Question

A solution is made by mixing 50.0 g of acetone (CH₃COCH₃) and 50.0 g of methanol (CH₃OH). What is the vapor pressure of this solution at 25⁰C? What is the composition of the vapor expressed as a mole fraction? Assume ideal solution and gas behavior. (At 25⁰C the vapor pressures of pure acetone and pure methanol are 271 torr and 143 torr, respectively.) The actual vapor pressure of this solution is 161 torr. Explain any discrepancies.

Short answer

In a closed chamber at some temperature when vapors are formed, the force applied by them on the liquid surface is known as vapor pressure. For any solution, the vapor pressure of acetone and methanol are 257.34 torr and 253.14 torr.

Step-by-step solution

Determine the Mole fraction

It is given that the mass of acetone and methanol is 50g each and molecular weight will be 58g and 32g respectively. So, the number of moles will be

Number of moles of acetone = $\frac{\text{Mass}}{\text{Molecularweight}}\text{=}\frac{50}{58}\text{=0.8620}$

Number of moles of methanol =$\frac{\text{Mass}}{\text{Molecularweight}}\text{=}\frac{50}{32}\text{=1.5625}$

Mole fraction of acetone = $\frac{\text{Molesofsolute}}{\text{Totalmoles}}=\frac{0.8620}{2.4245}\text{=0.3555}$

Mole fraction of methanol = $\frac{\text{Molesofsolute}}{\text{Totalmoles}}=\frac{1.5624}{2.4245}\text{=0.6444}$

Determine the vapour pressure

According to Raoult’s law, the vapor pressure of methanol is given by:

${\text{P}}_{\text{Solution}}{\text{=X}}_{\text{solvent}}\times {{\text{P}}^{°}}_{\text{Solvent}}$

Thus, the vapor pressure of acetone is:

$$\begin{gathered} {\text{P}}_{\text{Solution}}{\text{-P}}_{\text{PureSolution}}{\text{=X}}_{\text{acetone}}\times {{\text{P}}^{°}}_{\text{acetone}} \\ {\text{P}}_{\text{Solution}}\text{-161=0.3555}\times \text{271} \\ {\text{P}}_{\text{Solution}}\text{=96.34+161=257.34torr} \end{gathered}$$

The vapor pressure of methanol will be:

$\begin{array}{rcl}& & {\text{P}}_{\text{Solution}}{\text{=X}}_{\text{methanol}}\times {{\text{P}}^{°}}_{\text{methanol}}\\ & & {\text{P}}_{\text{Solution}}\text{-161=0.6444}\times \text{143}\\ & & \text{=92.14+161=253.14torr}\end{array}$

Therefore, the total vapor pressure of the solution will be

${\text{P}}_{\text{Solution}}{\text{=P}}_{\text{acetone}}{\text{+P}}_{\text{methanol}}\text{=257.34+253.14=510.48torr.}$

Vapour pressure composition

The composition of vapor pressure of the solution in terms of mole fraction will be:

$$\begin{gathered} {\text{X}}_{\text{acetone}}\text{=}\frac{{\text{P}}_{\text{Solution}}}{{{\text{P}}^{°}}_{\text{acetone}}}\text{=}\frac{257.34}{271}\text{=0.94} \\ {\text{X}}_{\text{methanol}}\text{=}\frac{{\text{P}}_{\text{Solution}}}{{{\text{P}}^{°}}_{\text{methanol}}}\text{=}\frac{253.14}{143}\text{=1.77} \end{gathered}$$