Chemistry Test

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Chemistry Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

CsBr has B.C.C. structure with edge length \(4.3\) Å.The shortest inter ionic distance in between \(\mathrm{Cs^+}\)and \(\mathrm{Br^-}\)is:

A
\(4.3\) Å

B
\(7.44\) Å

C
\(1.86\) Å

D
\(3.72\) Å

Solution

Given,

CsBr has B.C.C. structure with edge length (\(a\)) = \(4.3\) Å

As we know,

Shortest inter ionic distance in between \(\mathrm{Cs^+}\) and \(\mathrm{Br^-}\) in B.C.C. lattice \(=\frac{1}{2}\) of body diagonal

\(=\frac{1}{2} \times \sqrt{3} {a}\)

\(=\frac{\sqrt{3}}{2} \times 4.3\)

\(=3.72\) Å
Question 2
1 / -0

The value of \(n\) in the molecular formula \(\mathrm{Be}_{\mathrm{n}} \mathrm{Al}_{2} \mathrm{Si}_{6} \mathrm{O}_{18}\) is:

A
\(2\)

B
\(4\)

C
\(6\)

D
\(3\)

Solution

We have to find the value of \(n\) in \(\mathrm{Be}_{\mathrm{n}} \mathrm{Al}_{2} \mathrm{Si}_{6} \mathrm{O}_{18}\).
\(\mathrm{n}=?\)
Charge on \(\mathrm{Al}=+3\)
Charge on \(\mathrm{Si}=+4\)
Charge on \(\mathrm{O}=-2\)
Charge on \(\mathrm{Be}=+2\)
\(\Rightarrow\mathrm{n} \times \mathrm{2}+\mathrm{2}(+\mathrm{3})+\mathrm{6}(+\mathrm{4})+\mathrm{1 8}(-\mathrm{2})=\mathrm{0}\)
\(\Rightarrow n \times 2+6+24-36=0\)
\(\Rightarrow \mathrm{n} \times 2-6=0\)
\(\Rightarrow \mathrm{n}=\mathrm{3}\)
Question 3
1 / -0

Amine is not formed in the reaction:

I. Hydrolysis of RCN

II. Reduction of \(R C H=N O H\)

III. Hydrolysis of RNC

IV. Hydrolysis of \(R C O N H_{2}\)

The correct answer is:

A
I, II and IV

B
I and IV

C
II and III

D
I, II and III

Solution

In I and IV, amine is not formed.

I. Hydrolysis of\(R C N ; R N C \rightarrow R C O O H+N H_{3}\)

II. Reduction of \(R C H=N O H\)

\(RCH=NOH \rightarrow RCH_{2} NH_{2}+H_{2} O\)

III. Hydrolysis of RNC

\(R N C+2 H_{2} O \stackrel{\Delta}{\rightarrow} R N H_{2}+H C O O H\)

IV. Hydrolysis of \(RCONH_{2}\)

\(RCONH_{2} \rightarrow {RCOOH}+{NH}_{3}\)
Question 4
1 / -0

Chromatography is based on:

A
Physical adsorption

B
Chemical adsorption

C
Hydrogen bond

D
Adsorption

Solution

Chromatography is based on physical adsorption.

Chromatography is based on the principle of distributing the components of a mixture of organic compounds between two phases. Now since the process does not involve formation of chemical covalent bond but involves the formation of weak wander Waals fond. Thus, chromatography is physical adsorbtion.
Question 5
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Boiling point of water at \(750 \mathrm{~mm} \mathrm{Hg}\) is \(99.63^{\circ} \mathrm{C}\). How much sucrose is to be added to \(500 \mathrm{~g}\) of water such that it boils at \(100^{\circ} \mathrm{C} .\) Molal elevation constant for water is \(0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\).

