Chemistry Test

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Chemistry Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is a set of essential amino acids?

A
Glycine, Alanine, Tyrosine, Glutamic acid

B
Glycine, Alanine, Threonine, Lysine

C
Tyrosine, Glutamic acid, Valine, Leucine

D
Leucine, Threonine, Alanine, Tyrosine

Solution

Humans can produce 10 of the 20 amino acids.The 10 amino acids that we can produce are alanine, asparagine, aspartic acid, cysteine, glutamic acid, glutamine, glycine, proline, serine, and tyrosine. Tyrosine is produced from phenylalanine, so if the diet is deficient in phenylalanine, tyrosine will be required as well.
Question 2
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For the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\), if percentage of \(\mathrm{N}_{2} \mathrm{O}_{4}\) are \(25 \%, 50 \%, 75 \%\) and \(100 \%\), then the sequence of observed vapour densities will be

A
\(d_{1}>d_{2}>d_{3}>d_{4}\)

B
\(d_{4}>d_{3}>d_{2}>d_{1}\)

C
\(d_{1}=d_{2}=d_{3}=d_{4}\)

D
\(d_{1}>d_{3}>d_{2}>d_{4}\)

Solution

When dissociation increases, the percentage of nitrogen dioxide, will also increase. While vapour density is decreased, it means vapour density is inversely proportional to the dissociation.

Therefore,

The dissociation vapour density will be high in following order:

\(d_{25\%}>d_{50\%}>d_{75\%}>d_{100\%}\)

Therefore, the required sequence will be:

\(d_{1}>d_{2}>d_{3}>d_{4}\)
Question 3
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Ethylidene chloride is a/an ___________.

A
Vicinal dihalide

B
Gem-dihalide

C
Allylic halide

D
Vinylic halide
Solution
Ethylidene chloride is a gem-dihalide.
- Ethylidene chloride is also known as 1,1 Dichloroethane.
- It has 2 chlorine atoms on the same carbon atom, thus, has 1,1 relationship.
- Therefore, it is geminal-dihalide.
- Gem-dihalide is a compound that have two halogen atoms on the same carbon atom.
Below given the structure of Ethylidene chloride:
Question 4
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Which of the following conditions is not correct for ideal solution

A
No change in volume on mixing

B
No change in enthalpy on mixing

C
It obeys Raoults law

D
Ionisation of solute should occurs to a small extent

Solution

In an ideal solution, no change in volume on mixing, no change in enthalpy on mixing and it obeys Raoults law but ionisation of solute should not occur to a small extent.
Question 5
1 / -0

If \(20 mL\) of ethanol (density \(=0.7893 g / mL\) ) is mixed with \(40 mL\) water (density= \(0.9971 g / mL\) ) at \(25^{\circ} C\), the final solution has density of \(0.9571 g / mL\). Calculate the molality of alcohol in the final solution. (as nearest integer)

A
\(7.9\)

B
\(8.6\)

C
\(9.4\)

D
\(7.2\)

Solution

\(C _{2} H _{5} OH \rightarrow V _{1}=20 mL , d _{1}=0.7893 g / mL\)

\(m _{1}=15.786 g\)

\(H _{2} O \rightarrow V _{2}=40 mL , d _{2}=0.9971 g / mL\)

\(m _{2}=39.884 g\)

\(T\) otal mass \(=55.65 g\)

\(d _{\text {sol. }}=0.9571 g / mL\)

\(V _{\text {sol. }}=58.14 mL\)

\(\%\) change \(=\frac{60-58.14}{60} \times 100=3.1 \%\)

\(m =\frac{15.766 \times 1000}{46 \times 39.884}=8.6\)
Question 6
1 / -0

The order of basic strength in gas phase of the following compound would be

A
iv) < i) < iii) < ii)

B
i) < iv) < iii) < ii)

C
iv) < i) < ii) < iii)

D
ii) < iii) < i) < iv)

Solution

In decreasing order of basic strength in the gas phase,

In the gas phase, there is no hydrogen bonding. Therefore stabilization due to hydrogen bonding is not there. Thus, the only effect to determine the strength is the inductive effect. The +I effect increases with an increase in the alkyl group. Therefore the basic strength will be the highest in (C₂H₅)₃N and least in NH₃.

Therefore the increasing order of basic strength in the gas phase will be. iv) < i) < ii) < iii)
Question 7
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\(C u^{+}\)ion is not stable in aqueous solution because:

A
Second ionizationenthalpy of copper is less than the first ionizationenthalpy.

B
Large value of second ionization enthalpy of copper is compensated by much more negative hydration energy of \(C u_{(a q)}^{2+}\).

C
Hydration energy of \(C u_{(a q)}^{2+}\) is much more negative than that of \(C u_{(a q)}^{2}\).

D
Many copper (I) compounds are unstable in aqueous solution and undergo a disproportionation reaction..
Solution
In an aqueous medium, \(\mathrm{Cu}^{2+}\) is more stable than \(\mathrm{Cu}^{+}\).
- This is because although energy is required to remove one electron from \(\mathrm{Cu}^{+}\)to \(\mathrm{Cu}^{2+}\), high hydration energy of \(\mathrm{Cu}^{2+}\) compensates for it.
- Therefore, \(\mathrm{Cu}^{+}\) ion in an aqueous solution is unstable.
- It disproportionates to give \(\mathrm{Cu}^{2+}\) and \(\mathrm{Cu}\).
Question 8
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A mixture is known to contain \(N O_{ {3}}^-\) and \(N O_{ {2}}^-\). Before performing ring test for \(N O_{ {3}}^-\) the aqueous solution should be made free of \(N O_{ {2}}^-\). This is done by heating aqueous extract with:

A
Conc. \(\mathrm{HNO}_{3}\)

B
Dil. \(\mathrm{HNO}_{3}\)

C
Urea

D
Zinc Dust

Solution

In the mixture of nitrite and nitrate ions, the nitrite ions can be removed by heating with urea and supluric acid.

