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Published on: 22/06/2021
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Questions + Answers key
Take MCQ Chemistry Test1.
What are the advantages of Brownian movement?
2.
3.
Define buffer Index
4.
What are the general properties of f-block elements? (Lanthanides and Actinides)
(i) Electronic configuration
(ii) Oxidation state
(iii) Radii of tripositive ions.
5.
A reaction is of second order in A and first order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of A three times?
(iii) How is the rate affected when the concentration of both A and B is doubled?
6.
Why do noble gases form compounds with fluorine and oxygen only?
7.
If NaCI is doped with 10-3 mol % of SrCl2 What is the concentration of cation valencies?
8.
How is aluminum chloride prepared from aluminum?
9.
Before reduction, the ore is first converted into the oxide of metal of interest. Give reason
10.
Aqueous copper sulphate solution (blue) gives
(i) a green precipitate with aqueous potassium fluoride.
(ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.
1.
Brownian movement enables us,
(i) to calculate Avogadro number.
(ii) to confirm kinetic theory which considers the ceaseless rapid movement of molecules that increases with increase in temperature.
(iii) to understand the stability of colloids: As the particles are in continuous rapid movement they do not come close and hence not get condensed.
(iv) That is Brownian movement does not allow the particles to be acted on by force of gravity.
2.
3.
Buffer index β, as a quantitative measure of the buffer capacity. It is defined as the number of gram equivalents of acid or base added to 1litre of the buffer solution to change its pH by unity.
\(\beta =\frac { dB }{ d(pH) } \)
Here,
dB = number of gram equivalents of acid / base added to one litre of buffer solution
d(pH) = The change in the pH after the addition of acid / base.
4.
| Properties | Lanthanides | Actinides |
|---|---|---|
| Electronic configuration | [54Xe]4f1-14 5d16s2 | [Rn] 5f0,1-14 6d0,1-27s2 |
| Oxidation state | Common: +3 Uncommon: +2, +4 |
Common: +4 Uncommon: +2, +3, +5, +6 |
| Radii | M3+gradually decrease in size on moving from La to Lu Lanthanide contraction | M3+ and M4+ ions decrease in size on moving from Ac to Lr. Actinide contraction. |
5.
A reaction is second order in A and first order in B
(i) Differential rate equation,
Rate = \(\frac { -d\left[ R \right] }{ dt } =k{ \left[ A \right] }^{ 2 }\left[ B \right] \)
(ii) When the concentration of A is increased three times, (i.e) 3A, then
Rate = k[3A]2 [B]
= 9k [A]2[B] = 9 (initial rate)
This shows the rate will increase 9 times to the initial time.
(iii) When concentration of both A and B is doubled then,
Rate = k [2A]2 [2B] = 8k [A]2 [B] = 8 (initial rate)
This shows that rate will increase 8 times to the initial rate.
6.
(i) Both fluorine and oxygen have very high electron affinities and can easily cause the excitation of the electrons from 5p orbital to 5d orbital of xenon.
(ii) The unpaired electrons thus formed can take up electrons from oxygen or fluorine to form compounds.
(iii) Thus xenon has low ionisation energy and it can form compounds with strong oxidising agents (high E.A) like F2 and O2·
7.
Doping of NaCI with 10-3 mol% SrCl2 means that 100 moles of NaCl are doped with 10-3 mol SrCl2.
∴ 1mole of NaCl is doped with SrCl2
\(=\frac { { 10 }^{ -3 } }{ 100 } \times 6.02\times { 10 }^{ 23 }=6.02\times 10^{ 18 }\)
8.
(i) When aluminium metal or aluminium hydroxide is treated with hydrochloric acid, aluminium trichloride is formed.
(ii) The reaction mixture is evaporated to obtain hydrated aluminium chloride.
2Al + 6HCl ⟶ 2AICl3 + 3H2
Al(OH)3 + 3HCl ⟶ AlCl3 + 3H2O
9.
(i) In the concentrated ore, the metal exists in positive oxidation state and hence it is to be reduced to its elemental state.
(ii) From the principles of thermodynamics; that the reduction of oxide is easier when compared to reduction of other compounds of metal and hence, before reduction, the are is first converted into the oxide of metal of interest.
10.
Aqueous copper sulphate (blue) is [Cu(H2O)4] SO4.
[Cu(H2O)4]SO4 ⟶ [Cu(H2O)4]2+ + SO42-
[Cu(H2O4)]2+ is a liable complex in which H2O ligand get easily replaced by F- ions of KF and by Cl- ions of KCI.
(i) [Cu(H2O)4]2+(aq) + 4F- ⟶ \(\underset { Green \ ppt }{ { \left[ Cu{ F }_{ 4 } \right] }^{ 2- } } +4{ H }_{ 2 }O\)
(ii) [Cu(H2O)4]2+(aq) +4Cl-(aq) ⟶ \(\underset { Bright \ green\ ppt }{ { \left[ Cu{ F }_{ 4 } \right] }^{ 2- } } +4{ H }_{ 2 }O\)
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