Chemistry, asked by moni7960, 1 year ago

At 300 K the vapour pressure of an ideal solution containing 1 mole of liquid A and 2 moles of liquid B is 500 mm of Hg. The vapour pressure of the solution increases by 25 mm of Hg, if one more mole of B is added to the above ideal solution at 300 K. Then the vapour pressure of A in its pure state is
(a) 300 mm of Hg
(b) 400 mm of Hg
(c) 500 mm of Hg
(d) 600 mm of Hg

Answers

Answered by qwsuccess
1

Given:

  • Number of moles of A (na) = 1
  • Initial number of moles of B (n1b) = 2
  • Initial vapour pressure of solution (Pi) = 500 mm of Hg
  • Final number of moles of B (n2b) = 3
  • Final vapour pressure of solution (Pf) = 525 mm of Hg

To find:

The vapour pressure of A in its pure state.

Solution:

  • Let the vapour pressures of A and B in their pure states be Pa and Pb respectively.
  • Let initial mole fractions of A and B be X1a and X1b respectively. So, X1a = na/(na+n1b) = 1/3 and X1b = 1- X1a = 2/3
  • Let final mole fractions of A and B be X2a and X2b respectively. So, X2a = na/(na+n2b) = 1/4 and X2b = 1- X2a = 3/4
  • Now according to Raoult's law, P = Pa*Xa + Pb*Xb
  • Pa/3 + 2Pb/3 = 500 and Pa/4 + 3Pb/4 = 525
  • Solving the above two equations we get, Pa = 300 mm of Hg

Answer:

The vapour pressure of A in its pure state = 300 mm of Hg.

Answered by handgunmaine
1

Given :

Inital condition :

Moles of A is 1 mole .

Moles of B is 2 moles .

Pressure of system , P_i=500\ mm

Final condition :

Moles of A is 1 mole .

Moles of B is 3 moles .

Pressure of system , P_f=525\ mm

To find :

The vapour pressure of A in its pure state .

Solution :

Let , the vapour pressure of pure A and pure B is p^o_A\ and \ p^o_B .

We know , pressure of system is given by :

P=X_A.p^o_A+X_B.p^o_B

Here , X is the mole fraction .

For initial condition :

500=\dfrac{1}{3}p^o_A+\dfrac{2}{3}p^o_B\\\\1500=p^o_A+2p^o_B            ...... ( 1 )

For final condition :

525=\dfrac{1}{4}p^o_A+\dfrac{3}{4}p^o_B\\\\2100=p^o_A+3p^o_B            ...... ( 2 )

Subtracting equation 1 by 2 .

We get :

p^0_B=600\ mm

Therefore ,

p^0_A=1500-2\times 600\\\\p^0_A=300\ mm

The vapour pressure of A in its pure state is 300 mm of Hg .

Therefore , option ( a ) is correct .

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Vapour Pressure

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