Question

calculate the partial vapour pressure of c2h4br2 at 85°c for an ideal solution with mole fraction of 0.25. vapour pressure of pure c2h4br2 at 85°c is 170mm hg.​

Answer 1

Answer:

Vapour pressure of the solution = 42.5 mm Hg

Explanation:

Given:

• Partial vapour pressure of pure C₂H₄Br₂ at 85 °C = 170 mm Hg
• Mole fraction of the solution = 0.25

To Find:

• Partial vapour pressure of the solution

Solution:

Given that the solution is ideal.

Hence it obeys Raoult's law for the entire range of concentration.

By Raoult's law we know that, the partial vapour pressure of each of the volatile components in the solution is directly proportional to its mole fraction.

That is,

$\sf p_1\propto x_1$

$\sf p_1=(p_1)^0\: x_1$

where p₁⁰ is the vapour pressure of the pure component, x₁ is the mole fraction of the solution and p₁ is the vapour pressure of the solution.

Substituting the data we get,

$\sf p_1=0.25\times 170$

$\sf \implies 42.5\:mm\:Hg$

Hence the vapour pressure of the solution at the given temperature is 42.55 mm Hg.

Answer 2

Answer:

Given :-

• At 85°C for an ideal solution with mole fraction of 0.25.
• Vapour pressure of pure C₂H₄Br₂ at 85°C is 170 mm hg.

To Find :-

• What is the partial vapour pressure of C₂H₄Br₂.

Formula Used :-

$\longmapsto \sf\boxed{\bold{\pink{p_1 =\: x_1 \times {p}_{1}^{0}}}}$

where,

• p₁ = Partial vapour pressure of the solution
• p₁⁰ = Vapour pressure of the pure Components
• x₁ = Mole fraction of the solution

Solution :-

Given :

• Mole fraction of the solution = 170 mm hg
• Vapour pressure of the pure Components = 0.25

According to the question by using the formula we get,

$\dashrightarrow \sf p_1 =\: 170 \times 0.25$

$\dashrightarrow \sf p_1 =\: 170 \times \dfrac{25}{100}$

$\dashrightarrow \sf p_1 =\: \dfrac{4250}{100}$

$\dashrightarrow \sf p_1 =\: \dfrac{425\cancel{0}}{10\cancel{0}}$

$\dashrightarrow \sf p_1 =\: \dfrac{425}{10}$

$\dashrightarrow \sf\bold{\red{p_1 =\: 42.5\: mm\: hg}}$

$\therefore$ The partial vapour pressure of C₂H₄Br₂ is 42.5 mm hg .