Chapter 13, Problem 13.119P

(a)

Interpretation Introduction

Interpretation:

The total molarity of ions for each solution is to be calculated.

Concept introduction:

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by $M$ and its unit is $\text{mol/L}$.

The formula to calculate the molarity of the solution is as follows:

$\text{Molarity}\left(M\right)=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{volume }\left(\text{L}\right)\text{of}\text{}\text{solution}}$        (1)

Answer to Problem 13.119P

The total molarity of ions of solutions A, B and C are $3.2M$, $2.0M$ and $1.2M$.

Explanation of Solution

The formula to calculate the total molarity of ions of the solution is as follows:

$\text{Molarity of ions}=\frac{\left(\text{Number of spheres}\right)\left(\frac{0.010\text{mol}\text{ions}}{1\text{sphere}}\right)}{\text{Volume of solution}}$        (2)

Solution A has a total of 8 spheres.

Substitute 8 spheres for the number of spheres and $\text{25 mL}$ for the volume of solution in equation (2) to calculate the molarity of ions of solution A.

$\begin{array}{c}\text{Molarity of ions of solution A}=\frac{\left(\text{8}\text{spheres}\right)\left(\frac{0.010\text{mol}\text{ions}}{1\text{sphere}}\right)}{\text{25 mL}}\left(\frac{\text{1 mL}}{{10}^{-3}\text{L}}\right)\\ =3.2M\end{array}$

Solution B has a total of 10 spheres.

Substitute 10 spheres for the number of spheres and $\text{50 mL}$ for the volume of solution in equation (2) to calculate the molarity of ions of solution B.

$\begin{array}{c}\text{Molarity of ions of solution B}=\frac{\left(\text{10}\text{spheres}\right)\left(\frac{0.010\text{mol}\text{ions}}{1\text{sphere}}\right)}{\text{50 mL}}\left(\frac{\text{1 mL}}{{10}^{-3}\text{L}}\right)\\ =2.0M\end{array}$

Solution C has a total of 12 spheres.

Substitute 12 spheres for the number of spheres and $\text{100 mL}$ for the volume of solution in equation (2) to calculate the molarity of ions of solution C.

$\begin{array}{c}\text{Molarity of ions of solution C}=\frac{\left(\text{12}\text{spheres}\right)\left(\frac{0.010\text{mol}\text{ions}}{1\text{sphere}}\right)}{\text{100 mL}}\left(\frac{\text{1 mL}}{{10}^{-3}\text{L}}\right)\\ =1.2M\end{array}$

Conclusion

The total molarity of ions of solutions A, B and C are $3.2M$, $2.0M$ and $1.2M$.

(b)

Interpretation Introduction

Interpretation:

The highest molarity of the solute is to be determined.

Concept introduction:

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by $M$ and its unit is $\text{mol/L}$.

The formula to calculate the molarity of the solution is as follows:

$\text{Molarity}\left(M\right)=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{volume }\left(\text{L}\right)\text{of}\text{}\text{solution}}$        (1)

Answer to Problem 13.119P

Solution A has the highest molarity.

Explanation of Solution

The formula to calculate the molarity of the compound is as follows:

$\text{Molarity of compound}=\frac{\left[\begin{array}{l}\left(\text{Number of spheres}\right)\left(\frac{0.010\text{mol}\text{ions}}{1\text{sphere}}\right)\\ \left(\frac{1\text{mol}\text{compound}}{\text{Moles of dissociated ions}}\right)\end{array}\right]}{\text{Volume of solution}}$        (3)

Solution A has a total of 8 spheres and the moles of dissociated ions are two (one positive and one negative charge).

Substitute 8 spheres for the number of spheres, $2\text{mol}$ for the moles of dissociated ions and $\text{25 mL}$ for the volume of solution in equation (3) to calculate the molarity of compound in solution A.

$\begin{array}{c}\text{Molarity of compound in solution A}=\frac{\left(\text{8 spheres}\right)\left(\frac{0.010\text{mol}\text{ions}}{1\text{sphere}}\right)\left(\frac{1\text{mol}}{\text{ 2 mol}}\right)}{\text{25 mL}}\left(\frac{1\text{mL}}{{10}^{-3}\text{L}}\right)\\ =1.6M\end{array}$

Solution B has a total of 10 spheres and the moles of dissociated ions are two (one positive and one negative charge).

