Chemistry, asked by sharmavsv890, 9 months ago

vapour pressure of liquid Ais greater than that of liquid B at 298k .which will have higher value of Norma boiling point​

Answers

Answered by Anonymous
0

Answer:

ANSWER

It is given that,

P

A

0

=300 mm of Hg

P

B

0

=450 mm of Hg

P

Total

=405 mm of Hg

From Raoults law, we have

P

total

=P

A

+P

B

P

total

=P

A

0

X

A

+P

B

0

(1−X

A

)

P

total

=P

A

0

X

A

+P

B

0

−P

B

0

X

A

P

total

=(P

A

0

−P

B

0

)X

A

+P

B

0

405=(300−450)X

A

+450

−45=−150X

A

X

A

=0.3

Therefore,

X

B

=1−X

A

=1−0.3

=0.7

P

A

=P

A

0

X

A

=300×0.3

=90mmofHg

P

B

=P

B

0

×X

B

=450×0.7

=315mmofHg

Now in the vapour phase:

Mole fraction of liquid A,

A=

(P

A

+P

B

)

P

A

=

90+315

90

=0.22

Thus mole fraction of liquid A is 0.22

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