The reaction takes place as follows:

\(2 \mathrm{NaNO}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{HNO}_{2}\)

\(2 \mathrm{HNO}_{2}+\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{N}_{2}+3 \mathrm{H}_{2} \mathrm{O}\)

In the above reaction, we can see that the products, nitrogen and carbon dioxide are in gaseous form and thus, do not stay in the solution. This reaction is exothermic and thus, will occur spontaneously without the supply of any extra energy.
Question 9
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Identify the species in which the metal atom is in +6 oxidation state.

A
\(M n O_{ {4}}^-\)

B
\([C r(C N) 6]^{3-}\)

C
\(N i F_{6}^{2-}\)

D
\({CrO}_{2} {Cl}_{2}\)

Solution

(A) Let the oxidation number of \(\mathrm{Mn}\) be \(\mathrm{x}\).

We know that:

Oxidation number of \( \mathrm{O} =-2\)

\(\mathrm{x}+4(-2)=-1\)

or, \(\mathrm{x}-8=-1\)

or, \(\mathrm{x}=-1+8\)

or, \(\mathrm{x} =+7\)

The oxidation number of \(\mathrm{Mn}\) in \(\left[\mathrm{MnO}_{4}\right]^{-}\)ion is \(+7\).

(B)Let the oxidation number of \(\mathrm{Cr}\) be \(\mathrm{x}\).

In \([\mathrm{Cr}(\mathrm{CN})_{6}]^{3-}\), the charge is \(-3\).

We know that:

Oxidation number of \(\mathrm{CN}=-1\)

Therefore,

\(\mathrm{x}+6\left(\mathrm{C} \mathrm{N} \right)=-3\)

\(\mathrm{x}+6(-1)=-3\)

\(\mathrm{x}=+3\)

Therefore, oxidation state of \(\mathrm{Cr}\) is \(+3\).

(C)Let the oxidation number of \(\mathrm{Ni}\) be \(\mathrm{x}\).

We know that :

Oxidation number of \(\mathrm{F}=-1\)

Then,

\(\mathrm{x+6(-1)=-2}\)

\(\mathrm{x-6=-2}\)

\(\mathrm{x=-2+6}\)

\(\mathrm{x=+4}\)

Therefore, oxidation state of \(\mathrm{Ni}\) is \(+4\).

(D) Let the oxidation state of central metal atom \(C r\) be \(x\).

We know that:

Oxidation number of\(\mathrm{O}=-2, \mathrm{Cl}=-1\)

Then,

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

\(x=4+2\)

\(x=+6\)

\({CrO}_{2} {Cl}_{2}\) is the species in which the central metal atom is in +6 oxidation state.
Question 10
1 / -0

On oxidation with a mild oxidising agent like \(Br _{2} / H _{2} O\), the glucose is oxidised to:

A
Saccharic acid

B
Glucaric acid

C
Gluconic acid

D
Valeric acid

Solution

On oxidation with a mild oxidising agent like \(Br _{2} / H _{2} O\), the glucose is oxidised to six carbon carboxylic acid (gluconic acid). This indicates that the carbonyl group is present as an aldehydic group.

\(CHO -( CHOH )_{4} CH _{2} OH \stackrel{ Br _{2} \text { water }}{\longrightarrow} COOH -\underset{\text { Gluconic acid }}{( CHOH )_{4}}- CH _{2} OH\)

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Chemistry Test - 26

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Chemistry Test - 26

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Question 1

Glycine, Alanine, Tyrosine, Glutamic acid

Glycine, Alanine, Threonine, Lysine

Tyrosine, Glutamic acid, Valine, Leucine

Leucine, Threonine, Alanine, Tyrosine

Solution

Question 2

\(d_{1}>d_{2}>d_{3}>d_{4}\)

\(d_{4}>d_{3}>d_{2}>d_{1}\)

\(d_{1}=d_{2}=d_{3}=d_{4}\)

\(d_{1}>d_{3}>d_{2}>d_{4}\)

Solution

Question 3

Vicinal dihalide

Gem-dihalide

Allylic halide

Vinylic halide

Solution

Question 4

No change in volume on mixing

No change in enthalpy on mixing

It obeys Raoults law

Ionisation of solute should occurs to a small extent

Solution

Question 5

\(7.9\)

\(8.6\)

\(9.4\)

\(7.2\)

Solution

Question 6

iv) < i) < iii) < ii)

i) < iv) < iii) < ii)

iv) < i) < ii) < iii)

ii) < iii) < i) < iv)

Solution

Question 7

Second ionizationenthalpy of copper is less than the first ionizationenthalpy.

Large value of second ionization enthalpy of copper is compensated by much more negative hydration energy of \(C u_{(a q)}^{2+}\).

Hydration energy of \(C u_{(a q)}^{2+}\) is much more negative than that of \(C u_{(a q)}^{2}\).

Many copper (I) compounds are unstable in aqueous solution and undergo a disproportionation reaction..

Solution

Question 8

Conc. \(\mathrm{HNO}_{3}\)

Dil. \(\mathrm{HNO}_{3}\)

Urea

Zinc Dust

Solution

Question 9

\(M n O_{ {4}}^-\)

\([C r(C N) 6]^{3-}\)

\(N i F_{6}^{2-}\)

\({CrO}_{2} {Cl}_{2}\)

Solution

Question 10

Saccharic acid

Glucaric acid

Gluconic acid

Valeric acid

Solution

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