Chapter 13, Problem 13.118P

Interpretation Introduction

Interpretation:

The freezing point, boiling point and osmotic pressure of glucose are to be calculated.

Concept introduction:

The freezing point is the temperature at which both the solid and liquid phases coexist in equilibrium. It is the temperature at which the vapor pressure of the substance in the liquid state becomes equal to the vapor pressure in a solid state.

The formula to calculate the change in freezing point is as follows:

$\Delta T_{\text{f}}=i{k}_{\text{f}}m$        (1)

Here,

$\Delta T_{\text{f}}$ is the change in freezing point.

$i$ is van’t Hoff factor.

$k_{\text{f}}$ is the freezing point depression constant.

$m$ is the molality of the solution.

The boiling point of the substance is the temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure and the liquid changes into a vapor. Liquids can change into vapors at temperatures below the boiling point through evaporation. It is the process that occurs on the liquid surface due to which it changes into vapors. Both boiling and freezing points are colligative properties because these depend on the number of moles of solute particles that are present in the substance.

The formula to calculate the change in boiling point is as follows:

$\Delta T_{\text{b}}=i{k}_{\text{b}}m$        (2)

Here,

$\Delta T_{\text{b}}$ is the change in boiling point.

$i$ is van’t Hoff factor.

$k_{\text{b}}$ is the boiling point elevation constant.

$m$ is the molality of the solution.

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:

$\prod =iMRT$        (3)

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.

The conversion factor to convert $°\text{C}$ to $\text{K}$ is as follows:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\text{K}$

Answer to Problem 13.118P

The freezing point, boiling point and osmotic pressure of glucose are $-1.1\text{}°\text{C}$, $100.32°\text{C}$ and $14\text{ atm}$.

Explanation of Solution

Consider the mass of the solution to be $100\text{g}$.

The formula to calculate the mass of glucose is as follows:

$\text{Mass of glucose}=\left(\text{Mass of solution}\right)\left(\frac{\text{mass }\%\text{of glucose}}{\text{mass }\%\text{of solution}}\right)$        (4)

Substitute $100\text{g}$ for the mass of solution, $\text{10 }\%$ for the mass % of glucose and $\text{100 }\%$ for the mass % of the solution in equation (4).

$\begin{array}{c}\text{Mass of glucose}=\left(\text{100}\text{g}\right)\left(\frac{\text{10 }\%}{\text{100 }\%}\right)\\ =10\text{g}\end{array}$

The formula to calculate the moles of glucose is as follows:

$\text{Moles of glucose}=\frac{\text{Given mass of glucose}}{\text{Molar mass of glucose}}$        (5)

Substitute $10\text{g}$ for the given mass of glucose and $180.16\text{g/mol}$ for the molar mass of glucose in equation (5).

$\begin{array}{c}\text{Moles of glucose}=\left(10\text{g}\right)\left(\frac{1\text{mol}}{180.16\text{g/mol}}\right)\\ =0.055506217\text{mol}\end{array}$

The formula to calculate the mass of glucose solution is as follows:

$\text{Mass of glucose solution}=\text{Mass of glucose}+\text{Mass of water}$        (6)

Rearrange equation (6) to calculate the mass of water as follows:

$\text{Mass of water}=\text{Mass of glucose solution}-\text{Mass of glucose}$        (7)

Substitute $100\text{g}$ for the mass of glucose solution and $10\text{g}$ for the mass of glucose in equation (7).

$\begin{array}{c}\text{Mass of water}=\text{100 g}-\text{10 g}\\ =\text{90 g}\end{array}$

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

$\text{Density of solution}=\frac{\text{Mass of solution}}{\text{Volume of solution}}$        (8)

Rearrange equation (8) to calculate the volume of the solution as follows:

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}$        (9)

Substitute $100\text{g}$ for the mass of solution and $1.039\text{g/mL}$ for the density of the solution in equation (9).

