EXPERIMENT TO DETERMINE LOWERING OF VAPOUR PRESSURE
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The apparatus consists of two sets of bulbs. The first set of three bulbs is filled with solution to half of their capacity and second set of another three bulbs is filled with the pure solvent. Each set is separately weighed accurately. Both sets are connected to each other and then with the accurately weighed set of guard tubes filled with anhydrous calcium chloride or some other dehydrating agents like P2O5, conc. H2SO4 etc. The bulbs of solution and pure solvent are kept in a thermostat maintained at a constant temperature.
A current of pure dry air is bubbled through. The air gets saturated with the vapours in each set of bulbs. The air takes up an amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of solution, i.e. p0–ps. The two sets of bulbs are weighed again. The guard tubes are also weighed.
Loss in mass in the solution bulbs ∝ps
Loss in mass in the solvent bulbs ∝(p0–ps)
Total loss in both sets of bulbs ∝[ps+(p0–ps)]∝p0
Total loss in mass of both sets of bulbs is equal to gain in mass of guard tubes.
Thus, p0−psp0=Loss in mass in solvent bulbsTotal loss in mass in both sets of bulbs=Loss in mass in solvents bulbsGain in mass of guard tubes
Further we know from Raoult’s law
p0−psp0=wAmAwAmA+wBmB
The above relationship is used for calculation of molecular masses of non-volatile solutes.
A current of pure dry air is bubbled through. The air gets saturated with the vapours in each set of bulbs. The air takes up an amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of solution, i.e. p0–ps. The two sets of bulbs are weighed again. The guard tubes are also weighed.
Loss in mass in the solution bulbs ∝ps
Loss in mass in the solvent bulbs ∝(p0–ps)
Total loss in both sets of bulbs ∝[ps+(p0–ps)]∝p0
Total loss in mass of both sets of bulbs is equal to gain in mass of guard tubes.
Thus, p0−psp0=Loss in mass in solvent bulbsTotal loss in mass in both sets of bulbs=Loss in mass in solvents bulbsGain in mass of guard tubes
Further we know from Raoult’s law
p0−psp0=wAmAwAmA+wBmB
The above relationship is used for calculation of molecular masses of non-volatile solutes.
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