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If x is amount of adsorbate and m is amount of adsorbent,which of the following relations is not named to adsorption process?

x/m=f(t) at constant p

p=f(t) at constant (x/m)

x/m=p x T

x/m=f(p) at constant T

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Explanation

x =p x T is the incorrect relation.The correct relation is x∝P .
m m T

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The Langmuir adsorption isotherm is deduced using the assumption:

The adsorption takes place in multilayers

The adsorption sites are equivalent in their ability to adsorb the particles

The heat of adsorption varies with coverage

The adsorbed molecules interact with each other

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Explanation

(b) The main points of Langmuir's theory of adsorption are as:
(i) Adsorption takes place on the surface of the solid only till the whole of the surface is completely covered with a unimolecular layer of the adsorbed gas.
(ii) Adsorption consist of two opposing processes (a) condensation, and (b) evaporation.
(iii) The rate of condensation depend upon the uncovered surface of the adsorbent available for condensation.

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A plot of log x/m verses log p for the adsorption of a gas on a solid gives a straight line with slope equal to:-

-log k

1/n

log k

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The Brownian movement of colloidal particles is because of –

Convection currents in the fluid

Unequal bombardments by the molecules of the dispersion medium on colloidal particles

Setting of dispersed phase under gravity

Thermal gradient in the medium

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Explanation

(B). Unequal bombardments of solvent particles on colloidal particles cause zig-zag motion.

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When ${FeCl}_{3}$ solution is added to $NaOH$ a negatively charged sol is obtained. It is due to the

Presence of basic group

Preferential adsorption of ${OH}^{-}$ ions

Self dissociation

Electron capture by sol particles

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Graph between log $\frac{x}{m}$ and log P is a straight line inclined at an angle $θ$ - 45°. When pressure of 0.5 atm and log k = 0.699, the amount of solute adsorbed per g of adsorbent will be

1 g/g adsorbent

1.5 g/g adsorbent

2.5 g/g adsorbent

0.25 g/g adsorbent

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Explanation

$(C) . \frac{x}{m} = K \cdot P^{1 / n} since \log k = 0.699, hence k = 5 slope = 1 / n = \tan 45 ° = 1 thus, \frac{x}{m} = 5 \times 0.5 = 2.5 g / g adsorbent$

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The stabilization of the dispersed phase in a lyophobic sol is due to

The viscosity of the medium

The surface tension of the medium

Liking for the medium

The formation of an electrical layer between the two phases

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Which of the following reactions leads to the formation of a substance in colloidal state?

$2 {HNO}_{3} + 3 H_{2} S \overset{}{\to} 3 S + 3 H_{2} O + 2 NO$

${SnCl}_{2} + {HgCl}_{2} \overset{}{\to} {SnCl}_{4} + Hg$

$2 Mg + {CO}_{2} \overset{}{\to} 2 MgO + C$

$Cu + {CuCl}_{2} \overset{HCl}{\to} {Cu}_{2} {Cl}_{2}$

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Explanation

(A). Due to oxidation of $H_{2} S$ by ${HNO}_{3}$ colloidal particles of sulphur $(S_{8})$ are formed. In other

cases, particles of colloidal size are not formed.

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At 400K energy of activation of a reaction is decreased by 0.8 kcal in presence of catalyst. Hence rate will be

Increased by 2.71 times

Increased by 1.18 times

Decreased by 2.72 times

Increased by 6.26 times

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Explanation

$(A) . K_{1} (no catalyst) = {Ae}^{- E_{a} / RT} K_{2} (catalyst) = {Ae}^{- E_{a} - 0.8 / RT} \frac{K_{2}}{K_{1}} = e^{0.8 / RT} = e^{0.8 / 0.002 \times 400} = 2.72$

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$1 mol$ of $[AgI] I^{-}$ can be coagulated by

$1 mol of Pb {({NO}_{3})}_{2}$

$\frac{1}{2} mol of Pb {({NO}_{3})}_{2}$

$\frac{2}{3} mol of Pb {({NO}_{3})}_{2}$

$None of these$

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