How does <PERSON>-Walker method work?
From searching online I found this:
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 $\displaystyle P_2O_5$, conc. $\displaystyle H_2SO_4$ etc. 346
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. $\displaystyle p_0 – p_s$. The two sets of bulbs are weighed again. The guard tubes are also weighed.
Loss in mass in the solution bulbs $\displaystyle \propto p_s $
Loss in mass in the solvent bulbs $\displaystyle \propto (p_0 – p_s) $
Total loss in both sets of bulbs $\displaystyle \propto [p_s +(p_0 – p_s)] \propto p_0 $
Total loss in mass of both sets of bulbs is equal to gain in mass of guard tubes.
Thus, $\displaystyle \frac{p_0-p_s}{p_0} = \frac{\text{Loss in mass in solvent bulbs}}{\text{Total loss in mass in both sets of bulbs}} = \frac{\text{Loss in mass in solvents bulbs}}{\text{Gain in mass of guard tubes}}$
Further we know from <PERSON>’s law
$\displaystyle \frac{p_0-p_s}{p_0} = \frac{\frac{w_A}{m_A}}{\frac{w_A}{m_A} + \frac{w_B}{m_B}}$
The above relationship is used for calculation of molecular masses of non-volatile solutes.
For very dilute solutions, the following relationship can be applied.
$\displaystyle \frac{p_0-p_s}{p_0} = \frac{\text{Loss in mass of solvent bulbs}}{\text{Gain in mass of guard tubes}}= \frac{w_Am_B}{w_Bm_A}$
My questions:
* Why "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"?
* How can we add $\propto p_s$ and $\propto p_0-p_s$ when we don't know what the proportionality constants are or even if they are equal?
* It would be better if you could explain in detail the process because I'm not getting it properly from here.For your information the text after "From Raoult's Law" and about "the setup" is purely clear to me. 707