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Vapour pressure of pure water at 25 celsius is 80 torr and that of an aqueous solution os 75 torr

Vapour pressure of pure water at 25 celsius is 80 torr and that of an aqueous solution os 75 torr
 

Question Image
Grade:12

1 Answers

Anish
105 Points
one month 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 = 
0
 
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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