Question

100 g of liquid A (mol.mass of =140g/ mol)was dissolved in 1000g of liquid B (molar mass of =180g/mol).the vapour pressur of pure liquid B was found to be 500 atmosphere. calculate the vapour pressure of pure liquid A and its vapour pressure in the solution, if the total vapour pressure of the solution is 475 atmosphere

Answer 1

Number of Moles of Liquid A, nA = 100 / 140 = 0.714

Number of Moles of Liquid B, nB = 1000 / 180 = 5.556

Then Mole fraction of A = nA / nA + nB = 0.714 / 0.714 + 5.556

= 0.114

Now Mole of fraction of B = 1 - 0.114 = 0.886

Now ptotal  = pA + pB

Or ptotal = p°AXA + p°BxB

OR

475 = p°A X 0.114 + 500 X 0.886

OR

p°A = 280.7 torr

Therefore vapour pressure of pure A = 280.7 torr

Vapour pressure of A in solution = 280.7 x 0.114

= 32 torr

Now

pA  = p°AXA

Or

p°A  = pA   /  XA

⇒ 32 / 0.114

= 280.7 torr

Hence, the vapour pressure of pure liquid A is 280.7 torr.

Answer 2

Answer:

1. The vapour pressure of pure liquid A is 280.7 atm.
2. The vapour pressure of A in solution is 31.99 atm.

Explanation:

Given:

Weight of liquid A= 100g

∴ The number of moles of A (nA)=100/140=0.714

Weight of liquid B= 1000g

∴ The number of moles of B (nB)=1000/180=5.556

Vapour pressure of pure liquid B(p°B)=500 atm

Total vapour pressure of the solution (ptotal)=475 atm

To find: Vapour pressure of pure liquid A and its vapour pressure in the solution=?
Solution:
Mole fraction of A(xA)=$\frac{nA}{nA+nB}$=0.114

Mole fraction of B(xB)=$\frac{nB}{nA+nB}$=0.886

We know that,

ptotal=p°AxA+p°BxB

475=p°A*0.114+500*0.886

∴p°A=280.7 atm

Therefore, the vapour pressure of pure liquid A is 280.7 atm

pA=p°AxA

   =280.7*0.114

   =31.99 atm.

Therefore, the vapour pressure of A in solution is 31.99 atm.