Chapter 13, Problem 53E

Interpretation Introduction

(a)

The molar concentration of $\text{1}\text{.50 g NaCl}$ in $\text{100}\text{.0 mL}$ 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 53E

The molar concentration of $\text{1}\text{.50 g NaCl}$ in $\text{100}\text{.0 mL}$ of solution is $\text{0}\text{.257 }M$.

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 $\text{NaCl}$.

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

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

The given value of $w$, ${\text{M}}_{\text{o}}$ and volume of solution is $\text{1}\text{.50 g}$, $\text{58}\text{.44 }\text{g}/\text{mol}$ and $\text{100}\text{.0}\text{mL}$ respectively.

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

$\begin{array}{rll}M& =& \frac{w\left(g\right)\times 1000}{{\text{M}}_{\text{o}}\left(g/mol\right)\times \text{volume of solution }\left(\text{in mL}\right)}\\ & =& \frac{\text{1}\text{.50 g}\times 1000}{\text{58}\text{.44 }\text{g}/\text{mol}\times \text{100}\text{.0 mL}}\\ & =& \text{0}\text{.257 }M\end{array}$

Therefore, the molar concentration of $\text{1}\text{.50 g NaCl}$ in $\text{100}\text{.0 mL}$ of solution is $\text{0}\text{.257 }M$.

Conclusion

The molar concentration of $\text{1}\text{.50 g NaCl}$ in $\text{100}\text{.0 mL}$ of solution is calculated as $\text{0}\text{.257 }M$.

Interpretation Introduction

(b)

Interpretation:

The molar concentration of $\text{1}{\text{.50 g K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ in $\text{100}\text{.0 mL}$ 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 53E

The molar concentration of $\text{1}{\text{.50 g K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ in $\text{100}\text{.0 mL}$ of solution is $\text{0}\text{.0510 }M$.

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 ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$.

• ${\text{M}}_{\text{o}}$is the molar mass of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$.

• $M$ is the molarity of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$.

The given value of $w$, ${\text{M}}_{\text{o}}$ and volume of solution is $\text{1}\text{.50 g}$, $\text{294}\text{.18 }\text{g}/\text{mol}$ and $\text{100}\text{.0}\text{mL}$ respectively.

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

$\begin{array}{rll}M& =& \frac{w\left(g\right)\times 1000}{{\text{M}}_{\text{o}}\left(g/mol\right)\times \text{volume of solution }\left(\text{in mL}\right)}\\ & =& \frac{\text{1}\text{.50 g}\times 1000}{\text{294}\text{.18 }\text{g}/\text{mol}\times \text{100}\text{.0 mL}}\\ & =& \text{0}\text{.0510 }M\end{array}$

Therefore, the molar concentration of $\text{1}{\text{.50 g K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ in $\text{100}\text{.0 mL}$ of solution is $\text{0}\text{.0510 }M$.

Conclusion

The molar concentration of $\text{1}{\text{.50 g K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ in $\text{100}\text{.0 mL}$ of solution is calculated as $\text{0}\text{.0510 }M$.

Interpretation Introduction

(c)

Interpretation:

The molar concentration of $\text{5}{\text{.55 g CaCl}}_2$ in $\text{125}\text{.0 mL}$ 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 53E

The molar concentration of $\text{5}{\text{.55 g CaCl}}_2$ in $\text{125}\text{.0 mL}$ of solution is $\text{0}\text{.400 }M$.

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 ${\text{CaCl}}_2$.

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

• $M$ is the molarity of ${\text{CaCl}}_2$.

The given value of $w$, ${\text{M}}_{\text{o}}$ and volume of solution is $\text{5}\text{.55 g}$, $\text{110}\text{.98 }\text{g}/\text{mol}$ and $\text{125}\text{.0}\text{mL}$ respectively.

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

$\begin{array}{rll}M& =& \frac{w\left(g\right)\times 1000}{{\text{M}}_{\text{o}}\left(g/mol\right)\times \text{volume of solution }\left(\text{in mL}\right)}\\ & =& \frac{\text{5}\text{.55 g}\times 1000}{\text{110}\text{.98 }\text{g}/\text{mol}\times \text{125}\text{.0 mL}}\\ & =& \text{0}\text{.400 }M\end{array}$

Therefore, the molar concentration of $\text{5}{\text{.55 g CaCl}}_2$ in $\text{125}\text{.0 mL}$ of solution is $\text{0}\text{.400 }M$.

Conclusion

The molar concentration of $\text{5}{\text{.55 g CaCl}}_2$ in $\text{125}\text{.0 mL}$ of solution is calculated as $\text{0}\text{.400 }M$.

Interpretation Introduction

(d)

Interpretation:

The molar concentration of $\text{5}{\text{.55 g Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ in $\text{125}\text{.0 mL}$ 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 53E

The molar concentration of $\text{5}{\text{.55 g Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ in $\text{125}\text{.0 mL}$ of solution is $\text{0}\text{.313 }M$.

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 ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$.

• ${\text{M}}_{\text{o}}$is the molar mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$.

• $M$ is the molarity of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$.

The given value of $w$, ${\text{M}}_{\text{o}}$ and volume of solution is $\text{5}\text{.55 g}$, $\text{142}\text{.04 }\text{g}/\text{mol}$ and $\text{125}\text{.0}\text{mL}$ respectively.

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

$\begin{array}{rll}M& =& \frac{w\left(g\right)\times 1000}{{\text{M}}_{\text{o}}\left(g/mol\right)\times \text{volume of solution }\left(\text{in mL}\right)}\\ & =& \frac{\text{5}\text{.55 g}\times 1000}{\text{142}\text{.04 }\text{g}/\text{mol}\times \text{125}\text{.0 mL}}\\ & =& \text{0}\text{.313 }M\end{array}$

Therefore, the molar concentration of $\text{5}{\text{.55 g Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ in $\text{125}\text{.0 mL}$ of solution is $\text{0}\text{.313 }M$.

Conclusion

The molar concentration of $\text{5}{\text{.55 g Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ in $\text{125}\text{.0 mL}$ of solution is calculated as $\text{0}\text{.313 }M$.