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