Select Chapter Topics:
Q. No.On heating, some solid substances change from solid to vapour state without passing through liquid state. The technique used for the purification of such solid substances based on the above principle is known as :
1. Sublimation
2. Distillation
3. Chromatography
4. Crystallization
1. Sublimation
2. Distillation
3. Chromatography
4. Crystallization
Subtopic: Qualitative Analysis of Organic Compounds |
79%
From NCERT
NEET - 2024
Please attempt this question first.
Hints
Please attempt this question first.
Match List I with List II.
Choose the correct answer from the options given below:
1. A-IV, B-III, C-II, D-I
2. A-III, B-IV, C-II, D-I
3. A-III, B-IV, C-I, D-II
4. A-I, B-IV, C-II, D-III
| List I (Molecule) |
List II (Number and types of bond/s between two carbon atoms) |
||
| A. | ethane | I. | \(\mathrm{one ~\sigma-bond~ and~ two~ \pi-bonds}\) |
| B. | ethene | II. | \(\text { two } \pi \text {-bonds }\) |
| C. | carbon molecule, \(\mathrm{C} _2\) |
III. | \(\text { one } \sigma \text {-bond }\) |
| D. | ethyne | IV. | \(\mathrm{one ~\sigma-bond ~and ~one~ \pi-bond}\) |
1. A-IV, B-III, C-II, D-I
2. A-III, B-IV, C-II, D-I
3. A-III, B-IV, C-I, D-II
4. A-I, B-IV, C-II, D-III
Subtopic: Hybridisation & Structure of Carbon Compounds |
73%
From NCERT
NEET - 2024
Please attempt this question first.
Hints
Please attempt this question first.
A compound with a molecular formula of C6H14 has two tertiary carbons. Its IUPAC name is:
1. 2-Methyl pentane
2. 2, 3-Dimethylbutane
3. 2, 2-Dimethylbutane
4. n-Hexane
1. 2-Methyl pentane
2. 2, 3-Dimethylbutane
3. 2, 2-Dimethylbutane
4. n-Hexane
Subtopic: Nomenclature |
69%
From NCERT
NEET - 2024
Please attempt this question first.
Hints
Please attempt this question first.
Subtopic: Reaction Intermediates ; Preparation & Properties |
71%
From NCERT
NEET - 2024
Please attempt this question first.
Hints
Please attempt this question first.
The method used for quantitative estimation of halogens is:
| 1. | Dumas method | 2. | Carius method |
| 3. | Kjeldahl's method | 4. | Combustion method |
Subtopic: Qualitative Analysis of Organic Compounds |
73%
From NCERT
NEET - 2024
Please attempt this question first.
Hints
Please attempt this question first.
The technique used to purify liquids having high boiling points and getting decomposed at or below their boiling points is:
1. Fractional distillation
2. Steam distillation
3. Simple distillation
4. Distillation under reduced pressure
1. Fractional distillation
2. Steam distillation
3. Simple distillation
4. Distillation under reduced pressure
Subtopic: Qualitative Analysis of Organic Compounds |
63%
From NCERT
NEET - 2024
Please attempt this question first.
Hints
Please attempt this question first.
Subtopic: Qualitative Analysis of Organic Compounds |
75%
From NCERT
NEET - 2024
Please attempt this question first.
Hints
Please attempt this question first.
Which of the following options is the correct IUPAC name for the given compound?
1. 4-Ethyl-1-fluoro-2-nitrobenzene
2. 4-Ethyl-1-fluoro-6-nitrobenzene
3. 3-Ethyl-6-fluoro-1-nitrobenzene
4. 1-Ethyl-4-fluoro-3-nitrobenzene
1. 4-Ethyl-1-fluoro-2-nitrobenzene
2. 4-Ethyl-1-fluoro-6-nitrobenzene
3. 3-Ethyl-6-fluoro-1-nitrobenzene
4. 1-Ethyl-4-fluoro-3-nitrobenzene
Subtopic: Nomenclature |
56%
From NCERT
NEET - 2024
Please attempt this question first.
Hints
Please attempt this question first.
A steam volatile organic compound which is immiscible with water has a boiling point of \(250^\circ \text{C}\). During steam distillation, a mixture of this organic compound and water will boil:
1. Above \(100^\circ \text{C}\) but below \(250^\circ \text{C}\).
2. Above \(250^\circ \text{C}\).
3. At \(250^\circ \text{C}\).
4. Close to but below \(100^\circ \text{C}\).
1. Above \(100^\circ \text{C}\) but below \(250^\circ \text{C}\).
2. Above \(250^\circ \text{C}\).
3. At \(250^\circ \text{C}\).
4. Close to but below \(100^\circ \text{C}\).
Subtopic: Qualitative Analysis of Organic Compounds |
From NCERT
NEET - 2024
Please attempt this question first.
Hints
Please attempt this question first.
On complete combustion, \(0.3 \mathrm{~g}\) of an organic compound gave \(0.2 \mathrm{~g}\) of \(\mathrm{CO}_2~ \mathrm{and}~ 0.1 \mathrm{~g}\) of \(\mathrm{H}_2 \mathrm{O}\). The percentage composition of carbon and hydrogen in the compound, respectively is:
1. \(4.07 \% ~\mathrm{and}~ 15.02 \%\)
2. \(18.18 \% ~\mathrm{and}~ 3.70 \%\)
3. \(15.02 \% ~\mathrm{and}~ 4.07 \%\)
(4. \(3.70 \% ~\mathrm{and}~ 18.18 \%\)
1. \(4.07 \% ~\mathrm{and}~ 15.02 \%\)
2. \(18.18 \% ~\mathrm{and}~ 3.70 \%\)
3. \(15.02 \% ~\mathrm{and}~ 4.07 \%\)
(4. \(3.70 \% ~\mathrm{and}~ 18.18 \%\)
Subtopic: Quantitative Analysis of Organic Compounds |
72%
From NCERT
NEET - 2024
Please attempt this question first.
Hints
Select Chapter Topics:
© 2024 GoodEd Technologies Pvt. Ltd.