Chapter 13, Problem 58QP

Interpretation Introduction

Interpretation:

The amount of sucrose (in grams) required in the given amount of water to prepare a solution with a vapor pressure less than that of pure water at $20°\text{C}$, is to be calculated.

Concept introduction:

Raoult’s law describes the relationship between the partial vapor pressure of a solution and the solvent. This law states that the partial pressure of the solvent in the solution $\left(P_1\right)$ is equivalent to the product of mole fraction of the solvent $\left({\chi }_1\right)$, and the partial pressure of pure solvent $\left(P_1^{°}\right)$. It is expressed as follows:

$P_1={\chi }_1{P}_1^{°}$

Mole fraction of a component is defined as the ratio of number of moles of that component to the total number of moles of all the components present in the mixture. The relationship between mole fraction $\left(\chi \right)$ and moles $\left(n\right)$ of a component is as follows:

${\chi }_1=\frac{n_1}{n_1+n_2+\mathrm{....}}$

Answer to Problem 58QP

Solution: $1.35\times {10}^3\text{ g}$

Explanation of Solution

Given information:

The amount of water is, $\left(m_{{\text{H}}_{\text{2}}\text{O}}\right)=552\text{ g}$.

The vapor pressure of pure water is, $\left(P_{{\text{H}}_{\text{2}}\text{O}}^0\right)$ at $20°\text{C}$ $=17.5\text{ mm Hg}$.

The vapor pressure of the solutionis, $\left(P_{\text{sol}}^{}\right)$ at $20°\text{C}$ $=P_{{\text{H}}_{\text{2}}\text{O}}^0-2.0\text{ mm Hg}$.

Molar mass of water $\left({\text{MM}}_{{\text{H}}_{\text{2}}\text{O}}\right)=18\text{ g/mol}$.

Molar mass of sucrose $\left({\text{MM}}_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}\right)=342\text{ g/mol}$.

The moles of water can be evaluated as follows:

$\begin{array}{c}{n}_{{\text{H}}_{\text{2}}\text{O}}=\frac{552\text{ g}}{18\text{ g/mol}}\\ =30.67\text{ mol}\end{array}$

Therefore, the moles of water are $30.67\text{ mol}$.

Now, the mole fraction of water in the solution is calculated by the expression as follows:

$P_{\text{sol}}={\chi }_{{\text{H}}_{\text{2}}\text{O}}{P}_{{\text{H}}_{\text{2}}\text{O}}^{°}$

Substitute the values of $P_{\text{sol}}$ and $P_{{\text{H}}_{\text{2}}\text{O}}^{°}$ in the above equation to calculate the mole fraction of water,

$\begin{array}{c}{P}_{{\text{H}}_{\text{2}}\text{O}}^0-2.0\text{ mm Hg}={\chi }_{{\text{H}}_{\text{2}}\text{O}}{P}_{{\text{H}}_{\text{2}}\text{O}}^{°}\\ 17.5\text{ mm Hg}-2.0\text{ mm Hg}={\chi }_{{\text{H}}_{\text{2}}\text{O}}\left(17.5\text{ mm Hg}\right)\\ {\chi }_{{\text{H}}_{\text{2}}\text{O}}=\frac{15.5\cancel{\text{mm Hg}}}{17.5\cancel{\text{mm Hg}}}\\ {\chi }_{{\text{H}}_{\text{2}}\text{O}}=0.886\end{array}$

The moles fraction of water can be rewritten by the expression as follows:

${\chi }_{{\text{H}}_{\text{2}}\text{O}}=\frac{n_{{\text{H}}_{\text{2}}\text{O}}}{n_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}+n_{{\text{H}}_{\text{2}}\text{O}}}$

Substitute the values of mole fraction and the number of moles of water in the above equation,

$\begin{array}{c}0.886=\frac{30.67\text{ mol}}{n_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}+30.67\text{ mol}}\\ n_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}=\frac{30.67\text{ mol}}{0.886}-30.67\text{ mol}\\ n_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}\text{=3}\text{.95 mol}\end{array}$

Thus, the number of moles of sucrose is $\text{3}\text{.95 mol}$.

The amount of sucrose required in the solution is calculated as follows:

$\begin{array}{c}{n}_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}=\frac{m_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}}{{\text{MM}}_{{}_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}}}\\ \text{3}\text{.95 mol}=\frac{m_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}}{342\text{ g/mol}}\\ m_{{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}=1.35\times {10}^3\text{ g}\end{array}$

Conclusion

Hence, the amount of sucrose (in grams) required in $552\text{ g}$ of water to prepare a given solution at $20°\text{C}$ is, $1.35\times {10}^3\text{ g}$.