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Grade: 12th pass

                        

Ostwald walker process I need explanation regarding this topic

4 months ago

Answers : (1)

ROSHAN MUJEEB
askIITians Faculty
828 Points
							Measurement of relative lowering in vapour pressure (Ostwald and Walker method):

It 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 w ith the accurately weighed set of guard tubes filled with anhydrous calcium chloride or some other dehydrating agents like P2O5, conc. H2SO4etc. 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, P0ps/p0= Loss in mass in solvent bulbs/Total loss in mass in both sets of bulbs

= Loss in mass in solvents bulbs/Gain in mass of guard tubes

Further we know from Raoult’s law

P0ps/P0= wA/mA/wA/mA + wB/mB

The above relationship is used for calculation of molecular masses of non-volatile solutes.

For very dilute solutions, the following relationship can be applied.

P0ps/P0=Loss in mass of solvent bulbs/Gain in mass of guard tubes = wAmB/wBmA
4 months ago
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