Question

Two liquids X and Y form an ideal solution. The mole fraction of X in vapour phase is 0.2 while in liquid phase is 0.8 . If the total pressure in liquid phase is 800 torr then what are the original vapour pressure of X and Y??

Answer 1

Answer:

Explanation:

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Answer 2

Answer:

The vapour pressure of liquid X = $128torr$.

The vapour pressure of liquid Y = $3488torr$.

Explanation:

Given,

The total pressure, $P_{total}$ = $800torr$

The mole fraction of X in the vapour phase, $m_{X}$ = $0.2$

The mole fraction of X in the liquid phase, $n_{X}$ = 0.8

The original vapour pressure of X and Y =?

As we know,

• The mole fraction of X in the liquid phase + the mole fraction of Y in the liquid phase = 1
• $0.8 + n_{Y}$ = 1
• $n_{Y}$ = $0.2$

Now, in the vapour phase;

• $P_{X} = m_{X}P_{total}$

• Here, $P_{}$ = pressure of liquid X

• $P_{X} = 0.2\times 800$ = $160 torr$

Now, in the liquid phase;

• $P_{X} = P\textdegree _{X} \times n_{X}$

• Here, $P\textdegree _{X}$ = vapour pressure of liquid X

• $160 = P\textdegree _{X} \times 0.8$
• $P\textdegree _{X} = 128 torr$

As we know,

• $P_{total}= P\textdegree_{X}n_{X} + P\textdegree_{Y}n_{Y}$
• $800= 128\times 0.8 + P\textdegree_{Y}\times 0.2$
• $P\textdegree _{Y} = 3488torr$

Hence, the vapour pressure of liquid X = $128torr$

And the vapour pressure of liquid Y = $3488torr$.