the vapour pressure of pure water at 25 degree Celsius is 19.2 mmhg if 38 gram of glucose is added to 17 6.4 gram of water at 25 degree Celsius the vapour pressure of resulting solution will be
Answers
Your example answer:-
Explanation:
For solutions that contain non-volatile solutes, the vapor pressure of the solution can be determined by using the mole fraction of the solvent and the vapor pressure of the pure solvent at the same temperature.
P
sol
=
χ
solvent
⋅
P
∘
solvent
, where
P
sol
is the vapor pressure of the solution
χ
solvent
is the mole fraction of the solvent
P
∘
solvent
is the vapor pressure of the pure solvent
In your case, you know that the vapor pressure of pure water at
25
∘
C
is equal to
23.8
torr. This means that all you have to do is determine the mole fraction of water in the solution.
As you know, mole fraction is defined as the number of moles of a component of a solution divided by the total number of moles present in that solution.
Use glucose and water's respective molar masses to determine how many moles of each you have
18.0
g
⋅
1 mole glucose
180.0
g
=
0.100 moles glucose
and
95.0
g
⋅
1 mole water
18.015
g
=
5.273 moles water
The total number of moles present in the solution will be
n
total
=
n
glucose
+
n
water
n
total
=0.100
+5.273
=5.373 moles
This means that the mole fraction of water will be
χ
water
=
5.273
moles
5.373
moles
=
0.9814
Finally, the vapor pressure of the solution will be
P
sol
=0.9814⋅
23.8 torr=23.4 torr