Surface Chemistry
(Given log 3 = 0.4771)
Explanation:
${x \over m} \propto {P^{{1 \over n}}}$
$ \Rightarrow $ ${x \over m}$ = kP$^{{1 \over n}}$ ....(1)
taking log both sides, we get,
log${x \over m}$ = logk + ${1 \over n}$ logP
Here in graph between log${x \over m}$ and logP, slope is ${1 \over n}$ and intercepts = log k.
Given, log k = 0.4771 $ \Rightarrow $ k = 3
Slope ${1 \over n}$ = 2
put in eq. (1), ${x \over m}$ = 3 $ \times $ (4)2 = 48
To resist the coagulation of 100 cc gold sol; 1 cc of $10 \% \mathrm{NaCl}$ is added to it in the presence of $10^{-4} \mathrm{~g}$ gelatin. The gold number of gelatin is
0.00001
0.0001
0.001
0.01
Which of the following is true for spontaneous adsorption?
$\Delta G>0, \Delta S>0, \Delta H<0$
$\Delta G>0, \Delta S>0, \Delta H>0$
$\Delta G<0, \Delta S<0, \Delta H<0$
$\Delta G<0, \Delta S>0, \Delta H<0$
Which of the following statements is correct for chemisorption?
Chemisorption is reversible.
Multilayer adsorption takes place.
Adsorption increases with temperature.
The heat of adsorption is generally less than $40 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
The gold numbers of gelatin, haemoglobin and sodium acetate are $5 \times 10^{-3}, 5 \times 10^{-2}$ and $7 \times 10^{-1}$, respectively. The protective actions will be in the order
Gelatin < haemoglobin < sodium acetate
Gelatin > haemoglobin > sodium acetate
Haemoglobin $>$ gelatin $>$ sodium acetate
Sodium acetate $>$ gelatin $>$ haemoglobin
${x \over m} = k{p^{0.5}}$
Adsorption of the gas increase with :
| Column I | Column II |
|---|---|
| (A) V2O5 | (i) Polyethylene |
| (B) TiCl4/Al(Me)3 | (ii) ethanal |
| (C) PdCl2 | (iii) H2SO4 |
| (D) Iron Oxide | (iv) NH3 |
| Gas | H2 | CH4 | CO2 | SO2 |
|---|---|---|---|---|
| Critical
Temperature / K |
33 | 190 | 304 | 630 |
On the basis of data given above, predict which of the following gases shows least adsorption on a definite amount of charcoal ?
(n and k are constants and n > 1)
( ${x \over m}$ is the mass of the gas adsorbed per gram of adsorbent)
(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used.
(b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used.
(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude.
(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.
The given graphs/data I, II, III and IV represent general trends observed for different physisorption and chemisorption processes under mild conditions of temperature and pressure. Which of the following choice(s) about I, II, III and IV is(are) correct?




Among the electrolytes Na$_2$SO$_4$, CaCl$_2$, Al$_2$(SO$_4$)$_3$ and NH$_4$Cl, the most effective coagulating agent for Sb$_2$S$_3$ sol is
Among the following, the surfactant that will form micelles in aqueous solution at the lowest molar concentration at ambient condition is:
Statement 1 : Micelles are formed by surfactant molecules above the critical micellar concentration (CMC).
Statement 2 : The conductivity of a solution having surfactant molecules decreases sharply at the CMC.
Nitrogen gas is absorbed on 20% surface sites. On heating N$_2$ gas evolved from sites and was collected at 0.01 atm and 298 K in a container of volume 2.46 cm$^3$. Find out the number of surface sites occupied per molecule of N, if the density of surface sites is $6.023\times10^{14}/\mathrm{cm^3}$ and surface area is 1000 cm$^2$.
Explanation:
Given Data:
P = 0.01 atm
T = 298 K
V = 2.46 cm$^3$
Density of surface sites $=6.023\times10^{14}/\mathrm{cm^3}$
Surface area (A) = 1000 cm$^2$
Goal:
We want to find out how many surface sites are covered by each nitrogen (N$_2$) molecule.
Step 1: Find the Total Number of Surface Sites
We are given the number of sites per cm$^3$ and the total surface area. Multiply these to get the total number of sites:
Total number of surface sites = Density $\times$ Total surface area
$=6.023\times10^{14}\mathrm{~cm}^{-3}\times1000\mathrm{~cm^2}$
$=6.023\times10^{17}$ sites
Step 2: Calculate the Number of Sites Occupied by N$_2$
Nitrogen covers just 20% of the sites. So, multiply the total sites by 20% (or 0.20):
Sites occupied by N$_2$ = $\frac{20}{100}\times6.023\times10^{17}$
$=1.2046\times10^{17}$ sites
Step 3: Find Number of N$_2$ Molecules Released (Using Gas Laws)
To get the number of N$_2$ molecules, first calculate how many moles of N$_2$ gas are collected using the ideal gas law:
$n = \frac{PV}{RT}$
Where R = 0.0821 L-atm/K·mol; V = 2.46 cm$^3$ = 2.46 × 10$^{-3}$ L
Now, plug in the values:
$n = \frac{0.01\,\text{atm} \times 2.46 \times 10^{-3}\,\text{L}}{0.0821\, \text{L-atm/(K·mol)} \times 298\,\text{K}}$
$n = 1.01 \times 10^{-7}$ moles
Now, multiply the number of moles by Avogadro's number (6.023 × 10$^{23}$) to get molecules:
Total N$_2$ molecules released = $n \times N_A$
$= 1.01\times 10^{-7}\times 6.023\times10^{23} = 6.08\times10^{16}$ molecules
Step 4: Number of Surface Sites Occupied by Each N$_2$ Molecule
Now divide the number of sites occupied by the number of N$_2$ molecules:
$\text{Number of sites per N}_2 = \frac{1.2046\times10^{17}}{6.08\times10^{16}} = 1.98 \approx 2$
Final Answer
Each nitrogen molecule occupies about 2 surface sites.
Explanation:
The number of moles of acetic acid in 100 mL (before adding charcoal) = 0.05 mol
The number of moles of acetic acid in 100 mL (after adding charcoal) = 0.049 mol
The number of moles of acetic acid adsorbed on the surface of charcoal = 0.05 - 0.049 = 0.001 mol
The number of molecules of acetic acid adsorbed on the surface of charcoal = 0.001 $\times$ 6.02 $\times$ 1023 = 6.02 $\times$ 1020
Given that the surface area of charcoal = 3.01 $\times$ 102
m2
, so the area occupied by single acetic acid molecule on the
surface of charcoal is
= ${{3.01 \times {{10}^2}} \over {6.02 \times {{10}^{20}}}}$ = 5 $\times$ 10-19 m2