A

\(131.67 \mathrm{~g}\)

B

\(141.67 \mathrm{~g}\)

C

\(151.67 \mathrm{~g}\)

D

\(121.67 \mathrm{~g}\)

Solution

Here, elevation of boiling point \(\Delta \mathrm{T}_{\mathrm{b}}=(100+273)-(99.63+273)\)
\(=0.37 \mathrm{~K}\)
Mass of water, \(w_{1}=500 \mathrm{~g}\)
Molar mass of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\)
\(\mathrm{M}_{2}=11 \times 12+22 \times 1+11 \times 16=342 \mathrm{~g} \mathrm{~mol}^{-1}\)
Molal elevation constant, \(\mathrm{K}_{\mathrm{b}}=0.52 ~\mathrm{k} ~\mathrm{kg}~ \mathrm{mol}^{-1}\)
We know that:
\(\Delta T_{b}=\frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}\)
\(\Rightarrow w_{2}=\frac{\Delta T_{b} \times M_{2} \times w_{1}}{K_{b} \times 1000}\)
\(=\frac{(0.37 \times 342 \times 500)}{(0.52 \times 1000)}\)
\(=121.67 \mathrm{~g}\) (approximately)
so, \(121.67 \mathrm{~g}\) of sucrose is to be added.
Question 6
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The oxidation state of sulphur in \(N a_{2} S_{4} O_{6}\) is:

A
\(+6\)

B
\(+\frac{3}{2}\)

C
\(+\frac{5}{2}\)

D
\(-2\)

Solution

Oxidation number of \(N a=+1\)

Oxidation number of \(O=-2\)

Let oxidation number of \(S=x\)

\(\therefore 2(O N\) of \(N a)+4(O N\) of \(S)+6(O N\) of \(O)=0\)

\(2(+1)+4 x+6(-2)=0\)

\(+2+4 x-12=0\)

\(4 x=+12-2\)

\(x=+\frac{10}{4}\)

\(\Rightarrow x=+\frac{5}{2}\)
Question 7
1 / -0

The value of n in the molecular formula Be_nAl₂Si₆O₁₈ is:

A
3

B
5

C
7

D
9

Solution

Given,

Molecular formula =Be_nAl₂Si₆O₁₈

The oxidation states of each element of given molecular formula:

Be = +2

Al = +3

Si = +4

O = -2

(2n) + (3 × 2) + (4 + 6) + (−2 × 18) = 0

⇒2n + 30 − 36 = 0

⇒2n = 6

⇒n = 3
Question 8
1 / -0

The materials used for making non-stick utensils of kitchens is:

A
PVC

B
Polyether

C
Teflon

D
Rayon

Solution

The materials used for making non-stick utensils of kitchens is Teflon.
Question 9
1 / -0

Which of the following antibiotic contains a nitro group attached to the aromatic nucleus in its structure?

A
Tetracyclin

B
Penicillin

C
Streptomycin

D
Chloramphenicol

Solution

Among the given antibiotics, only chloramphenicol i.e., option D contains a nitro group attached to aromatic ring. Its structure is as follows:
Question 10
1 / -0

The atom bomb is based on the principle of:

A
Nuclear fusion

B
Nuclear fission

C
Radioactivity

D
None of these

Solution

The atom bomb is based on the principle ofnuclear fission.

The nucleus of a heavy atom (such as uranium, plutonium, or thorium), when bombarded with low-energy neutrons, can be split apart into lighter nuclei. This process is called nuclear fission.Fission reactions can further be classified into controlled and uncontrolled fission reactions.In controlled fission, the chain reaction is controlled and only a controlled amount of reaction is allowed, nuclear reactors in nuclear power plants are one example of the controlled fission reaction.And for an uncontrolled fission chain reaction, is allowed to happen unless fission material is over, atomic bomb is one example of an uncontrolled fission reaction.

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Chemistry Test - 9

Result

Chemistry Test - 9

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Question 1

\(4.3\) Å

\(7.44\) Å

\(1.86\) Å

\(3.72\) Å

Solution

Question 2

\(2\)

\(4\)

\(6\)

\(3\)

Solution

Question 3

I, II and IV

I and IV

II and III

I, II and III

Solution

Question 4

Physical adsorption

Chemical adsorption

Hydrogen bond

Adsorption

Solution

Question 5

\(131.67 \mathrm{~g}\)

\(141.67 \mathrm{~g}\)

\(151.67 \mathrm{~g}\)

\(121.67 \mathrm{~g}\)

Solution

Question 6

\(+6\)

\(+\frac{3}{2}\)

\(+\frac{5}{2}\)

\(-2\)

Solution

Question 7

3

5

7

9

Solution

Question 8

PVC

Polyether

Teflon

Rayon

Solution

Question 9

Tetracyclin

Penicillin

Streptomycin

Chloramphenicol

Solution

Question 10

Nuclear fusion

Nuclear fission

Radioactivity

None of these

Solution

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