Question

In a binary homogeneous solution liquid
A has mole fraction 0.8 and vapour
pressure of liq. A and B in pure state are
100 mm Hg and 50 mm Hg respectively
then on decrease in pressure at what
pressure last trace of liquid disappear?​

Answer 1

The last liquid of solution will disappear at P = 141.4 mm Hg

Explanation:

• Let P be the total pressure. Let x be the number of moles of A in the liquid state.
• The number of moles of the vapor state = 10−x.
• Let y be the number of moles of B in the liquid state
• The number of moles of B in vapor state will be 10−y.

Hence, x + y = 10

P × (10 − x + 10 − y) (10 − x) = 200 × (x + y) (x)

P × (10 − x + 10 − y) (10−y) = (x + y) 100 × (y)

These equations are solved for x, y, and P.

x = 4.142

y = 5.858

P = 141.4 mm Hg

Thus the last liquid of solution will disappear at P = 141.4 mm Hg

Answer 2

Aɴsᴡᴇʀ:-

The last liquid of solution will disappear at P = 141.4 mm Hg

Explanation:

Let P be the total pressure. Let x be the number of moles of A in the liquid state.

The number of moles of the vapor state = 10−x.

Let y be the number of moles of B in the liquid state

The number of moles of B in vapor state will be 10−y.

Hence, x + y = 10

P × (10 − x + 10 − y) (10 − x) = 200 × (x + y) (x)

P × (10 − x + 10 − y) (10−y) = (x + y) 100 × (y)

These equations are solved for x, y, and P.

x = 4.142

y = 5.858

P = 141.4 mm Hg

Thus the last liquid of solution will disappear at P = 141.4 mm Hg