Chemistry Test

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Chemistry Test - 27

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Weekly Quiz Competition

Question 1
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In the reaction \(4 \mathrm{Fe}+3 \mathrm{O}_{2} \rightarrow 4 \mathrm{Fe}^{3+}+6 O^{2-}\) which of the following statements is incorrect?

A
It is a redox reaction

B
Metallic iron is a reducing agent

C
\(\mathrm{Fe}^{3+}\) is an oxidising agent

D
Metallic iron is reduced to \(F e^{3+}\)

Solution

Given reaction,

\(4 \mathrm{Fe}+3 \mathrm{O}_{2} \longrightarrow 4 \mathrm{Fe}^{+3}+6 \mathrm{O}^{2-}\)

Clearly, given reaction is a redox reaction in which, Fe gets oxidised to \(\mathrm{Fe}^{+3}\) by losing of three electrons, and \(O_{2}\), gets reduced to \(O^{2-}\).

Thus, Metalliciron \((F e)\) is a reducing agent and \(\mathrm{Fe}^{+3}\) is an oxidising agent.
Question 2
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Calculate e.m.f of the following cell at \(298 \mathrm{~K}\) :

\(2 \mathrm{Cr}(\mathrm{s})+3 \mathrm{Fe}^{2+}(0.1 \mathrm{M}) \rightarrow 2 \mathrm{Cr}^{3+}(0.01 \mathrm{M})+3 \mathrm{Fe}(\mathrm{s})\)

Given: \(\mathrm{E}^{\circ}\left(\mathrm{Cr}^{3+} \mid \mathrm{Cr}\right)=-0.74 \mathrm{VE}^{\circ}\left(\mathrm{Fe}^{2+} \mid \mathrm{Fe}\right)=-0.44 \mathrm{~V}\)

A
\(.31 V\)

B
\(.27 V\)

C
\(.34 V\)

D
\(.43 V\)

Solution

The balanced chemical reaction can be written as follows:

\(2 {Cr}({s})+3 {Fe}^{2}+(0.1 {M}) \rightarrow 2 {Cr}^{3}+(0.01 {M})+3 {Fe}({s})\)

Given.

\(E_{\frac{{Cr}^{3+}} {{Cr}}^{o}}^{o}=-0.74 V\)

\(E_{\frac{{Fe}^{2+}} {{Fe}}^{o}}=-0.44 {~V}\)

The cell can be represented as follows:

\(C r\left|C r^{3+}(0.01 M)\right|\left|F e^{2+}(0.1 M)\right| F e(s)\)

\(E_{\text {Cell }}^{o}\) can be calculated as follows:

\(E_{{Cell}}^{o}=E_{\frac{{Fe}^{2+}} {{Fe}}^{o}}-E_{\frac{{Cr}^{3+}} {{Cr}}^{o}}\)

\(E_{{Cell}}^{o}=-0.44-(-0.74)=0.30 {~V}\)

\(E_{{Cell}}=E_{{Cell}}^{o}-\frac{2.303 R T}{n F} \log \left(\frac{\left[C r^{3}\right]^{2}}{\left[F e^{2+}\right]^{3}}\right)\)

Putting all the values of the constants, we get,

\(E_{\text {Cell }}=0.30-\frac{0.0591}{6} \log \left(\frac{(0.01)^{2}}{(0.1)^{2}}\right)\)

\({E}_{{Cell}}=0.30+0.01\)

\({E}_{{Cell}}=0.31 {~V}\)
Question 3
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The helical structure of protein is stabilized by:

A
Hydrogen bonds

B
Disulphide bonds

C
Peptide bonds

D
None of these

Solution

The helical structure of protein is stablised by hydrogen bonds between amide group of the same peptide chain. These bond are formed by -NH- group of one unit and oxygen of carbonyl group of the other unit.

Question 4

1 / -0

Match the following :

Test/Method	Reagent
(i) Lucas Test	(a) \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_2 \mathrm{Cl} /\) aq. \(\mathrm{KOH}\)
(ii) Dumas method	(b) \(\mathrm{HNO}_3 / \mathrm{AgNO}_3\)
(iii) Kjeldahl’s method	(c) \(\mathrm{CuO} / \mathrm{CO}_2\)
(iv) Hinsberg test	(d) \(\mathrm{Conc} . \mathrm{HCl}\) and \(\mathrm{ZnCl}_2\)(I
	(e) \(\mathrm{H}_2 \mathrm{SO}_4\)

(i)-(b), (ii)-(d), (iii)-(e), (iv)-(a)

(i)-(d), (ii)-(c), (iii)-(e), (iv)-(a)

(i)-(b), (ii)-(a), (iii)-(c), (iv)-(d)

(i)-(d), (ii)-(c), (iii)-(b), (iv)-(e)

Solution

Lucas reagent – Conc. HCl/ZnCl2

Dumas method – CuO/CO2

Kjeldahl’s method – H2SO4

Hinsberg test – C6H5SO2Cl/aq. KOH

Question 5
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L-isomer of tetrose \(\mathrm{X}\left(\mathrm{C}_4 \mathrm{H}_8 \mathrm{O}_4\right)\) gives positive schiff's test and has two chiral carbons. On acetylation. ' \(\mathrm{X}\) ' yields triacetate. ' \(\mathrm{X}\) ' undergoes following reactions

