Chapter 11, Problem 11.100AP

Interpretation Introduction

Interpretation: The value of van’t Hoff factor for each given salt is to be calculated.

Concept introduction: Van’t Hoff factor is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

Colligative properties are basically the properties of the solutions after the process of dissolution. It is already defined above that the vapor pressure of solvent changes after the dissolution.

For ionic compound the value of Van’t Hoff factor is equal to the number of ions.

For covalent compound the value of Van’t Hoff factor is equal one.

To determine: The value of van’t Hoff factor for each given salt.

Answer to Problem 11.100AP

Solution

• For the compound $\text{LiCl}$ the value of van’t Hoff factor is $\underset{\_}{1.87}$.
• For the compound $\text{HCl}$ the value of van’t Hoff factor is $\underset{\_}{1.91}$.
• For the compound $\text{NaCl}$ the value of van’t Hoff factor is $\underset{\_}{1.89}$.

Explanation of Solution

Explanation

Given

The given concentration of compound $\text{LiCl}$ is $5.0\text{g/kg}$.

The given concentration of compound $\text{HCl}$ is $5.0\text{g/kg}$.

The given concentration of compound $\text{NaCl}$ is $5.0\text{g/kg}$.

The given experimentally measured value of $\Delta T_{\text{f}}$ for $\text{LiCl}$ is $0.410°\text{C}$.

The given experimentally measured value of $\Delta T_{\text{f}}$ for $\text{HCl}$ is $0.486°\text{C}$.

The given experimentally measured value of $\Delta T_{\text{f}}$ for $\text{NaCl}$ is $0.299°\text{C}$.

The given value of $K_{\text{f}}$ for water is $1.86°\text{C/m}$.

The depression in freezing point $\left(\Delta T_{\text{f}}\right)$ is the difference between the freezing point of pure solvent and freezing point of the solvent after the addition of the solute.

The depression in freezing point $\left(\Delta T_{\text{f}}\right)$ of the aqueous solution is calculated by the formula,

$\Delta T_{\text{f}}=i\times K_{\text{f}}\times m$

Where,

• $K_{\text{f}}$ is the freezing point depression constant.
• $m$ is the molality.
• $i$ is the van’t Hoff factor.

To calculate the value of van’t Hoff factor the above formula is rearranged as,

$i=\frac{\Delta T}{K_{\text{f}}\times m}$ (1)

Calculation for the compound $\text{LiCl}$,

The molar mass of $\text{LiCl}$ is $42.5\text{g/mol}$.

The mass of solvent is $1\text{kg}$.

The moles of $\text{LiCl}$ is calculated by the formula,

$n=\frac{\text{Mass}\text{of}\text{LiCl}}{\text{Molar}\text{mass}}$

Substitute the values in the above formula.

$\begin{array}{l}n=\frac{\text{Mass}\text{of}\text{LiCl}}{\text{Molar}\text{mass}}\\ n=\frac{5.0\text{g}}{42.5\text{g/mol}}\\ n=0.118\text{mol}\end{array}$

Molality of $\text{LiCl}$ is calculated by the formula,

$\text{Molality}=\frac{\text{Moles}\text{of}\text{LiCl}}{\text{Volume}\text{of}\text{solvent}\left(\text{kg}\right)}$

Substitute the values in the above formula.

$\begin{array}{l}\text{Molality}=\frac{\text{Moles}\text{of}\text{LiCl}}{\text{Mass}\text{of}\text{solvent}\left(\text{kg}\right)}\\ \text{Molality}=\frac{0.118\text{mol}}{1\text{kg}}\\ \text{Molality}=0.118\text{mol/kg}\\ \text{Molality}=0.118\text{m}\end{array}$

Now, substitute the values in equation (1) for the compound $\text{LiCl}$.

$\begin{array}{l}i=\frac{\Delta T}{K_{\text{f}}\times m}\\ i=\frac{0.410°\text{C}}{1.86°\text{C/m}\times 0.118\text{m}}\\ i=\underset{\_}{1.87}\end{array}$

Calculation for the compound $\text{HCl}$,

The molar mass of $\text{HCl}$ is $36.5\text{g/mol}$.

The mass of solvent is $1\text{kg}$.

The moles of $\text{HCl}$ is calculated by the formula,

$n=\frac{\text{Mass}\text{of}\text{HCl}}{\text{Molar}\text{mass}}$

Substitute the values in the above formula.

$\begin{array}{l}n=\frac{\text{Mass}\text{of}\text{HCl}}{\text{Molar}\text{mass}}\\ n=\frac{5.0\text{g}}{36.5\text{g/mol}}\\ n=0.137\text{mol}\end{array}$

Molality of $\text{HCl}$ is calculated by the formula,

$\text{Molality}=\frac{\text{Moles}\text{of}\text{HCl}}{\text{Volume}\text{of}\text{solvent}\left(\text{kg}\right)}$

Substitute the values in the above formula.

$\begin{array}{l}\text{Molality}=\frac{\text{Moles}\text{of}\text{HCl}}{\text{Mass}\text{of}\text{solvent}\left(\text{kg}\right)}\\ \text{Molality}=\frac{0.137\text{mol}}{1\text{kg}}\\ \text{Molality}=0.137\text{mol/kg}\\ \text{Molality}=0.137\text{m}\end{array}$

Now, substitute the values in equation (1) for the compound $\text{HCl}$.

$\begin{array}{l}i=\frac{\Delta T}{K_{\text{f}}\times m}\\ i=\frac{0.486°\text{C}}{1.86°\text{C/m}\times 0.137\text{m}}\\ i=\underset{\_}{1.91}\end{array}$

Calculation for the compound $\text{NaCl}$,

The molar mass of $\text{NaCl}$ is $58.5\text{g/mol}$.

The mass of solvent is $1\text{kg}$.

The moles of $\text{NaCl}$ is calculated by the formula,

$n=\frac{\text{Mass}\text{of}\text{NaCl}}{\text{Molar}\text{mass}}$

Substitute the values in the above formula.

$\begin{array}{l}n=\frac{\text{Mass}\text{of}\text{NaCl}}{\text{Molar}\text{mass}}\\ n=\frac{5.0\text{g}}{58.5\text{g/mol}}\\ n=0.085\text{mol}\end{array}$

Molality of $\text{NaCl}$ is calculated by the formula,

$\text{Molality}=\frac{\text{Moles}\text{of}\text{NaCl}}{\text{Volume}\text{of}\text{solvent}\left(\text{kg}\right)}$

Substitute the values in the above formula.

$\begin{array}{l}\text{Molality}=\frac{\text{Moles}\text{of}\text{NaCl}}{\text{Mass}\text{of}\text{solvent}\left(\text{kg}\right)}\\ \text{Molality}=\frac{0.085\text{mol}}{1\text{kg}}\\ \text{Molality}=0.085\text{mol/kg}\\ \text{Molality}=0.085\text{m}\end{array}$

Now, substitute the values in equation (1) for the compound $\text{NaCl}$.

$\begin{array}{l}i=\frac{\Delta T}{K_{\text{f}}\times m}\\ i=\frac{0.299°\text{C}}{1.86°\text{C/m}\times 0.085\text{m}}\\ i=\underset{\_}{1.89}\end{array}$

Conclusion

• For the compound $\text{LiCl}$ the value of van’t Hoff factor is $\underset{\_}{1.87}$.
• For the compound $\text{HCl}$ the value of van’t Hoff factor is $\underset{\_}{1.91}$.
• For the compound $\text{NaCl}$ the value of van’t Hoff factor is $\underset{\_}{1.89}$.