Caffeine + Water
(https://www.wallpaperflare.com/search?wallpaper=caffeine)
“Pure caffeine (trimethyl xanthine) occurs as a white powder or as silky needles, which melt at 238 °C (460 °F); it
sublimes at 178 °C (352 °F) at atmospheric pressure. It is very soluble in hot water; upon cooling, the solution
deposits crystals of caffeine monohydrate. Caffeine is generally less soluble in organic solvents than in hot water.
It is odorless but has a bitter taste.” (https://www.britannica.com/science/caffeine)
Type of Solid
It is a molecular solid with low conductivity. Caffeine is soluble in water because water can hydrogen bondto
caffeine at six different places.
Intermolecular Forces
The intermolecular forces that are present in caffeine are London dispersion forces and Dipole-dipole forces.
Effect in Boiling Point
Adding a non-volatile substance such as caffeine to a liquid has the effect of raising the boiling point abovethat of
the pure liquid(water).
Effect in Freezing Point
The effect of adding a solute(caffeine) to a solvent(water) has the opposite effect on the freezing point of asolution
as it does on the boiling point. So the solution will have a lower freezing point than a pure solvent.
Equation to calculate the increase in the boiling point:
ΔTb=kb⋅m⋅i
Where:
ΔTb= the amount the boiling point increases.
kb= the boiling point elevation constant which depends on the solvent (water = 0.51°C/m)
m= the molality of the solution.
i= the number of particles formed when that compound dissolves
Given:
kb=
0.51°C/m
m= ?
i= 1
𝑚𝑚 =
𝑚𝑚 =
2.33 𝑔𝑔 𝐶𝐶8𝐻𝐻10𝑁𝑁4𝑂𝑂2
15.0 𝑔𝑔 𝐻𝐻2𝑂𝑂
•
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑘𝑘𝑘𝑘 𝑜𝑜𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶8𝐻𝐻10𝑁𝑁4𝑂𝑂2
194.19 𝑔𝑔 𝐶𝐶8𝐻𝐻10𝑁𝑁4𝑂𝑂2
𝛥𝛥𝛥𝛥𝑏𝑏 = (0.51
•
1000 𝑔𝑔 𝐻𝐻2𝑂𝑂
1 𝑘𝑘𝑘𝑘 𝐻𝐻2𝑂𝑂11
= 0.799 𝑚𝑚 𝑜𝑜𝑜𝑜 0.80𝑚𝑚
°𝐶𝐶
) (0.80 𝑚𝑚)(1) = 0.41°𝐶𝐶
𝑚𝑚
Therefore the increase in boiling point of water is 𝛥𝛥𝛥𝛥𝑏𝑏 = 0.41°𝐶𝐶
Equation to calculate the decrease in the freezing point:
ΔTf=kf⋅m⋅i
Where:
ΔTf= the amount the freezing temperature decreases.
kf= the freezing point depression constant which depends on the solvent (water = 1.86°C/m).
m= the molality of the solution.
i= the number of particles formed when that compound dissolves
Given:
ΔTf= ?
kf= 1.86°C/m
m= 0.80 m
i= 1
𝛥𝛥𝛥𝛥𝑓𝑓 = 𝑘𝑘𝑓𝑓 • 𝑚𝑚 • 𝑖𝑖
°𝐶𝐶
𝛥𝛥𝛥𝛥𝑓𝑓 = (1.86 ) (0.80 𝑚𝑚)(1) = 1.49°𝐶𝐶
𝑚𝑚
Therefore the decrease in freezing point of water is 𝛥𝛥𝛥𝛥𝛥𝛥 = 1.49°𝐶𝐶