Q13-108P - Chemistry: Molecular Nature Of Matter And Change

Question

The U.S. food and Drug administration lists dichloromethane  and carbon tetrachloride among the many cancer-causing chlorinated organic compounds. What are the partial pressures of these substances in the vapor above a solution of 1.60 mol of CH₂Cl₂ and 1.10 mol of CCl₄ at $23.5^{\circ }C$? the vapor pressure of pure CH₂Cl₂ and CCl₄ at $23.5^{\circ }C$are 352 torr and 118 torr, respectively.(assume ideal behavior)

Step-by-Step Solution

Answer

Partial vapor pressure of CH₂Cl₂ is 207.68 torr and partial vapor pressure of CCl₄ is 47.2 torr.

Step 1: Raoult’s law

Raoult’s law states that a solvent’s partial vapor pressure in a solution is equal or identical to the vapor pressure of pure solvent multiplied by its mole fraction in the solution.

 $P_{solution}={\chi }_{solvent}\times P_{solvent}$

P_solution is partial vapor pressure of the solution, X_solvent is mole fraction of the solvent and P_solvent is the partial vapor pressure of pure solvent.

Step 2: calculation of mole fraction

 1.60mol of CH₂Cl₂ and 1.10mol of CCl₄ are present in solution. So, the total moles present in solution is $1.60+1.10=2.70$.

Mole fraction of CH₂Cl₂ can be calculated as

$$\begin{gathered} X_{C{H}_{2}C{l}_2}=\frac{moles of C{H}_{2}C{l}_2}{Total moles of solvent} \\ =\frac{1.60}{2.70} \\ =0.59 \end{gathered}$$

Mole fraction of CCl₄ can be calculated as

$$\begin{gathered} X_{CC{l}_4}=\frac{moles of CC{l}_4}{Total moles of solvent} \\ =\frac{1.10}{2.70} \\ =0.40 \end{gathered}$$

Step 2: Calculation of partial vapor pressure

Partial vapor pressure of pure CH₂Cl₂  and CCl₄ is 352 torr and 118 torr respectively

Hence, partial vapor pressure of CH₂Cl₂

$$\begin{gathered} P_{C{H}_{2}C{l}_2}=X_{C{H}_{2}C{l}_2}\times Partial vapor pressure of pure C{H}_{2}C{l}_2 \\ =0.59\times 352torr \\ =207.68torr \end{gathered}$$

And the partial vapor pressure of CCl₄

$$\begin{gathered} P_{CC{l}_4}=X_{CC{l}_4}\times Partial vapor pressure of pure CC{l}_4 \\ =0.40\times 118 torr \\ =47.2 torr \end{gathered}$$