Question

1‑Propanol (P∘1=20.9 Torr at 25 ∘C) and 2‑propanol (P∘2=45.2 Torr at 25 ∘C) form ideal solutions in all proportions. Let x1 and x2 represent the mole fractions of 1‑propanol and 2‑propanol in a liquid mixture, respectively, and y1 and y2 represent the mole fractions of each in the vapor phase. For a solution of these liquids with x1=0.220, calculate the composition of the vapor phase at 25 ∘C.

Answer 1

Answer:

The composition of 1-propanol in vapor phase = 11.54 %

The composition of 2-propanol in vapor phase = 88.46 %

Explanation:

Mole fraction of components in liquid phase:

1-propanol = $x_1=0.220$

2-propanol = $x_2$

$x_2=1-x_1=1-0.22=0.78$

Partial pressure of 1-propanol =$p_1=20.9 Torr$

Partial pressure of 2-propanol =$p_2=45.2 Torr$

According to Raoults law:

$P=x_1\times p_1+x_2\times p_2$

$P=0.22\times 20.9 Torr+0.78\times 45.2 Torr=39.854$

Mole fraction of components in vapor phase:

1-propanol = $y_1=\frac{x_1\times p_1}{P}$

$=y_1=\frac{0.22\times 20.9 Torr}{39.854 Torr}=0.1154$

The composition of 1-propanol in vapor phase:

100 × 0.1154= 11.54 %

2-propanol = $y_2$

$=y_2=\frac{0.78\times 45.2 Torr}{39.854 Torr}=0.8846$

The composition of 2-propanol in vapor phase:

100 × 0.8846 = 88.46 %

Answer 2

The composition of the vapor phase of propane 1 is 11.54%.

The composition of the vapor phase of propane 2  is  88.46%.

What is the vapor phase?

Vapor refers to a gas phase at a temperature below the critical temperature of the chemical, where the same substance can also exist in the liquid or solid-state.

Calculation:

Given, the pressure of 1‑Propanol is 20.9 Torr

The pressure of 2-propane is 45.2 Torr

Mole fraction of x1 = 0.220

Mole fraction of x2 =?

x2 = 1 – x1

x2 = 1 – 0.220 = 0.78

Now, from Raoult's law:

$\bold{R = x1\times p1 + x2\times p2}$

$\bold{R = 0.220 \times 20.9 + 0.78 \times 45.2 = 39.854}$

Mole fraction of the components:

$\bold{Mole\;fraction\;of\;Propane1 =\dfrac{ partial\;pressure}{Total\;pressure}}$$\bold{Mole\;fraction\;of\;Propane1 =\dfrac{0.220\times20.9}{39.854}= 0.1154}$

$\bold{Mole\;fraction\;of\;Propane2 =\dfrac{ partial\;pressure}{Total\;pressure}}$

$\bold{Mole\;fraction\;of\;Propane1 =\dfrac{0.78\times 45.2}{39.854}= 0.8846}$

The composition of the vapor for propane 1

$\bold{0.1154 \times 100 = 11.54\%}$

The composition of the vapor for propane 2

$\bold{0.8846 \times 100 = 88.46\%}$

Thus, the composition of the vapor phase for propane1 and propane 2 is 11.54 and 88.46 respectively.

Learn more about vapor phase, here: