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The incorrect relationship according to Raoult's law for two miscible liquid mixture is :
CHEMISTRY · Single correct · CBSE · Chemistry · Solutions · Solutions
Problem
The incorrect relationship according to Raoult's law for two miscible liquid mixture is : (1)
P
T
=
P
A
0
+
(
P
B
0
−
P
A
0
)
X
B
P_T = P_A^0 + (P_B^0 - P_A^0) X_B
P
T
=
P
A
0
+
(
P
B
0
−
P
A
0
)
X
B
(2)
P
A
i
+
P
B
i
=
P
T
P_A^i + P_B^i = P_T
P
A
i
+
P
B
i
=
P
T
(3)
P
T
=
P
B
0
+
(
P
A
0
−
P
B
0
)
X
A
P_T = P_B^0 + (P_A^0 - P_B^0) X_A
P
T
=
P
B
0
+
(
P
A
0
−
P
B
0
)
X
A
(4)
P
T
=
P
A
0
+
(
P
A
0
−
P
B
0
)
X
A
P_T = P_A^0 + (P_A^0 - P_B^0) X_A
P
T
=
P
A
0
+
(
P
A
0
−
P
B
0
)
X
A
Related theory
Raoult's Law
Solutions
Vapour Pressure of Liquid Solutions
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