$\begin{array}{c}\text{Volume of solution}=\left(100\text{g}\right)\left(\frac{1\text{mL}}{1.039\text{g}}\right)\left(\frac{1\text{L}}{{10}^3\text{mL}}\right)\\ =0.096246\text{L}\end{array}$

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

$\text{Molarity}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{volume }\left(\text{L}\right)\text{of}\text{}\text{solution}}$        (10)

Substitute $0.055506217\text{mol}$ for the amount of solute and $0.096246\text{L}$ for the volume of solution in equation (10) to calculate the molarity of glucose.

$\begin{array}{c}\text{Molarity of glucose}=\frac{0.055506217\text{mol}}{0.096246\text{L}}\\ =0.57671M\end{array}$

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

$\text{Molality}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{mass}\left(\text{kg}\right)\text{of}\text{}\text{solvent}}$        (11)

Substitute $0.055506217\text{mol}$ for the amount of solute and $90\text{g}$ for the mass of solvent in equation (11) to calculate the molality of glucose.

$\begin{array}{c}\text{Molality of glucose}=\left(\frac{0.055506217\text{mol}}{90\text{g}}\right)\left(\frac{{10}^3\text{g}}{1\text{kg}}\right)\\ =0.6167357m\end{array}$

Glucose is a non-electrolyte so its van’t Hoff factor is 1.

Substitute 1 for $i$, $0.6167357m$ for $m$ and $1.86°\text{C/}m$ for $k_{\text{f}}$ in equation (1).

$\begin{array}{c}\Delta T_{\text{f}}=\left(1\right)\left(1.86°\text{C/}m\right)\left(0.6167357m\right)\\ =1.1471°\text{C}\end{array}$

The formula to calculate the freezing point of glucose is as follows:

$\text{Freezing}\text{point}\text{}\text{of }\text{}\text{glucose}=\text{Freezing}\text{point}\text{}\text{of }\text{}{\text{pure H}}_2\text{O}-\Delta T_{\text{f}}$        (12)

Substitute $0°\text{C}$ for the freezing point of pure water and $1.1471°\text{C}$ for $\Delta T_{\text{f}}$ in equation (12).

$\begin{array}{c}\text{Freezing}\text{point}\text{}\text{of }\text{}\text{glucose}=0°\text{C}-1.1471°\text{C}\\ =-1.1471°\text{C}\\ =-1.1°\text{C}\end{array}$

Substitute 1 for $i$, $0.6167357m$ for $m$ and $0.512°\text{C/}m$ for $k_{\text{b}}$ in equation (2).

$\begin{array}{c}\Delta T_{\text{b}}=\left(1\right)\left(0.512°\text{C/}m\right)\left(0.6167357m\right)\\ =0.3157687°\text{C}\end{array}$

The formula to calculate the boiling point of glucose is as follows:

$\text{Boiling}\text{point}\text{}\text{of }\text{}\text{glucose}=\text{Boiling}\text{point}\text{}\text{of }\text{}{\text{pure H}}_2\text{O}+\Delta T_{\text{b}}$        (13)

Substitute $100°\text{C}$ for the boiling point of pure water and $0.3157687°\text{C}$ for $\Delta T_{\text{b}}$ in equation (12).

$\begin{array}{c}\text{Boiling}\text{point}\text{}\text{of }\text{}\text{glucose}=100°\text{C}+0.3157687°\text{C}\\ =100.3157687°\text{C}\\ =100.32°\text{C}\end{array}$

Substitute $0.57671M$ for $M$, 1 for $i$, $0.0821\text{L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ and $20°\text{C}$ for $T$ in equation (3).

$\begin{array}{c}\prod =\left(1\right)\left(0.57671M\right)\left(0.0821\text{L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(20°\text{C}+\text{273}\text{.15}\right)\text{K}\\ =13.8729\text{atm}\\ \text{=14 atm}\end{array}$

Conclusion

The freezing point, boiling point and osmotic pressure of glucose are $-1.1\text{}°\text{C}$, $100.32°\text{C}$ and $14\text{ atm}$.