Chapter 7, Problem 7.69E

Calculate the osmolarity for the following solutions:

a. A $0.25\text{M}$ solution of $\text{KCl}$, a strong electrolyte

b. A solution containing $15.0\text{g}$ of urea $\left({\text{CH}}_{\text{4}}{\text{N}}_{\text{2}}\text{O}\right)$, a nonelectrolyte, per $500.\text{mL}$

c. A solution containing $50.0\text{mL}$ of ethylene glycol $\left({\text{C}}_2{\text{H}}_6{\text{O}}_2\right)$, a nonelectrolyte with a density of $1.11\text{g}/\text{mL}$, per $250.\text{mL}$

Interpretation Introduction

(a)

Interpretation:

The osmolarity for a $0.25\text{M}$ solution of $\text{KCl}$, a strong electrolyte is to be calculated.

Concept introduction:

Solutes which give conducting solutions on dissolution are called electrolytes. Those which dissociate completely in the solution are known as strong electrolytes. Solutes which do not give conducting solutions are called nonelectrolytes. The properties which depend on the number of solute particles are known as colligative properties. Some of these properties are boiling point, freezing point and osmotic pressure.

Answer to Problem 7.69E

The osmolarity for a $0.25\text{M}$ solution of $\text{KCl}$, a strong electrolyte is $0.5\text{M}$.

Explanation of Solution

The formula to calculate osmolarity is given below as,

$\text{Osmolarity}=n\text{M}$

Where,

• $n$ is the number of ions in the solution.

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

Since $\text{KCl}$ is a strong electrolyte, it will dissociate in the solution completely and the value of $n$ for $\text{KCl}$ is $2$.

Substitute the value of $n$ and $\text{M}$ in the above equation as follows.

$\begin{array}{rll}\text{Osmolarity}& =& n\text{M}\\ & =& 2\times 0.25\text{M}\\ & =& 0.5\text{M}\end{array}$

Conclusion

The osmolarity for a $0.25\text{M}$ solution of $\text{KCl}$, a strong electrolyte is $0.5\text{M}$.

Interpretation Introduction

(b)

Interpretation:

The osmolarity for a solution containing $15.0\text{g}$ of urea $\left({\text{CH}}_{\text{4}}{\text{N}}_{\text{2}}\text{O}\right)$, a nonelectrolyte, per $500.\text{mL}$ is to be calculated.

Concept introduction:

Solutes which give conducting solutions on dissolution are called electrolytes. Those which dissociate completely in the solution are known as strong electrolytes. Solutes which do not give conducting solutions are called nonelectrolytes. The properties which depend on the number of solute particles are known as colligative properties. Some of these properties are boiling point, freezing point and osmotic pressure.

Answer to Problem 7.69E

The osmolarity for a solution containing $15.0\text{g}$ of urea $\left({\text{CH}}_{\text{4}}{\text{N}}_{\text{2}}\text{O}\right)$, a nonelectrolyte, per $500.\text{mL}$ is $0.5\text{M}$.

Explanation of Solution

The formula to calculate number of moles of solutes is given below as,

$\text{Moles}\text{of}\text{solute}=\frac{\text{Given}\text{mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$

The molar mass of urea can be calculated as follows.

$\begin{array}{rll}{\text{CH}}_{\text{4}}{\text{N}}_{\text{2}}\text{O}& =& \left(1\times \text{C}\right)+\left(4\times \text{H}\right)+\left(2\times \text{N}\right)+\left(1\times \text{O}\right)\\ & =& \left(1\times 12\right)+\left(4\times 1\right)+\left(2\times \text{14}\right)+\left(1\times 16\right)\\ & =& 60\text{g}/\text{mol}\end{array}$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Moles}\text{of}{\text{CH}}_{\text{4}}{\text{N}}_{\text{2}}\text{O}& =& \frac{\text{Given}\text{mass}\text{of}{\text{CH}}_{\text{4}}{\text{N}}_{\text{2}}\text{O}}{\text{Molar}\text{mass}\text{of}{\text{CH}}_{\text{4}}{\text{N}}_{\text{2}}\text{O}}\\ & =& \frac{15\text{g}}{60\text{g}/\text{mol}}\\ & =& 0.25\text{mol}\end{array}$

The formula to calculate molarity is given below as,

$\text{M}=\frac{\text{moles}\text{of}\text{solute}}{\text{liters}\text{of}\text{solution}}$

Substitute the value of number of moles of solute and volume of solution in the above equation as follows.

$\begin{array}{rll}\text{M}& =& \frac{\text{moles}\text{of}{\text{CH}}_{\text{4}}{\text{N}}_{\text{2}}\text{O}}{\text{liters}\text{of}{\text{CH}}_{\text{4}}{\text{N}}_{\text{2}}\text{O}\text{solution}}\\ & =& \frac{0.25\text{mol}}{500\text{mL}\times \frac{\text{1}\text{L}}{1000\text{mL}}}\\ & =& 0.5\text{M}\end{array}$

The formula to calculate osmolarity is given below as,

$\text{Osmolarity}=n\text{M}$

Where,

• $n$ is the number of ions in the solution.

