derive an expression showing the relationship between molar mass of solute and lowering of vapour pressure
Answers
Answer:
The expression for the relative lowering of vapour pressure is
P
0
ΔP
=
P
0
P
0
−P
=X
2
(1)
Here,
P
0
ΔP
and
P
0
P
0
−P
represents relative lowering of vapour pressure and X
2
represents mole fraction of solute.
But the mole fraction of solute
X
2
=
n
1
+n
2
n
2
=
M
2
W
2
+
M
1
W
1
M
2
W
2
(2)
Here, W
1
and M
1
are the mass and molar masses of solvent and W
2
and M
2
are the mass and molar masses of solute. n
1
and n
2
are the number of moles of solute and solvent respectively.
For dilute solutions n
1
>>n
2
So n
2
may be neglected in comparison with n
1
.
Hence, equation (2) becomes
X
2
=
n
1
n
2
=
M
1
W
1
M
2
W
2
(3)
From equation (1) and equation (3)
P
0
ΔP
=
P
0
P
0
−P
=
M
1
W
1
M
2
W
2
This gives the relationship between relative lowering of vapour pressure and molar mass of solute.