In Freundlich adsorption isotherm,
$$\frac{x}{m} = k{p^{\frac{1}{n}}}$$
Where, $$x =$$ amount of adsorbent
$$m=$$ amount of adsorbate
The value of $$n$$ is always greater than 1. So, the value of $$\frac{1}{n}$$ lies between 0 and 1 in all cases.
Adsorption is the ability of a substance to concentrate or hold gases, liquids upon its surface. Solids adsorb greater amounts of substances at lower temperature. In general adsorption decreases with increasing temperature.
35.
In an experiment, addition of $$4.0\,ml$$ of $$0.005\,M$$ $$BaC{l_2}$$ to $$16.0\,ml$$ of arsenius sulphide sol just causes the complete coagulation in 2 hrs. The
flocculating value of the effective ion is
$$A{s_2}{S_3}$$ sol is negatively charged owing to preferential adsorption of $${S^{2 - }}\,ions.$$ Cation would be the effective ion in coagulation. Flocculating value = minimum milli $$mol$$ of the effective ion per litre of sol $$ = \frac{{4 \times 0.005 \times {{10}^3}}}{{4 + 16}} = 1.0$$
36.
The values of colligative properties of colloidal solution are of small order in comparison to those shown by true solutions of same concentration because of colloidal particles
39.
Under the influence of an electric field, the particles in a sol migrate towards cathode. The
coagulation of the same sol is studied using $$NaCl,N{a_2}S{O_4}$$ and $$N{a_3}P{O_4}$$ solutions. Their coagulating values will be in maximum for :
The sol particles migrate towards cathode. So they are positively charged. Hence, anions would be effective in coagulation. Greater is the valence of effective ion, smaller will be its coagulating value.
40.
Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, $$x$$ is the mass of the gas adsorbed on mass m of the adsorbent at pressure $$p.$$ $$\frac{x}{m}$$ is proportional to :
In Freundlich adsorption isotherm the extent of adsorption $$\left( {\frac{x}{m}} \right)$$ of a gas on the surface of a solid is related to the pressure of the gas $$(P)$$ which can be formulated as :
$$\eqalign{
& \frac{x}{m} = k{\left( p \right)^{\frac{1}{n}}} \cr
& \Rightarrow \,\log \,\frac{x}{m} = \log k + \frac{1}{n}\,\log p \cr} $$
In the given plot, the slope between $$\log \,\frac{x}{m}$$ versus
$$\eqalign{
& \log \,p = \frac{2}{4} = \frac{1}{2} \cr
& \therefore \,\,\frac{x}{m} \propto {p^{\frac{1}{2}}} \cr} $$