Question

The total pressure exerted by an ideal binary mixture of A and B is [Given: P = 40 mm Hg, P = 80 mm Hg
and mole fraction of A in vapour phase is 1/3.
first correct answer to be marked brainliest.​

Answer 1

Answer:

200/3 mm Hg

Explanation:

From Raoult's Law

PT = P°aXa + P°bXb

= P°aXa + P°(1-Xa)

= P°aXa + P°b - P°bXa

= P°b + (P°a - P°b)Xa

So, PT = 80 + (40 - 80)1/3

= 80 - 40/3

= 240-40/3

= 200/3

Answer 2

Answer:

The total pressure exerted by an ideal binary mixture of A and B with the given parameters would be $\frac{200}{3}$ mm Hg.

Explanation:

From our understanding of the question, this is based on Raoult's Law of Partial Pressures.

The law states that the partial vapour pressure of a given solvent in the solution/mixture mathematically equals the product of the vapour pressure of the solvent in its pure form in a solution (or mixture) is equal or identical to the vapour pressure of the pure solvent and the mole fraction.

Thus, it can be written as $P_s_o_l_u_t_i_o_n=X_s_o_l_v_e_n_tP^0_s_o_l_v_e_n_t$

We can thus calculate the total pressure exerted by a binary mixture as follows -

From Raoult's Law,

$P_T = P^0_AX_A + P^0_BX_B\\\\P_T = P^0_AX_A + P^0(1-X_A) \\\\P_T = P^0_AX_A + P^0_B - P^0_BX_A \\\\P_T = P^0_B + (P^0_A - P^0_B)X_A$

$P_T = 80 + \frac{80-40}{3} \\\\P_T = 80 - \frac{40}{3}\\\\P_T =\frac{240-40}{3}\\\\P_T = \frac{200}{3}$

Thus, the total pressure exerted is $\frac{200}{3}$ mm Hg.

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