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Chemistry Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following will have the lowest boiling point?

A
2-MethylButane

B
2-MethylPropane

C
2,2-Dimethylpropane

D
n-Pentane

Solution

Boiling point increases with increase in molecular mass. For the compounds with the same molecular mass, boiling point decreases with an increase in branching.

The molecular mass of 2-Methylbutane: 72 g mol^-1

The molecular mass of 2-Methylpropane: 58 g mol^-1

The molecular mass of 2, 2-Dimethylpropane: 72 g mol^-1

The molecular mass of 2-Methylbutane: 72 g mol^-1

2-Methylpropane has the lowest molecular mass among all of the given compounds.

Thus, 2-Methylpropane has the lowest boiling point among the given options.

Question 2

1 / -0

Match List - I with Match List - II.

List I	List II
(A)Glucose + HI	(I)Gluconic acid
(B)Glucose + Br₂ water	(II)Glucose pentacetate
(C)Glucose + acetic anhydride	(III)Saccharic acid
(D)Glucose +HNO₃	(IV)Hexane

Choose the correct answer from the options given below:

(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

(A)-(I), (B)-(III), (C)-(IV), (D)-(II)

Solution

List I	List II
(A)Glucose + HI	(IV)Hexane
(B)Glucose + Br₂ water	(I)Gluconic acid
(C)Glucose + acetic anhydride	(II)Glucose pentacetate
(D)Glucose +HNO₃	(III)Saccharic acid

Question 3
1 / -0

\(\mathrm{RNH}_{2}+\mathrm{CHCl}_{3}+\mathrm{KOH} \stackrel{\Delta}{\rightarrow} ?\)

The product of above reaction is:

A
\({RCl}\)

B
\(\mathrm{ROH}\)

C
\(RCN\)

D
\(RNC\)

Solution

Primary amines when treated with chloroform and ethanolic \(KOH\) gives carbylamines or isocyanides. The reaction occurs via carbene intermediate, Chloroform reacts with \(KOH\) to give chloro carbene: \(CCl_2\).

The general form of the reaction is:
Question 4
1 / -0

The number of primary, secondary, tertiary and quaternery carbon atoms in neopentane are respectively:

A
\(4, 1, 0, 0\)

B
\(4,3, 2,1\)

C
\(4, 0, 0, 1\)

D
\(4,0,1,1\)

Solution

The number of primary, secondary, tertiary and quaternery carbon atoms in neopentane are respectively \(4, 0, 0, 1\).

Neopentane has one carbon in the middle, which is attached to four methyl groups \(CH_{3}\). The carbon in the middle is attached to the methyl group from all sides, so it is a quaternery carbon, in the same way the rest of the carbons are attached to the same carbon, so they are primary carbon.

\(\begin{array}{lll}1^{\circ} & 2^{\circ} & 3^{\circ} & 4^{\circ} \\ 4 & 0 & 0&1\end{array}\)

Neopentane has \(4\) primary carbons, \(0\) secondary carbons,\(0\) tertiary carbons and \(1\) quaternery carbon.
Question 5
1 / -0

Which is the only alcohol that can be prepared by the indirect hydration of alkene?

A
Ethyl alcohol

B
Propyl alcohol

C
Isobutyl alcohol

D
Methyl alcohol

Solution

Alkene is treated with sulfuric acid to give alkyl sulphate esters. In the case of ethanol production, this step can be written:

\(\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{C}_{2} \mathrm{H}_{4} \rightarrow \mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{O}-\mathrm{SO}_{3} \mathrm{H}\)

Subsequently, this sulphate ester is hydrolyzed to regenerate sulphuric acid and release ethanol:

\(\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{O}-\mathrm{SO} \mathrm{O}_{3} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)

This two step route is called the "indirect process". Only in case of ethane it will produce ethyl alcohol while in case of other alkenes product is according to Markovnikov's Rule. Methyl alcohol can't be produced by indirect hydration of alkene.
Question 6
1 / -0

In which of the following reactions, \(H _{2} O _{2}\) behaves as a reducing agent?

A
\(Na _{2} SO _{3}( aq )+ H _{2} O _{2}( aq ) \rightarrow Na _{2} SO _{4}( aq )+ H _{2} O ( l )\)

B
\(P bO _{2}( s )+ H _{2} O _{2}( aq ) \rightarrow P bO ( s )+ H _{2} O ( l )+ O _{2}( g )\)

C
\(2 KI ( aq )+ H _{2} O _{2}( aq ) \rightarrow 2 KOH ( aq )+ I _{2}( s )\)

D
\(KNO _{2}( aq )+ H _{2} O _{2}( aq ) \rightarrow KNO _{3}( aq )+ H _{2} O ( l )\)

Solution

\(PbO _{2}( s )+ H _{2} O _{2}( aq ) \rightarrow PbO ( s )+ H _{2} O ( l )+ O _{2}( g )\)

In the above reaction, \(H _{2} O _{2}\) acts as a reducing agent because it undergoes oxidation to \(O _{2}\) and it reduces the \(P bO _{2}\) to \(P bO\).
Question 7
1 / -0

Resistance of a conductivity cell (cell constant \(129 \mathrm{~m}^{-1}\) ) filled with \(74.5 \mathrm{ppm}\) solution of \(\mathrm{KCl}\) is \(100 \Omega\) (labelled as solution 1). When the same cell is filled with \(\mathrm{KCl}\) solution of \(149 \mathrm{ppm}\), the resistance is \(50 \Omega\) (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. \(\frac{\wedge_1}{\wedge_2}=x \times 10^{-3}\).