' \(\mathrm{X}\) ' is

A

B

C

D

Solution
Question 6
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\(1.22 \mathrm{~g}\) of an organic acid is separately dissolved in \(100 \mathrm{~g}\) of benzene \(\left(\mathrm{K}_{\mathrm{b}}=2.6 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\) and \(100 \mathrm{~g}\) of acetone \(\left(\mathrm{K}_{\mathrm{b}}=1.7 \mathrm{Kkg} \mathrm{mol}^{-1}\right)\). The acid is known to dimerise in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by \(0.17^{\circ} \mathrm{C}\). The increase in boiling point of solution in benzene in \({ }^{\circ} \mathrm{C}\) is \(\mathrm{x} \times 10^{-2}\). The value of \(\mathrm{x}\) is....... (Nearest integer)

[Atomic mass: \(\mathrm{C}=12.0, \mathrm{H}=1.0, \mathrm{O}=16.0\) ]

A
13

B
15

C
-15

D
-13

Solution

With benzene as solvent, RCOOH dimerises.

\(2 \mathrm{RCOOH} \rightleftharpoons(\mathrm{RCOOH})_2\)

\(\therefore \mathrm{i}=\frac{1}{2}\)

\(\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \times \mathrm{K}_{\mathrm{b}} \times \mathrm{m}\)

where, \(\Delta \mathrm{T}_{\mathrm{b}}=\) boiling point elevation,

\(\mathrm{K}_{\mathrm{b}}=\) ebulliscopic constant

\(\mathrm{m}=\) molality

i = van't Hoff factor

\(\Delta \mathrm{T}_{\mathrm{b}}=\frac{1}{2} \times 2.6 \times \frac{1.22 / \mathrm{M}_{\mathrm{W}}}{100 / 1000} \ldots\) (i)

With acetone as solvent, no dimerisation.

\(\therefore \mathrm{I}=1 \)

\(\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \times \mathrm{K}_{\mathrm{b}} \times \mathrm{m} \)

\(0.17=1 \times 1.7 \times \frac{1.22 / \mathrm{M}_{\mathrm{W}}}{100 / 1000} . ..(\text{ii})\)

Eq. (i) divide by Eq. (ii),

\(\frac{\Delta \mathrm{T}_{\mathrm{b}}}{0.17}=\frac{\frac{1}{2} \times 2.6 \times \frac{1.22 / \mathrm{M}_{\mathrm{w}}}{100 / 1000}}{1 \times 1.7 \times \frac{1.22 / \mathrm{M}_{\mathrm{w}}}{100 / 1000}} \)

\( \Rightarrow \Delta \mathrm{T}_{\mathrm{b}}=\frac{0.26}{2}=13 \times 10^{-2} \)

\(\mathrm{x}=13\)
Question 7
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For a 1^st order reaction, if the concentration is doubled then the rate of reaction becomes:

A
Double

B
Half

C
Four times

D
Remain same

Solution

For a first-order reaction,

rate=K[A]

When the concentration of A is doubled,

New rate= K[2A]

= 2 × rate

The rate of reaction becomes double.
Question 8
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When \(\mathrm{KMnO}_{4}\) solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because:

A
\(\mathrm{CO}_{2}\) is formed as the product

B
Reaction is exothermic

C
\(\mathrm{MnO}_{4}^{-}\)catalyses the reaction

D
\(\mathrm{M n}^{2+}\) acts as autocatalyst
Solution
In titration of \(\mathrm{KMnO}_{4}\) and oxalic acid, \(\mathrm{KMnO}_{4}\) is an oxidizing agent and oxalic acid is a reducing agent.
- To initiate the reaction, a little amount of acid is required. So, til. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is used for this purpose.
- But the reaction is slower at acidic \(\mathrm{pH}\).
- As soon as reaction proceeds more and more \(\mathrm{Mn}^{2+}\) is
- \(\mathrm{Mn}^{2+}\) acts as an auto catalyst in this reaction. So, as the concentration of \(\mathrm{Mn}^{2+}\) increases, rate of reaction becomes faster with time.
The redox reaction can be written as:

\(5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+2 \mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+} \longrightarrow 10 \mathrm{CO}_{2}+2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}\)
Question 9
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Which of the following statement is correct?

A
Copper (II) metaborate is bluish green

B
Copper (II) metaborate is colourless

C
Copper (II) metaborate is light green

D
Copper (I) metaborate is dark green

Solution

Copper (II) metaborate is bluish green and colour of Copper (I) metaborate is blue in colour.

Blue cupric metaborate is reduced to colourless cuprous metaborate in a luminous flame

Cupric metaborate is obtained by heating boric anhydride and copper sulphate in a non luminous flame.
Question 10
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For the process to occur under adiabatic conditions, the correct conditionis:

A
\(\mathrm{\Delta T=0}\)

B
\(\mathrm{\Delta p=0}\)

C
\(\mathrm{q=0}\)

D
\(\mathrm{w=0}\)

Solution

For the process to occur under adiabatic conditions, the correct condition is\(\mathrm{q=0}\).

When a thermodynamic system undergoes a change in such a way that no exchange of heat takes place between Systems and surroundings, the process is known as an adiabatic process. During the adiabatic process, there is no exchange of heat that takes place between the system and the surroundings.