Chapter 15, Problem 11ALQ

Which of the following solutions contains the greatest number of particles? Support your answer.

.0 mL of 0 10 M sodium chloride

.0 mL of 0.10 M calcium chloride

.0 mL of 0.10 M iron(III) chloride

.0 mL of 0.10 M potassium bromide

.0 mL of 0.10 M sucrose (table sugar)

Interpretation Introduction

Interpretation:

The solution containing the greatest number of particles is to be predicted.

Concept Introduction:

There are many ways to determine the concentration of the solution. One of the most used methods is molarity. Molarity may be defined as the number of moles of the solute in one liter of the whole solution. Thus, the molarity can be calculated as,

$\text{M}=\frac{\text{moles}\text{of}\text{solute}\left(\text{mole}\right)}{\text{total}\text{volume}\text{of}\text{solution}\left(\text{L}\right)}$.

Answer to Problem 11ALQ

The correct option is (b).

Explanation of Solution

Reason for correct option:

(b) The molarity of the $300\text{mL}$ solution of calcium chloride is $0.10\text{M}$.The number of moles present in the given solution is calculated by the formula,

$\text{M}=\frac{\text{moles}\text{of}\text{solute}\left(\text{mole}\right)}{\text{total}\text{volume}\text{of}\text{solution}\left(\text{L}\right)}$

Substitute the values of molarity and volume in the given formula.

$\begin{array}{c}\text{moles}\text{of}\text{solute}\left(\text{mole}\right)=\text{M}\times \text{total}\text{volume}\text{of}\text{solution}\left(\text{L}\right)\\ =0.1(\text{mol}{\text{L}}^{\text{-1}})\times \frac{300}{1000}\text{L}\\ =0.03\text{mol}\end{array}$

The total number of moles in the given solution is $0.03\text{mol}$.

To calculate the number of particles in a solution the formula used is,

$\text{Moles}=\frac{\text{Number of particles}}{{\text{Avogadro number(N}}_{\text{A}}\text{)}}$

Avogadro number is equal to the number of particles present in one mole, that is, $6.022\times {10}^{23}$ particles.

Substitute the values of moles and Avogadro number in the given formula.

$\begin{array}{c}\text{Number of particles}=\text{Moles}\times {\text{Avogadro number(N}}_{\text{A}}\text{)}\\ =0.03\times 6.022\times {10}^{23}\\ =1.8\times {10}^{22}\end{array}$

One mole of ${\text{CaCl}}_2$ contains $1$ mole of ${\text{Ca}}^{2+}$ and $2$ moles of ${\text{Cl}}^{-}$. Thus, the total number of particles present in the solution is $3\times 1.8\times {10}^{22}$, that is, $5.4\times {10}^{22}$ particles.

Conclusion

Reasons for incorrect options:

(a) The molarity of the $400\text{mL}$ solution of sodium chloride is $0.10\text{M}$.The number of moles present in the given solution is calculated by the formula,

$\text{M}=\frac{\text{moles}\text{of}\text{solute}\left(\text{mole}\right)}{\text{total}\text{volume}\text{of}\text{solution}\left(\text{L}\right)}$

Substitute the values of molarity and volume in the given formula.

$\begin{array}{c}\text{moles}\text{of}\text{solute}\left(\text{mole}\right)=\text{M}\times \text{total}\text{volume}\text{of}\text{solution}\left(\text{L}\right)\\ =0.1(\text{mol}{\text{L}}^{\text{-1}})\times \frac{400}{1000}\text{L}\\ =0.04\text{mol}\end{array}$

The total number of moles in the given solution is $0.04\text{mol}$.

To calculate the number of particles in a solution the formula used is,

$\text{Moles}=\frac{\text{Number of particles}}{{\text{Avogadro number(N}}_{\text{A}}\text{)}}$

Avogadro number is equal to the number of particles present in one mole, that is, $6.022\times {10}^{23}$ particles.

Substitute the values of moles and Avogadro number in the given formula.

$\begin{array}{c}\text{Number of particles}=\text{Moles}\times {\text{Avogadro number(N}}_{\text{A}}\text{)}\\ =0.04\times 6.022\times {10}^{23}\\ =2.4\times {10}^{22}\end{array}$

One mole of $\text{NaCl}$ contains $1$ mole of ${\text{Na}}^{+}$ and $1$ mol of ${\text{Cl}}^{-}$. Thus, the total number of particles present in the solution is $2\times 2.4\times {10}^{22}$, that is, $4.8\times {10}^{22}$ particles.

(c) The molarity of the $200\text{mL}$ solution of iron $\text{(III)}$ chloride is $0.10\text{M}$.The number of moles present in the given solution is calculated by the formula,

$\text{M}=\frac{\text{moles}\text{of}\text{solute}\left(\text{mole}\right)}{\text{total}\text{volume}\text{of}\text{solution}\left(\text{L}\right)}$

Substitute the values of molarity and volume in the given formula.

