}\text{.434 }M$.

Conclusion

The molar concentration of $\text{20}{\text{.0 g MgBr}}_{\text{2}}$ in $\text{250}\text{.0 mL}$ is calculated as $\text{0}\text{.434 }M$.

Interpretation Introduction

(d)

Interpretation:

The molar concentration of $\text{20}{\text{.0 g Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ in $\text{250}\text{.0 mL}$ of solution is to be stated.

Concept introduction:

The molarity is defined as the number of moles present in one liter of solution. General expression is shown below.

$\text{Molarity}=\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}}$

Measuring unit of molarity is $\text{mol}/\text{L}$ or $\text{molar}\left(\text{M}\right)$.

Answer to Problem 54E

The molar concentration of $\text{20}{\text{.0 g Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ in $\text{250}\text{.0 mL}$ of solution is $\text{1}\text{.08 }M$.

Explanation of Solution

The molar concentration is calculated by the formula shown below.

$M=\frac{w\left(g\right)\times 1000}{{\text{M}}_{\text{o}}\left(g/mol\right)\times \text{volume of solution }\left(\text{in mL}\right)}$ … (1)

Where,

• $w$ is the mass of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$.

• ${\text{M}}_{\text{o}}$is the molar mass of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$.

• $M$ is the molarity of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$.

The given value of $w$, ${\text{M}}_{\text{o}}$ and volume of solution is $\text{20}\text{.0 g}$, $\text{73}\text{.89 }\text{g}/\text{mol}$ and $\text{250}\text{.0}\text{mL}$ respectively.

Substitute all the values of $w$, ${\text{M}}_{\text{o}}$ and volume of solution in equation 1 as shown below.

$\begin{array}{rll}M& =& \frac{w\left(g\right)\times 1000}{{\text{M}}_{\text{o}}\left(g/mol\right)\times \text{volume of solution }\left(\text{in mL}\right)}\\ & =& \frac{\text{20}\text{.0 g}\times 1000}{\text{73}\text{.89 }\text{g}/\text{mol}\times \text{250}\text{.0 mL}}\\ & =& \text{1}\text{.08 }M\end{array}$

Therefore, the molar concentration of $\text{20}{\text{.0 g Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ in $\text{250}\text{.0 mL}$ of solution is $\text{1}\text{.08 }M$.

Conclusion

The concentration of $\text{20}{\text{.0 g Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ in $\text{250}\text{.0 mL}$ of solution is calculated as $\text{1}\text{.08 }M$.