Substitute 10 spheres for the number of spheres, $2\text{mol}$ for the moles of dissociated ions and $\text{50 mL}$ for the volume of solution in equation (3) to calculate the molarity of compound in solution B.

$\begin{array}{c}\text{Molarity of compound in solution B}=\frac{\left(\text{10 spheres}\right)\left(\frac{0.010\text{mol}\text{ions}}{1\text{sphere}}\right)\left(\frac{1\text{mol}}{\text{ 2 mol}}\right)}{\text{50 mL}}\left(\frac{1\text{mL}}{{10}^{-3}\text{L}}\right)\\ =1.0M\end{array}$

Solution C has a total of 12 spheres and the moles of dissociated ions are three (two positive and one negative charge).

Substitute 12 spheres for the number of spheres, $3\text{mol}$ for the moles of dissociated ions and $\text{100 mL}$ for the volume of solution in equation (3) to calculate the molarity of compound in solution C.

$\begin{array}{c}\text{Molarity of compound in solution C}=\frac{\left(\text{12 spheres}\right)\left(\frac{0.010\text{mol}\text{ions}}{1\text{sphere}}\right)\left(\frac{1\text{mol}}{\text{ 3 mol}}\right)}{\text{100 mL}}\left(\frac{1\text{mL}}{{10}^{-3}\text{L}}\right)\\ =0.40M\end{array}$

Solution A has the highest molarity.

Conclusion

Solution A has the highest molarity.

(c)

Interpretation Introduction

Interpretation:

The lowest molality of solute is to be calculated.

Concept introduction:

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by $m$ and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

The formula to calculate the molality of the solution is as follows:

$\text{Molality}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{mass}\left(\text{kg}\right)\text{of}\text{}\text{solvent}}$        (4)

The density of the substance is defined as the mass per unit volume. It is represented by $\rho$ and its unit is ${\text{kg/m}}^3$.

The formula to calculate the density of the solution is as follows:

$\text{Density of solution}\left(\rho \right)=\frac{\text{Mass }\left(M\right)}{\text{Volume of solution}\left(V\right)}$        (5)

Answer to Problem 13.119P

Solution C has the lowest molality.

Explanation of Solution

Rearrange equation (5) to calculate the mass of the substance as follows:

$\text{Mass}\left(M\right)=\left(\text{Density of solution}\right)\left(\text{Volume of solution}\right)$        (6)

Consider the equal densities of all the three solutions to be $1\text{g/mL}$.

Substitute $1\text{g/mL}$ for the density of the solution and $\text{25 mL}$ for the volume of solution in equation (6) to calculate the mass of solvent in solution A.

$\begin{array}{c}\text{Mass of solvent in solution A}=\left(\frac{1\text{g}}{1\text{mL}}\right)\left(\text{25 mL}\right)\\ =25\text{g}\end{array}$

Substitute $1\text{g/mL}$ for the density of the solution and $\text{50 mL}$ for the volume of solution in equation (6) to calculate the mass of solvent in solution B.

$\begin{array}{c}\text{Mass of solvent in solution B}=\left(\frac{1\text{g}}{1\text{mL}}\right)\left(\text{50 mL}\right)\\ =50\text{g}\end{array}$

Substitute $1\text{g/mL}$ for the density of the solution and $\text{100 mL}$ for the volume of solution in equation (6) to calculate the mass of solvent in solution C.

$\begin{array}{c}\text{Mass of solvent in solution C}=\left(\frac{1\text{g}}{1\text{mL}}\right)\left(\text{100 mL}\right)\\ =100\text{g}\end{array}$

The formula to calculate the molality of the compound is as follows:

$\text{Molality of compound}=\frac{\begin{array}{l}\left(\text{Number of spheres}\right)\left(\frac{0.010\text{mol}\text{ions}}{1\text{sphere}}\right)\\ \left(\frac{1\text{mol}\text{compound}}{\text{Moles of dissociated ions}}\right)\end{array}}{\text{Mass of solvent}}$        (7)

Solution A has a total of 8 spheres and the moles of dissociated ions are two (one positive and one negative charge).

Substitute 8 spheres for the number of spheres, $2\text{mol}$ for the moles of dissociated ions and $\text{25 g}$ for the mass of solvent in equation (7) to calculate the molality of the compound in solution A.