The value of \(x\) is _______________. (Nearest integer)

Given, molar mass of \(\mathrm{KCl}\) is \(74.5 \mathrm{gmol}^{-1}\).

A
5000

B
8000

C
1000

D
1200

Solution

\(\frac{1}{A}=129 \mathrm{~m}^{-1}\)

\(\mathrm{KCl}\) solution \(1 \Rightarrow 74.5 \mathrm{ppm}, \mathrm{R}_1=100 \Omega\)

\(\mathrm{KCl}\) solution \(2 \Rightarrow 149 \mathrm{ppm}, \mathrm{R}_2=50 \Omega\)

Here, \(\frac{\mathrm{ppm}_1}{\mathrm{ppm}_2}=\frac{\mathrm{M}_1}{\mathrm{M}_2}=\left(\frac{\mathrm{w}_{1 / \mathrm{M}_0}}{\mathrm{~V}} \times \frac{\mathrm{V}}{\mathrm{w}_{2 / \mathrm{M}_0}}\right)\)

\(\frac{\Lambda_1}{\Lambda_2}=\frac{\mathrm{k}_1 \times \frac{1000}{\mathrm{M}_1}}{\mathrm{k}_2 \times \frac{1000}{\mathrm{M}_2}} \)

\(=\frac{\mathrm{k}_1}{\mathrm{k}_2} \times \frac{\mathrm{M}_1}{\mathrm{M}_2} \)

\(=\frac{50}{100} \times 2 \)

\(=\frac{\wedge_1}{\wedge_2}=1000 \times 10^{-3} \)

\( =1000\)
Question 8
1 / -0

Hydrocarbons having single bonds are known as:

A
Saturated hydrocarbon

B
Unsaturated hydrocarbon

C
Alkyne

D
None of above

Solution

Hydrocarbons having single bonds are known as saturated hydrocarbon.

Saturated hydrocarbons are known as an alkane. These are also known as paraffin.Unsaturated hydrocarbons are having double or triple bonds i.e., alkenes and alkynes.A Saturated hydrocarbon is a hydrocarbon in which all the carbon-carbon bonds are single bonds.
Question 9
1 / -0

Calculate the amount of benzoic acid \(\left({C}_{6} {H}_{5} {COOH}\right)\) required for preparing 250 \({mL}\) of \(0.15 {M}\) solution in methanol.

A
4.680 g

B
4.790 g

C
4.875 g

D
4.575 g

Solution

0.15 M solution of benzoic acid in methanol means,

1000 mL of solution contains 0.15 mol of benzoic acid.

Therefore, \(250 {~mL}\) of solution contains \(\frac{0.15 \times 250}{1000}\) mol of benzoic acid

\(=0.0375\) mol of benzoic acid

Molar mass of benzoic acid \(\left({C}_{6} {H}_{5} {COOH}\right)=7 \times 12+6 \times 1+2 \times 16=122 {~g} {~mol}^{-1}\)

Therefore, required benzoic acid \(=0.0375 {~mol} \times 122 {~g} {~mol}^{-1}=4.575 {~g}\)
Question 10
1 / -0

Which of the following is not a rule for calculating oxidation number?

A
For ions, oxidation number is equal to the charge on the ion.

B
The oxidation number of oxygen is \(-2\) in all of its compound.

C
The oxidation number of fluorine is \(-1\) in all of its compounds.

D
Oxidation number of hydrogen is \(+1\) except in binary hydrides of alkali metals and alkaline earth metals where it is \(-1\).

Solution

We know that fluorine is the highly electronegative atom in the entire periodic table. So it has always an oxidation number of -1 in all its compounds.

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Chemistry Test - 11

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Chemistry Test - 11

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Question 1

2-MethylButane

2-MethylPropane

2,2-Dimethylpropane

n-Pentane

Solution

Question 2

(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

(A)-(I), (B)-(III), (C)-(IV), (D)-(II)

Solution

Question 3

\({RCl}\)

\(\mathrm{ROH}\)

\(RCN\)

\(RNC\)

Solution

Question 4

\(4, 1, 0, 0\)

\(4,3, 2,1\)

\(4, 0, 0, 1\)

\(4,0,1,1\)

Solution

Question 5

Ethyl alcohol

Propyl alcohol

Isobutyl alcohol

Methyl alcohol

Solution

Question 6

\(Na _{2} SO _{3}( aq )+ H _{2} O _{2}( aq ) \rightarrow Na _{2} SO _{4}( aq )+ H _{2} O ( l )\)

\(P bO _{2}( s )+ H _{2} O _{2}( aq ) \rightarrow P bO ( s )+ H _{2} O ( l )+ O _{2}( g )\)

\(2 KI ( aq )+ H _{2} O _{2}( aq ) \rightarrow 2 KOH ( aq )+ I _{2}( s )\)

\(KNO _{2}( aq )+ H _{2} O _{2}( aq ) \rightarrow KNO _{3}( aq )+ H _{2} O ( l )\)

Solution

Question 7

5000

8000

1000

1200

Solution

Question 8

Saturated hydrocarbon

Unsaturated hydrocarbon

Alkyne

None of above

Solution

Question 9

4.680 g

4.790 g

4.875 g

4.575 g

Solution

Question 10

For ions, oxidation number is equal to the charge on the ion.

The oxidation number of oxygen is \(-2\) in all of its compound.

The oxidation number of fluorine is \(-1\) in all of its compounds.

Oxidation number of hydrogen is \(+1\) except in binary hydrides of alkali metals and alkaline earth metals where it is \(-1\).

Solution

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