Vapour-liquid equilibrium of a two-component ideal solution of trichloroethene ($\ce{C2HCl3}$) and trichloromethane ($\ce{CHCl3}$) is established at $\pu{25 °C}$. The mole fraction of $\ce{CHCl3}$ in the vapour phase is $0.73$. What is the mass fraction of $\ce{C2HCl3}$ in the liquid phase? Round your answer to two significant figures.

The vapour pressures of trichloroethene and trichloromethane at $\pu{25 °C}$ are:

$$\begin{align}P_\text{vap}(\ce{C2HCl3}) &= \pu{73.0 mmHg}\\[0.5em] P_\text{vap}(\ce{CHCl3}) &= \pu{199.1 mmHg}\end{align}$$

So, what I did was I found mole fraction of $\ce{C2HCl3}$ and then used the two mole fractions along with the vapour pressures to find the total pressure of the solution.

$$P_\text{vap}= \frac{0.73}{199.1}+\frac{0.27}{73} = \pu{165.053mmHg}$$

Then, from Raoult's Law I know that the mole fraction in liquid phase is equal to mole fraction in vapour phase, multiplied by vapour pressure, divided by total pressure. From that, I found the mole fraction of both things in liquid phase. I use the mole fraction to find mass of both, and then did mass of $\ce{C2HCl3}$ divided by the total mass that I calculated. I got an answer of

$\frac{15.68}{120}=0.13$ but it says that it's wrong. I'm not sure where I messed up?


2 Answers 2


There is actually small error in the calculation part of total pressure and in the Raoult’s law formulae.

According to Raoult’s law: $$ P_A= P^0 _ A \chi_A $$

i.e. partial pressure of a component in a solution($P_A$) is the product of the vapour pressure of the pure solvent A at that temperature($ P^0_A$) and it's mole fraction in liquid phase($\chi_A$).

Also, we have $$ P_A= P_{total}\chi^‘_A$$

where: $P_A$=partial pressure of component A

$P_{total}$=total vapour pressure of the solution

$\chi^{’}_A$=mole fraction of component A in vapour phase

Considering these equations along with Dalton’s law of partial pressure (total vapour pressure of a solution is the sum of vapour pressures of its individual components)$$P_{total}=P_A+P_B$$

we get: $$P^0_A\chi_A=P_{total}\chi^{‘}_A$$

Substituting the respective values:

$$ (199.1){\chi_{\ce{CHCl3}}}=P_{total}(0.73)$$ and $$ (73){\chi_{\ce{C2HCl3}}}=P_{total}(0.27)$$

Since $${\chi_{\ce{CHCl3}}}+{\chi_{\ce{C2HCl3}}}=1$$

we have two equation with two unknowns which upon solving we get ${\chi_{\ce{CHCl3}}}=0.4978$ and ${\chi_{\ce{C2HCl3}}}=0.5022$ and you shall continue in the same way to get the mass fraction.

Hope it helps.


As far as the question is concerned, you do not need the vapor pressure. In order to calculate the mass fraction of $\ce{C2HCl3}$, we first have to calculate the molar mass of $\ce{C2HCl3}$ and $\ce{CHCl3}$. $$M(\ce{C2HCl3}) = \pu{131.38 g mol-1}$$ $$M(\ce{CHCl3}) = \pu{119.37 g mol-1}$$ With these two calculated we can calculate the absolute mass relative to $\pu{1 mol}$, due to the fact of ${\pu{0.73 mol}}/{\pu{1 mol}}$ being relative to a total of $\pu{1 mol}$ of molecules. Now if we calculate the absolute masses, we get: $$M(\ce{C2HCl3}) = \pu{131.38 g mol-1}\times\pu{0.27\frac{mol}{mol}} = \pu{35.47 g mol-1}$$ $$M(\ce{CHCl3}) = \pu{119.37 g mol-1}\times\pu{0.73\frac{mol}{mol}} = \pu{87.14 g mol-1}$$ And the total mass in the system results in: $$m_\text{total}=\pu{35.47 g mol-1}+\pu{87.14 g mol-1}=\pu{122.61 g mol-1}$$ And with the total relative mass given, we can calculate the mass fraction of $\ce{C2HCl3}$, as follows: $$w(\ce{C2HCl3})=\frac{\pu{35.47 g mol-1}}{\pu{122.61 g mol-1}}=0.29$$ $$w(\ce{CHCl3})=\frac{\pu{87.14 g mol-1}}{\pu{122.61 g mol-1}}=0.71$$

  • $\begingroup$ Did you notice the mole fraction provided (0.73) is for the vapor phase and the mass fraction asked is for the liquid phase? $\endgroup$ Nov 9, 2021 at 19:18

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