Question

A solution is prepared by mixing 8.5 g of $\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$ and 11.95 g of $\text{CHC}{{\text{l}}_{\text{3}}}$ . If the vapor pressure of $\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$ and $\text{CHC}{{\text{l}}_{\text{3}}}$ at 298 K are 415 and 200 mm Hg respectively, the mole fraction of $\text{CHC}{{\text{l}}_{\text{3}}}$ in vapor form is:
(Molar mass of Cl is $35.5\text{ g mo}{{\text{l}}^{-1}}$ )

Answer 1

The mole fraction of a component in the vapor phase can be calculated by using the formula:
${{\text{y}}_{\text{CHC}{{\text{l}}_{\text{3}}}}}=\dfrac{{{\text{p}}_{\text{CHC}{{\text{l}}_{\text{3}}}}}}{{{\text{p}}_{\text{Total}}}}$
Where ${{\text{y}}_{\text{CHC}{{\text{l}}_{\text{3}}}}}$ is the mole fraction of $\text{CHC}{{\text{l}}_{\text{3}}}$ in vapor phase, ${{\text{p}}_{\text{CHC}{{\text{l}}_{\text{3}}}}}$ is the partial vapor pressure of $\text{CHC}{{\text{l}}_{\text{3}}}$ , and ${{\text{p}}_{\text{Total}}}$ is the total pressure of the solution.

Complete answer:
Raoult’s Law gives the relationship between the vapor pressure at a given temperature and mole fraction. According to this law given by Raoult, the vapor pressure (p) of a liquid component in an ideal mixture is directly proportional to its mole fraction (x). Mathematically,

& \text{p}\propto \text{x} \\\ & \Rightarrow \text{p}={{\text{p}}^{\text{o}}}\text{x} \\\ \end{aligned}$$ Where ${{\text{p}}^{\text{o}}}$ is the vapor pressure of the component in the pure state. For a mixture of $\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$ and $\text{CHC}{{\text{l}}_{\text{3}}}$, we can write: $$\begin{aligned} & {{\text{p}}_{\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}}=\text{p}_{\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}^{\text{o}}{{\text{x}}_{\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}} \\\ & {{\text{p}}_{\text{CHC}{{\text{l}}_{3}}}}=\text{p}_{\text{CHC}{{\text{l}}_{3}}}^{\text{o}}{{\text{x}}_{\text{CHC}{{\text{l}}_{3}}}} \\\ \end{aligned}$$ So, first, we need to find the mole fraction of each component present in the mixture. It can be calculated as follows: (1) – Find the number of moles. $$\begin{aligned} & {{\text{n}}_{\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}}=\dfrac{\text{Mass of C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}{\text{Molar mass of C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}=\dfrac{8.5\text{ g}}{85\text{ g mo}{{\text{l}}^{-1}}}=0.1\text{ moles} \\\ & {{\text{n}}_{\text{CHC}{{\text{l}}_{3}}}}=\dfrac{\text{Mass of CHC}{{\text{l}}_{3}}}{\text{Molar mass of CHC}{{\text{l}}_{3}}}=\dfrac{11.95\text{ g}}{119.5\text{ g mo}{{\text{l}}^{-1}}}=0.1\text{ moles} \\\ \end{aligned}$$ (2) – Find mole fraction. $$\begin{aligned} & {{\text{x}}_{\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}}=\dfrac{{{\text{n}}_{\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}}}{{{\text{n}}_{\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}}+{{\text{n}}_{\text{CHC}{{\text{l}}_{\text{3}}}}}} \\\ & {{\text{x}}_{\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}}=\dfrac{\text{0}\text{.1}}{\text{0}\text{.1}+\text{0}\text{.1}}=0.5 \\\ & \text{Similarly for CHC}{{\text{l}}_{\text{3}}},\text{ }{{\text{x}}_{\text{CHC}{{\text{l}}_{3}}}}=\dfrac{\text{0}\text{.1}}{\text{0}\text{.1}+\text{0}\text{.1}}=0.5 \\\ \end{aligned}$$ Now, let us find the partial pressure of each component. $$\begin{aligned} & {{\text{p}}_{\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}}=\text{p}_{\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}^{\text{o}}{{\text{x}}_{\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}}=415\times 0.5=207.5\text{ mm Hg} \\\ & {{\text{p}}_{\text{CHC}{{\text{l}}_{3}}}}=\text{p}_{\text{CHC}{{\text{l}}_{3}}}^{\text{o}}{{\text{x}}_{\text{CHC}{{\text{l}}_{3}}}}=200\times 0.5=100\text{ mm Hg} \\\ \end{aligned}$$ According to Dalton’s law of partial pressure, the total pressure is the sum of partial vapor pressures of each component, i.e. $$\begin{aligned} & {{\text{p}}_{\text{Total}}}={{\text{p}}_{\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}}}+{{\text{p}}_{\text{CHC}{{\text{l}}_{3}}}} \\\ & \therefore {{\text{p}}_{\text{Total}}}=\left( 207.5+100 \right)\text{ mm Hg}=307.5\text{ mm Hg} \\\ \end{aligned}$$ Let the mole fraction of $\text{CHC}{{\text{l}}_{\text{3}}}$ in vapor phase be ${{\text{y}}_{\text{CHC}{{\text{l}}_{\text{3}}}}}$. Then, $$\begin{aligned} & {{\text{y}}_{\text{CHC}{{\text{l}}_{\text{3}}}}}=\dfrac{{{\text{p}}_{\text{CHC}{{\text{l}}_{\text{3}}}}}}{{{\text{p}}_{\text{Total}}}} \\\ & \Rightarrow {{\text{y}}_{\text{CHC}{{\text{l}}_{\text{3}}}}}=\dfrac{\text{100}}{\text{307}\text{.5}} \\\ & \Rightarrow {{\text{y}}_{\text{CHC}{{\text{l}}_{\text{3}}}}}=\text{0}\text{.32} \\\ \end{aligned}$$ Hence, the mole fraction of $\text{CHC}{{\text{l}}_{\text{3}}}$ in vapor form is 0.32. **Note:** The mole fraction is simply a ratio of the amount of one component in a mixture and thus it has no unit. The mole fraction of any component is different in two different phases (vapor and liquid phase) because at a given temperature when both phases are present, a certain amount of component in the vapor phase will remain in equilibrium with some amount of it in the liquid phase.