Chapter 13, Problem 130AP

Interpretation Introduction

Interpretation:

The freezing point of the given glucose solution is to be calculated.

Concept introduction:

The osmotic pressure (π) of a solution is the product of the molarity, gas constant, and temperature. It is expressed as follows:

$\pi =M\text{R}T$

Here, $\pi$ is the osmotic pressure, $M$ is the molarity, $\text{R}$ is the gas constant, and $T$ is the temperature.

Molality is defined as the ratio of number of moles of the solute to the mass of the solvent (in kilograms). It is expressed as follows:

$m=\frac{n}{M_{solvent}}$

Here, $m$ is the molality, $n$ is the moles of the solute, and $M_{solvent}$ is the mass of the solvent (in kilograms).

The temperature at which change of liquid state to solid state occurs is called freezing point.

The freezing point depression is as follows:

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

Here, $\Delta T_{\text{f}}$ is the change in temperature of the freezing point in $\text{C}$, $m$ is the molality of the solution, and $K_{\text{f}}$ is themolal freezing point depression constant for the solvent.

Density is defined as the ratio of mass to volume. It is expressed as follows:

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

Here, $d$ is the density, $m$ is the mass and $V$ is the volume.

Answer to Problem 130AP

Solution: $-{0.737}^{\circ }\text{C}$

Explanation of Solution

Given information:

The osmotic pressure is $10.50\text{ atm}$.

The density is $1.16\text{ g/mL}$.

The gas constant R is $0.0821\text{ L}\text{.atm}{\text{.mol}}^{\text{-1}}{\text{. K}}^{\text{-1}}$.

The osmotic pressure is expressed as follows:

$\pi =M\text{R}T$

Here, $\pi$ is the osmotic pressure, $M$ is the molarity, $\text{R}$ is the gas constant, and $T$ is the temperature.

Rearrange the above equation for the calculation of molarity as follows:

$\text{M}=\frac{\pi }{\text{R}\text{T}}$

Substitute $10.50\text{ atm}$ for $\pi$, $0.0821\text{ L}\text{.atm}{\text{.mol}}^{\text{-1}}{\text{. K}}^{\text{-1}}$ for $\text{R}$ and $298\text{ K}$ for $T$ in the above equation as follows:

$\begin{array}{c}\text{M}=\frac{10.50atm}{(0.0821L\cdot \text{atm/mol}\cdot K)(298K)}\\ =0.429\text{M}\end{array}$

The density of the solution is expressed as follows:

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

Rearrange the above equation for the calculation of mass as follows:

$m=\left(d\right)\left(V\right)$

Consider the volume of the solution to be $1\text{ L }\left(1000\text{ mL}\right)$.

The mass of $1000\text{ mL}$ of a solution is calculated bysubstituting $1.16\text{ g/mL}$ for d and $1000\text{ mL}$ for $V$ in the above equation as follows:

$\begin{array}{c}m=\left(d\right)\left(V\right)\\ =\left(\frac{1.16\text{g}}{1\text{mL}}\right)\left(1000\text{mL}\right)\\ =1160\text{g}\text{solution}\end{array}$

Number of moles is calculated as follows:

$n=\frac{\text{mass}}{\text{molar mass}}$

Rearrange the above equation for the calculation of mass of glucose as follows:

$\text{mass}=\text{ number of moles }\times \text{molar mass}$

Substitute, $0.429mol$ for number of moles and $180.2g/mol$ for molar mass in the above equation as follows:

$\begin{array}{l}\text{mass}=0.429mol\text{glucose}\times \frac{180.2g\text{glucose}}{1mol\text{glucose}}\\ \text{ = 77}\text{.31g}\end{array}$

The mass of the solvent is the difference between the mass of the solution and the mass of the solute. It is calculated as follows:

${\text{mass of H}}_{\text{2}}\text{O}=\text{mass of solution}-\text{mass of solute}$

Substitute, $1160g$ for the mass of solution and $\text{77}\text{.31g}$ for the mass of solute in the above equation as follows:

$\begin{array}{c}mass{H}_{2}O=1160g-\text{77}\text{.31g}\\ =1083g\\ =1.083kg\end{array}$

The molality of the solution is calculated by the expression given below:

$m=\frac{n}{M_{solvent}}$

Substitute $0.429\text{ mol}$ for $n_{urea}$ and $\text{1}\text{.083 kg}$ for $M_{solvent}$ in the above equation as follows:

$\begin{array}{c}m=\frac{0.429mol}{1.083kg}\\ =0.396\text{m}\end{array}$

The freezing point depression is calculated by using the expression given below:

$\Delta T_f=K_f.m$

Here, $\Delta T_f$ is the depression in the freezing point, $K_f$ is the freezing point constant and $m$ is the molality.

Substituting the required values:

$\begin{array}{c}\Delta T_f=\left({1.86}^{\circ }C\right).\left(0.936\text{ m}\right)\\ ={0.737}^{\circ }C\end{array}$

Conclusion

The freezing point of the glucose solution is $-{0.737}^{\circ }\text{C}$.