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george8396
03.12.2019 •
Chemistry
Glycerol (c3h8o3, 92.1 g/mol) is a nonvolatile nonelectrolyte substance. consider that you have an aqueous solution that contains 34.4 % glycerol by mass. if the vapor pressure of pure water is 23.8 torr at 25oc, what is the vapor pressure of the solution at 25oc?
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Ответ:
The vapor pressure of the solution at 25°C is 26.01 Torr
Explanation:
This is a usual excersise of colligative properties. In this case we apply the vapor pressure lowering formula:
ΔP = Xst . P°
Where ΔP is the diferrence between Pressure of solution - Pressure of pure solvent.
And Xst the molar fraction.
P° is Pressure of pure solvent.
So the formula will be:
Pressure Solution - P° = Xst . P°
Pressure Solution - 23.8 Tor = Xst . 23.8 Torr
Xst : Mole fraction ( Moles of solute or solvent / Total moles)
34.4 % m/m means that in 100 g of solution I have 34.4 g of solute
If I have 34.4 g of solute and the mass of 100 g in solution, I can know the mass of solvent, and finally the moles.
100 g solution - 34.4 g solute = 65.6 g (mass of solvent)
Molar mass of water : 18 g/m
Moles of water: Mass of water / Molar mass
65.6 g / 18g/m = 3.64 moles
Moles of glycerol : Mass glycerol / Molar mass glycerol
34.4 g / 92.1 g/m = 0.373 moles
Total moles: moles of glycerol + moles of water
0.373 m + 3.64 m = 4.01 m
So Xst = 0.373 m / 4.01 m → 0.093
Pressure Solution - 23.8 Tor = 0.093 . 23.8 Torr
Xst HAVE NO UNITS
Pressure Solution = (0.093 . 23.8 Torr ) + 23.8 Tor
Pressure Solution = 26.01 Torr
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