Vapour pressure of pure water at 25 celsius is 80 torr and that of an aqueous solution os 75 torr
Simon , 3 Years ago
Grade 12
Physical Chemistry> Vapour pressure of pure water at 25 celsi...
1 Answers
Anish
Last Activity: 3 Years ago
Let M be the molecular weight of solute.
Number of moles of solute =
M
5.40
Number of moles of water =
18
90.0
=5
Mole fraction of solute =
M
5.40
+5
M
5.40
=
5.40+5M
5.40
Relative lowering in the vapour pressure =
P
0
P
0
−P
=
23.76
23.76−23.32
=0.01852
The relative lowering in the vapour pressure of the solution is equal to the mole fraction of solute.
0.01852=
5.40+5M
5.40
⟹5.40+5M=291.6
5M=286.2
⟹M=57.24 g/mol
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