Chemistry Test 211

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Chemistry Test 211

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Weekly Quiz Competition

Question 1
4 / -1

Referring to the following reactions the missing products A, B, C, and D respectively are

A
N₂, N₂, NO, NO₂

B
N₂, NH₃, N₂, NO

C
N₂, NH₃, N₂, NO₂

D
N₂, NH₃, NO₂, N₂O₄

Solution

(i) In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.

NH₄Cl(aq) + NaNO₂(aq) → N₂(g) + 2H2O(l) + NaCl(aq)

Small amounts of NO and HNO₃ are also formed as impurities, which can be removed by passing the gas through aqueous sulphuric acid containing potassium dichromate.

(ii) Dinitrogen can also be prepared by heating ammonium dichromate

(iii) Copper metal reacts with dilute HNO3HNO3 to form nitrogen (II) oxide (NO)

3Cu(s)+8HNO₃ (dilute) → 3Cu(NO₃)₂(aq) + 2NO(g) + 4H₂O(l)

(iv) Copper metal reacts with conc. HNO₃ to form nitrogen (IV) oxide or nitrogen dioxide (NO₂).

Cu(s) + 4HNO₃(conc.) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)
Question 2
4 / -1

Combustion of Hydrogen in a fuel cell at 300 K is represented as 2H₂(g) + O₂(g) → 2H₂O(g). If ∆H and ∆G are -241.60 kJ mol^-1 and -228.40 kJ mol^-1 respectively for H₂O then the value of ∆S for the above process is:

A
44 J K⁻¹

B
−88 J K⁻¹

C
+88 J K⁻¹

D
−44 J K⁻¹

Solution

For 1 mole of H₂O = −44 J K⁻¹

Hence, for 2 moles = −88 J K⁻¹
Question 3
4 / -1

At 25°C, pH range of phenolphthalein is.8-10. At 100°C, pH range of phenolphthalein would be

A
pH range remain unaffected by the temperature

B
pH range is altered to 7 - 9

C
pH range is altered to 7 - 11

D
pH range is altered to 8 - 11

Solution

The pH range of the indicator is pK_In ± 1. On increase in temperature pK_In decreases therefore pH range of indicator decreases.
Question 4
4 / -1

A: Hybridization of carbon is sp² in all its crystalline allotropes.

R: There are alternate double-single bonds in each allotrope of carbon.

A
Both Assertion & Reason are true and the reason is the correct explanation of the assertion.

B
Both Assertion & Reason are true but the reason is not the correct explanation of the assertion.

C
Assertion is true statement but Reason is false.

D
Both Assertion and Reason are false statements.

Solution

(i) In diamond, all C-atom are sp³ hybridized.

(ii) In graphite hybridization is sp²
Question 5
4 / -1

The standard reduction potential for Cu²⁺/Cu is +0.34 V. Calculate the reduction potential at pH = 14 for the above couple.

A
-0.22 V

B
+0.22 V

C
-0.44 V

D
+0.44 V

Solution
Question 6
4 / -1

Which of the following will have least hindered rotation about carbon-carbon bond?

A
Acetylene

B
Hexachloroethane

C
Ethane

D
Ethylene

Solution

Rotation around the C−C bond in ethylene C₂H₄ and acetylene C₂H₂ is not possible because of the presence of double and triple bond. Among ethane and hexachloroethane, ethane has the smallest sized group (H) bonded to carbons, hence there will be least hindered rotation about C−C bond.
Question 7
4 / -1

The N which contribute least to the basicity of the compound is :

A
N−9

B
N−3

C
N−1

D
N−7

Solution

N-9 lone pair is delocalized and participated in resonance, so it is not basic in nature.
Question 8
4 / -1

In the hydroboration - oxidation reaction of propene with diborane, H₂O₂ and NaOH, the organic compound formed is:

A
CH₃CH₂CH₂OH

B
CH₃CH₂OH

C
CH₃CHOHCH₃

D
(CH₃)₃COH

Solution

According to the question, The hydroboration oxidation reaction is an organic chemical reaction which is employed for the conversion of alkenes into alcohols that are neutral. This is done via a two-step process which includes a hydroboration step and an oxidation step.

Hydroboration oxidation reaction is a two-step reaction in which alkene reacts with BH₃, THF (tetrahydrofuran) and hydrogen peroxide in a basic medium to give alcohol. This is done by a net addition (across the entire double bond) of water.

Hydroboration oxidation reaction mechanism can be considered as an anti-markovnikov reaction where a hydroxyl group attaches itself to the carbon which is less substituted.

So, according to this reaction, propene give n-propyl-alcohol.
Question 9
4 / -1

The method of zone refining of metals is based on the principle of

A
greater mobility of the pure metal than that of impurity.

B
greater solubility of the impurity in the molten state than in the solid.

C
higher melting point of the impurity than that of the pure metal.

D
All above the correct

Solution

A movable heater is fitted around a rod of impure metal. As heater is moved from one end to another pure metal crystallises while impurities pass on to adjacent melted zone.
Question 10
4 / -1

The correct IUPAC name of the compound is

A
4−Bromo−5−chloro−3−iodohexane

B
3− Bromo−2−chloro−4−iodohexane

C
3− Bromo−4−iodo −2−chlorohexane

D
2− Bromo−3−bromo−4−iodohexane

Solution

The decreasing order of priority of prefix in numbering the carbon chain of an organic compound is

Bromo > Chloro > Iodo