CBSE chemistry

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Question
CBSEENCH12010137

How many atoms constitute one unit cell of a face-centered cubic crystal? 

Solution

Number of atoms in one face centred cubic unit cell can be determined:-

 

(i) 8 corners atoms × 1/8 per corner atom =   1 over 8 x 8= 1 atom

(ii) 6 face-centered atoms × 1/ 2 atom per unit cell =  1 half x 6= 3 atoms

 

∴ Total number of atoms per unit cell = 4 atoms

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Question
CBSEENCH12010138

Name the method used for the refining of Nickel metal?

Solution

Nickel is refined by mond’s process .In this process, nickel is heated in the presence of carbon monoxide to form nickel tetracarbonyl, which is a volatile complex.

  Ni space plus 4 CO space rightwards arrow with 330 minus 350 straight K on top space Ni left parenthesis CO right parenthesis subscript 4
NI left parenthesis CO right parenthesis subscript 4 rightwards arrow with 450 minus 470 straight K on top space Ni space plus CO


Then, the obtained nickel tetracarbonyl is decomposed by subjecting it to a higher temperature (450-470) to obtain nickel metal.

Question
CBSEENCH12010139

What is the covalency of nitrogen in N2O5

Solution

In N2O5, the covalency of N is restricted to 4 due to sp2 hybridization of nitrogen atom involving one 2s and three 2p orbitals.

Question
CBSEENCH12010143

Arrange the following in increasing order of their basic strength in aqueous solution: 

 CH3NH2, (CH3)3N, (CH3)2NH

Solution

With the increase in the alkyl group, the +I effect will increase which will increase the ease of donation of lone pair electron. But in water one other factor is controlling the strength of basicity.

 Amine will accept a proton and from cation will be stabilised in water by salvation (by hydrogen bonding) better the salvation by hydrogen bonding higher will be the basic strength.

 (CH3)3N<CH3NH2< (CH3)2NH

Question
CBSEENCH12010145

18 g of glucose, C6H12O6 (Molar Mass = 180 g mol-1) is dissolved in 1 kg of water in a saucepan. At what temperature will this solution boil? 

 (Kb for water = 0.52 K kg mol-1, boiling point of pure water = 373.15 K)

Solution

w1 = weight of solvent (H2O) = 1 kg

w2 = weight of solute (C6H12O6) = 18 gm

M2 = Molar mass of solute (C6 H12O6) = 180 g mol-1

Kb = 0.52 K Kg mol-1

straight T subscript straight b superscript 0 space equals 373.15 straight K

increment straight T subscript straight b space equals fraction numerator straight K subscript straight b space straight x space 1000 space straight x space straight w subscript 2 over denominator straight M subscript 2 space straight x space straight w subscript 1 end fraction space equals space fraction numerator 0.52 space straight x space 1000 space straight x space 18 over denominator 180 space straight x space 1000 end fraction equals 0.052 straight K

Therefore comma

increment straight T subscript straight b equals space straight T subscript straight b minus space straight T subscript straight b superscript 0 space
0.052 space straight K space equals space straight T subscript straight b minus 373.15

straight T subscript straight b equals 373.202 space straight K