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Chapter 27: Q1 P (page 767)

(a) What is the difference between absorption and adsorption?

(b) How is an inclusion different from an occlusion?

Short Answer

Expert verified

a) Absorption is a chemical process in which one substance dissolves in the bulk of the other substance.

For example when we spill water , we use a towel to wipe it. The towel absorbs the water.

Adsorption is a process in which particles of a substance gets attached to the surface of another substance.

For example black charcoal adsorbs impurities on its surface.

b) Impurities get trapped inside crystal lattice. The impurities are classified into inclusions and occlusions.

  • Inclusions- When the impurities occupy the lattice sites of ions of comparable sizes and charges randomly, then they are known as Inclusions.
  • Occlusions- When cluster of impurities are trapped inside a crystal structure, they are known as occlusions.

Step by step solution

01

Step 1: Define Absorption and adsorption

Absorption is the process in which a substance is dissolved in the bulk of another substance.

Adsorption is the process in which particles from a substance adhere to a surface of the adsorbent.

02

Difference between absorption and adsorption

The main difference between adsorption and absorption is that absorption takes place in the bulk of a substance while adsorption takes place only at the surface.

03

Step 3: Define Inclusion and occlusion

Inclusions- When the impurities occupy the lattice sites of ions of comparable sizes and charges randomly, then they are known as Inclusions.

Occlusions- When cluster of impurities are trapped inside a crystal structure, they are known as occlusions.

04

Step 4: Difference between inclusion and occlusion

Inclusion occupies random lattice sites substituting the ions of comparable sizes and charges whereas occlusions are cluster of impurities trapped inside a crystal structure.

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Most popular questions from this chapter

Consider a mixture of the two solids,

BaCl22H2O(FM244.26)andKCI(FM74.551), in an unknown ratio. (The notation BaCl22H2O means that a crystal is formed with two water molecules for each BaCl2) When the unknown is heated to 160°C for 1h, the water of crystallization is driven off:

BaCl22H2O(s)→160°BaCI(s)+2H2O(g)

A sample originally weighing 1.7839 g weighed 1.5623 g after heating. Calculate the weight percent Ba,K and Cl ofin the original sample.

1.475-g sample containing NH4Cl(FM53.491),K2CO3(FM 138.21), and inert ingredients was dissolved to give 0.100 L of solution. A 25.0-mL aliquot was acidified and treated with excess sodium tetraphenylborate,Na+B(C6H5)4-, to precipitateK+and

NH4+ions completely:

(C6H5)4B-+K+→(C6H5)4BK(s)FM358.33(C6H5)4B-+NH4+→(C6H5)4BNH4(s)FM337.27

The resulting precipitate amounted to 0.617 g. A fresh 50.0-mL aliquot of the original solution was made alkaline and heated to drive off all theNH3.

NH4++OH-→NH3(g)+H2O

It was then acidified and treated with sodium tetraphenylborate to give 0.554 g of precipitate. Find the weight percent ofNH4ClandK2CO3in the original solid.

Finely ground mineral (0.6324g)was dissolved in 25 mLof boiling 4M HCland diluted with 175mLH2Ocontaining two drops of methyl red indicator. The solution was heated to100oC,and50mL of warm solution containing2.0g(NH4)2C2O4 were slowly added to precipitateCaC2O4.Then6MNH3 was added until the indicator changed from red to yellow, showing that the liquid was neutral or slightly basic. After slow cooling for 1 h, the liquid was decanted and the solid transferred to a filter crucible and washed with cold10.1wt%(NH4)2C2O4 solution five times until noCl- was detected in the filtrate upon addition ofAgNO3 solution. The crucible was dried at 1 h and then at105°C in a furnace for 2 h.

Ca2++C2O42-→105°CCaC2O4+H2O(s)→500oCCaCO3(s)

FM 40.078 18.5467 g

The mass of the empty crucible was 18.2311 g and the mass of the crucible with CaCO3was 18.5467 g .

(a) Find the wt% Ca in the mineral.

(b) Why is the unknown solution heated to boiling and the precipitant solution, (NH4)2C2O4 also heated before slowly mixing the two?

(c) What is the purpose of washing the precipitate with0.1wt%(NH4)2C2O4?

(d) What is the purpose of testing the filtrate withAgNO3solution?

A 50.00-mL solution containingwas treated with excess AgNo3 to precipitate 0.2146 g of AgBr (FM 187.772). What was the molarity of NaBr in the solution?

Why is it less desirable to wash AgCl precipitate with aqueous NaNO3 than with HNO3 solution?
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