$\begin{array}{c}\text{Molarity of compound in solution A}=\frac{\left(\text{8 spheres}\right)\left(\frac{0.010\text{mol}\text{ions}}{1\text{sphere}}\right)\left(\frac{1\text{mol}}{\text{ 2 mol}}\right)}{\text{25 g}}\left(\frac{{10}^3\text{g}}{1\text{kg}}\right)\\ =1.6m\end{array}$

Solution B has a total of 10 spheres and the moles of dissociated ions are two (one positive and one negative charge).

Substitute 10 spheres for the number of spheres, $2\text{mol}$ for the moles of dissociated ions and $\text{50 g}$ for the mass of solvent in equation (7) to calculate the molality of the compound in solution B.

$\begin{array}{c}\text{Molarity of compound in solution B}=\frac{\left(\text{10 spheres}\right)\left(\frac{0.010\text{mol}\text{ions}}{1\text{sphere}}\right)\left(\frac{1\text{mol}}{\text{ 2 mol}}\right)}{\text{50 g}}\left(\frac{{10}^3\text{g}}{1\text{kg}}\right)\\ =1.0m\end{array}$

Solution C has a total of 12 spheres and the moles of dissociated ions are three (two positive and one negative charge).

Substitute 12 spheres for the number of spheres, $3\text{mol}$ for the moles of dissociated ions and $\text{100 g}$ for the mass of solvent in equation (7) to calculate the molality of the compound in solution C.

$\begin{array}{c}\text{Molarity of compound in solution C}=\frac{\left(\text{12 spheres}\right)\left(\frac{0.010\text{mol}\text{ions}}{1\text{sphere}}\right)\left(\frac{1\text{mol}}{\text{ 3 mol}}\right)}{\text{100 g}}\left(\frac{{10}^3\text{g}}{1\text{kg}}\right)\\ =0.40m\end{array}$

Solution C has the lowest molality.

Conclusion

Solution C has the lowest molality.

(d)

Interpretation Introduction

Interpretation:

The highest osmotic pressure is to be calculated.

Concept introduction:

The osmotic pressure is defined as the measure of the tendency of a solution to take in pure solvent via osmosis. It is defined as the minimum pressure that is to be applied to the solution to prevent the inward flow of the pure solvent across the semipermeable membrane. Osmosis occurs when two solutions have different concentrations of solute and are separated by a semipermeable membrane.

The formula to calculate the osmotic pressure of the solution is as follows:

$\prod =iMRT$        (8)

Here,

$\prod$ is the osmotic pressure.

$i$ is van’t Hoff factor.

$M$ is the molarity of the solution.

$R$ is universal gas constant.

$T$ is the absolute temperature.

Answer to Problem 13.119P

Solution A has the highest osmotic pressure.

Explanation of Solution

Consider the temperature to be $298\text{K}$.

Solution A breaks into two ions so its van’t Hoff factor is 2.

Substitute 2 for $i$, $1.6M$ for $M$, $298\text{K}$ for $T$ and $0.0821\text{}\text{L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ in equation (8).

$\begin{array}{c}\prod =\left(2\right)\left(1.6\text{}M\right)\left(0.0821\text{}\text{L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(298\text{K}\right)\\ =78.29056\text{atm}\\ =78\text{atm}\end{array}$

Solution B breaks into two ions so its van’t Hoff factor is 2.

Substitute 2 for $i$, $1.0\text{}M$ for $M$, $298\text{K}$ for $T$ and $0.0821\text{}\text{L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ in equation (8).

$\begin{array}{c}\prod =\left(2\right)\left(1.0\text{}M\right)\left(0.0821\text{}\text{L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(298\text{K}\right)\\ =48.9316\text{atm}\\ =49\text{atm}\end{array}$

Solution C breaks into three ions so its van’t Hoff factor is 3.

Substitute 3 for $i$, $0.40M$ for $M$, $298\text{K}$ for $T$ and $0.0821\text{}\text{L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ in equation (8).

$\begin{array}{c}\prod =\left(2\right)\left(0.40\text{}M\right)\left(0.0821\text{}\text{L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(298\text{K}\right)\\ =29.35896\text{atm}\\ =29\text{atm}\end{array}$

Solution A has the highest osmotic pressure.

Conclusion

Solution A has the highest osmotic pressure.