CBSE chemistry

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Question
CBSEENCH12010183

What is meant by ‘doping’ in a semiconductor?

Solution

Doping is the process of increasing the conductivities of the intrinsic semiconductors by adding suitable impurity.

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Question
CBSEENCH12010184

What is the role of graphite in the electrometallurgy of aluminium?

Solution

In the metallurgy of aluminium, steel cathode and graphite anode are used. The graphite anode is useful for the reduction of Al2O3 into Al.

2Al2O+ 3C  --->  4Al + 3CO2

                          (graphite)

Question
CBSEENCH12010185

Which one of  PCl subscript 4 superscript plus space and space PCl subscript 4 superscript minus is not likely to exist and why?

Solution

The oxidation state of P in   PCl subscript 4 superscript plusis +5 while in  PCl subscript 4 superscript minus is +3. As we know that the stability of the +5 oxidation state is highest on top of the group and decrease down the group and stability of +3 is low on the top of the group and increase down the group.

Therefore  PCl subscript 4 superscript minus  is more likely to exist.

Question
CBSEENCH12010189

Arrange the following in the decreasing order of their basic strength in aqueous solutions:

CH3NH2, (CH3)2NH, (CH3)3 N and NH3

Solution

With the increase in 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)2 NH > CH3 NH2 > (CH3)3 N > NH3

Question
CBSEENCH12010191

A 1.00 molar aqueous solution of trichloroacetic acid (CCl3COOH) is heated to its boiling point. The solution has the boiling point of 100.180C. Determine the van’t Hoff factor for trichloroacetic acid. (Kb for water = 0.512 kg mol-1)

Solution

Molality of solution = m = 1.00 m

 Boiling points of solution = Tb = 100.180C = 373.18 K

 Boiling point of water (solvent) = straight T subscript straight b superscript 0 = 100.00° C = 373 K
Elevation in boiling point = straight T subscript straight b superscript 0  - Tb

 Observed boiling point = 373.18 K - 373 K = 0.18 K

 Kb water = 0.512 K kg mol- 1

 ∴  incrementb= Kb x m

 = 0.512 x 1 = 0.512 K

 ∴ Calculated boiling point = 0.512 K
Van apostrophe straight t space Hoff space Factor space left parenthesis straight i right parenthesis space equals space fraction numerator Observed space Colligative space Property over denominator Calculated space colligative space Property end fraction
space space
space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 0.18 straight K over denominator 0.512 space straight K end fraction
space space straight i space equals space 0.35