Question

A 1 : 2 mixture (by mole) of ethanol and propanol exerts a vapour pressure of 250 mm Hg. The vapour pressure of propanol in solution is 200 mm Hg, then the vapour pressure of pure ethanol is (in mm Hg)


350


150


215


400

Answer 1

Answer:

See below.

Explanation:

We are given a 1 : 2 mixture of ethanol and propanol. Meaning, we have 1 mole of ethanol and 2 moles of propanol.

Mole fraction of ethanol,

$\eta _{ethanol}=\dfrac {1}{1+2}=\dfrac {1}{3}$

Mole fraction of propanol,

$\eta _{propanol}=\dfrac {2}{1+2}=\dfrac {2}{3}$

We know the equation,

$p_{so1}=p^{0}_{A}\eta _{A}+p^{0}_{B}\eta _{B}$

Substituing the values,

$\begin{aligned}250=\dfrac {1}{3}p_{methanol}+\dfrac {2}{3}\left( 200\right) \\ p_{methanol}=750-400\\ =350 \ mm \ of \ Hg\end{aligned}$

Answer 2

Given:

Ratio of ethanol to propanol = 1 : 2

Total Vapour pressure, P = 250 mm Hg

Vapour pressure of propanol, P°₁ = 200 mm Hg

To Find:

The vapour pressure of pure ethanol, i.e., P°₂.

Calculation:

- Mole fraction of propanol, X₁  = 2 / (2+1) = 2 / 3

- Mole fraction of ethanol, X₂ = 1 / (1+2) = 1 / 3

- We know that total vapour pressure is given as:

P = P°₁X₁ + P°₂X₂

⇒ 250 = 200 × (2/3) + P°₂ × (1/3)

⇒ P°₂ = 3 × {250 - (400/3)}

⇒ P°₂ = 350 mm Hg

- So the vapour pressure of pure ethanol is 350 mm Hg.

- Hence, the correct answer is option (a) 350 mm Hg.