$\begin{array}{c}\text{moles}\text{of}\text{solute}\left(\text{mole}\right)=\text{M}\times \text{total}\text{volume}\text{of}\text{solution}\left(\text{L}\right)\\ =0.1(\text{mol}{\text{L}}^{\text{-1}})\times \frac{200}{1000}\text{L}\\ =0.02\text{mol}\end{array}$

The total number of moles in the given solution is $0.02\text{mol}$.

To calculate the number of particles in a solution the formula used is,

$\text{Moles}=\frac{\text{Number of particles}}{{\text{Avogadro number(N}}_{\text{A}}\text{)}}$

Avogadro number is equal to the number of particles present in one mole, that is, $6.022\times {10}^{23}$ particles.

Substitute the values of moles and Avogadro number in the given formula.

$\begin{array}{c}\text{Number of particles}=\text{Moles}\times {\text{Avogadro number(N}}_{\text{A}}\text{)}\\ =0.02\times 6.022\times {10}^{23}\\ =1.2\times {10}^{22}\end{array}$

One mole of ${\text{FeCl}}_3$ contains $1$ mole of ${\text{Fe}}^{3+}$ and $3$ mole of ${\text{Cl}}^{-}$. Thus, the total number of particles present in the solution is $4\times 1.2\times {10}^{22}$, that is, $4.8\times {10}^{22}$ particles.

(d) The molarity of the $200\text{mL}$ solution of potassium bromide is $0.10\text{M}$.The number of moles present in the given solution is calculated by the formula,

$\text{M}=\frac{\text{moles}\text{of}\text{solute}\left(\text{mole}\right)}{\text{total}\text{volume}\text{of}\text{solution}\left(\text{L}\right)}$

Substitute the values of molarity and volume in the given formula.

$\begin{array}{c}\text{moles}\text{of}\text{solute}\left(\text{mole}\right)=\text{M}\times \text{total}\text{volume}\text{of}\text{solution}\left(\text{L}\right)\\ =0.1(\text{mol}{\text{L}}^{\text{-1}})\times \frac{200}{1000}\text{L}\\ =0.02\text{mol}\end{array}$

The total number of moles in the given solution is $0.02\text{mol}$.

To calculate the number of particles in a solution the formula used is,

$\text{Moles}=\frac{\text{Number of particles}}{{\text{Avogadro number(N}}_{\text{A}}\text{)}}$

Avogadro number is equal to the number of particles present in one mole, that is, $6.022\times {10}^{23}$ particles.

Substitute the values of moles and Avogadro number in the given formula.

$\begin{array}{c}\text{Number of particles}=\text{Moles}\times {\text{Avogadro number(N}}_{\text{A}}\text{)}\\ =0.02\times 6.022\times {10}^{23}\\ =1.2\times {10}^{22}\end{array}$

One mole of $\text{KBr}$ contains $1$ mole of ${\text{K}}^{+}$ and $1$ mol of ${\text{Br}}^{-}$. Thus, the total number of particles present in the solution is $2\times 1.2\times {10}^{22}$, that is, $2.4\times {10}^{22}$ particles.

(e) The molarity of the $800\text{mL}$ solution of sucrose is $0.10\text{M}$.The number of moles present in the given solution is calculated by the formula,

$\text{M}=\frac{\text{moles}\text{of}\text{solute}\left(\text{mole}\right)}{\text{total}\text{volume}\text{of}\text{solution}\left(\text{L}\right)}$

Substitute the values of molarity and volume in the given formula.

$\begin{array}{c}\text{moles}\text{of}\text{solute}\left(\text{mole}\right)=\text{M}\times \text{total}\text{volume}\text{of}\text{solution}\left(\text{L}\right)\\ =0.1(\text{mol}{\text{L}}^{\text{-1}})\times \frac{800}{1000}\text{L}\\ =0.08\text{mol}\end{array}$

The total number of moles in the given solution is $0.08\text{mol}$.

To calculate the number of particles in a solution the formula used is,

$\text{Moles}=\frac{\text{Number of particles}}{{\text{Avogadro number(N}}_{\text{A}}\text{)}}$

Avogadro number is equal to the number of particles present in one mole, that is, $6.022\times {10}^{23}$ particles.

Substitute the values of moles and Avogadro number in the given formula.

$\begin{array}{c}\text{Number of particles}=\text{Moles}\times {\text{Avogadro number(N}}_{\text{A}}\text{)}\\ =0.08\times 6.022\times {10}^{23}\\ =4.8\times {10}^{22}\end{array}$

The total number of particles present in the solution is $4.8\times {10}^{22}$ particles.