CBSE chemistry

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Question
CBSEENCH12010086

Give one example each of 'oil in water' and 'water in oil' emulsion.

Solution

Type of emulsion

Example

Oil in water

 Milk, vanishing cream

Water in oil

 Butter, cold cream, cod liver oil

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Question
CBSEENCH12010087

Which reducing agent is employed to get copper from the leached low-grade copper ore?

Solution

Copper can be obtained from low-grade ore through the process of leaching using acid or bacteria (leaching is a process in which ore is treated with a suitable reagent that dissolves ore but not the impurities).

The solution containing copper can be reduced with the help of reducing agents such as scrap iron or H2 to get copper metal.

Cu2+(aq) + Fe Cu(s) + Fe2+(aq)

 Cu2+ (aq) + H2(g)  Cu(s) + 2 H+(aq) 

Question
CBSEENCH12010088

Which of the following is a more stable complex and why?

 (i) [Co(NH3)6]3+

 (ii) [Co(en)3]3+

Solution

Chelating ligands form more stable complexes compared to non-chelating ligands. Since ethylene diamine is a bidentate ligand and forms stable chelate, [Co(en)3]3+ will be a more stable complex than [Co(NH3)6]3+

Question
CBSEENCH12010094

An element with density 11.2 g cm-3  forms a f.c.c. lattice with edge length of 4 x10-8

Calculate the atomic mass of the element. (Given:  NA = 6.022x 10-23 (mol-1

Solution

Density, d = 11.2 g cm-3

Edge length, a = 4x10-8 cm

Avogadro number, NA = 6.022x1023 mol-1

Number of atoms present per unit cell, Z (fcc) = 4


We know for a crystal system,

  straight d equals space fraction numerator straight z space straight x space straight m over denominator straight a cubed space straight x space straight N subscript straight A end fraction

straight m space equals fraction numerator straight d space straight x space straight a cubed space straight x space straight N subscript straight A over denominator straight Z end fraction

We space get comma

straight M space equals fraction numerator 11.2 space straight x space 64 space straight x space 10 to the power of negative 24 end exponent space straight x space 6.022 straight x space 10 to the power of 23 over denominator 4 end fraction equals space 107.91 space straight g

   

Thus, the atomic mass of the element is 107.91 g.

 

Question
CBSEENCH12010095

Examine the given defective crystal:
 

Answer the following questions:


(i) What type of stoichiometric defect is shown by the crystal?

(ii) How is the density of the crystal affected by this defect?

 (iii) What type of ionic substances shows such defect? 

Solution

(i) Schottky defect is shown by the mentioned crystal, as an equal number of cations and anions are missing in the crystal lattice.

(ii) This defect leads to decrease in density, as an equal number of the cations and anions are missing from the crystal lattice. A number of such defects in ionic solids are quite significant. For example, in NaCl, there are approximately 106 Schottky pairs per cm3 at room temperature.

(iii) This kind of defect is shown by that ionic substance in which the cations and anions are of almost similar sizes. 


Examples:  NaCl, KCl and CsCl.