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

Since urea is a nonelectrolyte, it will not dissociate in the solution and the value of $n$ for urea is $1$.

Substitute the value of $n$ and $\text{M}$ in the above equation as follows.

$\begin{array}{rll}\text{Osmolarity}& =& n\text{M}\\ & =& 1\times 0.5\text{M}\\ & =& 0.5\text{M}\end{array}$

Conclusion

The osmolarity for a solution containing $15.0\text{g}$ of urea $\left({\text{CH}}_{\text{4}}{\text{N}}_{\text{2}}\text{O}\right)$, a nonelectrolyte, per $500.\text{mL}$ is $0.5\text{M}$.

Interpretation Introduction

(c)

Interpretation:

The osmolarity for a solution containing $50.0\text{mL}$ of ethylene glycol $\left({\text{C}}_2{\text{H}}_6{\text{O}}_2\right)$, a nonelectrolyte with a density of $1.11\text{g}/\text{mL}$, per $250.\text{mL}$ is to be calculated.

Concept introduction:

Solutes which give conducting solutions on dissolution are called electrolytes. Those which dissociate completely in the solution are known as strong electrolytes. Solutes which do not give conducting solutions are called nonelectrolytes. The properties which depend on the number of solute particles are known as colligative properties. Some of these properties are boiling point, freezing point and osmotic pressure.

Answer to Problem 7.69E

The osmolarity for a solution containing $50.0\text{mL}$ of ethylene glycol $\left({\text{C}}_2{\text{H}}_6{\text{O}}_2\right)$, a nonelectrolyte with a density of $1.11\text{g}/\text{mL}$, per $250.\text{mL}$ is $3.58\text{M}$.

Explanation of Solution

The formula to calculate density is given below as,

$d=\frac{m}{V}$

Where,

• $m$ is the mass of the substance.

• $V$ is the volume of the substance.

• $d$ is the density of the substance.

Thus, the mass of $50.0\text{mL}$ of ethylene glycol $\left({\text{C}}_2{\text{H}}_6{\text{O}}_2\right)$ can be calculated as follows.

$\begin{array}{rll}d& =& \frac{m}{V}\\ m& =& d\times V\\ & =& 1.11\text{g}/\text{mL}\times 50.0\text{mL}\\ & =& 55.5\text{g}\end{array}$

The formula to calculate number of moles of solutes is given below as,

$\text{Moles}\text{of}\text{solute}=\frac{\text{Given}\text{mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$

The molar mass of ethylene glycol can be calculated as follows.

$\begin{array}{rll}{\text{C}}_2{\text{H}}_6{\text{O}}_2& =& \left(2\times \text{C}\right)+\left(6\times \text{H}\right)+\left(2\times \text{O}\right)\\ & =& \left(2\times 12\right)+\left(6\times 1\right)+\left(2\times 16\right)\\ & =& 62\text{g}/\text{mol}\end{array}$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Moles}\text{of}{\text{C}}_2{\text{H}}_6{\text{O}}_2& =& \frac{\text{Given}\text{mass}\text{of}{\text{C}}_2{\text{H}}_6{\text{O}}_2}{\text{Molar}\text{mass}\text{of}{\text{C}}_2{\text{H}}_6{\text{O}}_2}\\ & =& \frac{55.5\text{g}}{62\text{g}/\text{mol}}\\ & =& 0.895\text{mol}\end{array}$

The formula to calculate molarity is given below as,

$\text{M}=\frac{\text{moles}\text{of}\text{solute}}{\text{liters}\text{of}\text{solution}}$

Substitute the value of number of moles of solute and volume of solution in the above equation as follows.

$\begin{array}{rll}\text{M}& =& \frac{\text{moles}\text{of}{\text{C}}_2{\text{H}}_6{\text{O}}_2}{\text{liters}\text{of}{\text{C}}_2{\text{H}}_6{\text{O}}_2\text{solution}}\\ & =& \frac{0.895\text{mol}}{250\text{mL}\times \frac{\text{1}\text{L}}{1000\text{mL}}}\\ & =& 3.58\text{M}\end{array}$

The formula to calculate osmolarity is given below as,

$\text{Osmolarity}=n\text{M}$

Where,

• $n$ is the number of ions in the solution.

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

Since ethylene glycol is a nonelectrolyte, it will not dissociate in the solution and the value of $n$ for ethylene glycol is $1$.

Substitute the value of $n$ and $\text{M}$ in the above equation as follows.

$\begin{array}{rll}\text{Osmolarity}& =& n\text{M}\\ & =& 1\times 3.58\text{M}\\ & =& 3.58\text{M}\end{array}$

Conclusion

The osmolarity for a solution containing $50.0\text{mL}$ of ethylene glycol $\left({\text{C}}_2{\text{H}}_6{\text{O}}_2\right)$, a nonelectrolyte with a density of $1.11\text{g}/\text{mL}$, per $250.\text{mL}$ is $3.58\text{M}$.