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:

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 .

Answer 2

$\huge\bold{AnswEr}$

The phenomenon of the irregular movement of pollen grains suspended in water was observed by Robert Brown in 1827.

Brownian motion, the name given to this process, was not given a rigorous mathematical treatment until almost one hundred years later by N. Wiener in 1923.