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VITEEE 2025 Paper
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© examsnet.com
Question : 53 of 125
Marks:
+1
,
-0
Vapour pressure of solution containing 2 mol of liquid
A
A
A
(
P
A
o
=
80
P_{A}^{o}=80
P
A
o
​
=
80
torr ) and 3 mol of liquid
B
B
B
(
P
B
o
=
100
P_{B}^{o}=100
P
B
o
​
=
100
torr ) is 87 torr. We can conclude that
there is negative deviation from Raoult's law
boiling point is higher than that expected for ideal solution
molecular attractions between unlike molecules are stronger than those between like molecules
All of these statements are correct
Validate
Solution:
For ideal solution vapour pressure of solution
  
=
P
A
∘
X
A
+
P
B
∘
X
B
\;=P_A^{\circ} X_A+P_B^{\circ} X_B
=
P
A
∘
​
X
A
​
+
P
B
∘
​
X
B
​
  
=
80
×
  
2
5
+
100
×
  
3
5
=
92
  
 torrÂ
  
\;=80 \times \;\frac{2}{5}+100 \times \;\frac{3}{5}=92 \;\text{ torr }\;
=
80
×
5
2
​
+
100
×
5
3
​
=
92
 torrÂ
Since observed vapour pressure of solution < ideal vapour pressure, the solution shows negative deviation.
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