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201. Sublimation is a purification method used for substances that:
ⓐ. Decompose on heating and cannot be vaporized
ⓑ. Melt easily and dissolve in most solvents
ⓒ. Directly change from solid to vapor without passing through the liquid phase
ⓓ. Can only be purified by distillation
Correct Answer: Directly change from solid to vapor without passing through the liquid phase
Explanation: Sublimation works for solids that vaporize directly without melting. Compounds like camphor, naphthalene, and benzoic acid sublime easily. Since impurities do not vaporize, the vapor condenses as pure solid, allowing effective purification.
202. Which of the following compounds can be purified using sublimation?
ⓐ. Glucose
ⓑ. Urea
ⓒ. Sodium chloride
ⓓ. Camphor
Correct Answer: Camphor
Explanation: Camphor is a classic sublimable substance. It turns into vapor upon gentle heating and re-solidifies upon cooling. Ionic compounds like NaCl and non-volatile solids like glucose do not sublime, so sublimation is ineffective for them.
203. Sublimation separates a solid from impurities that are:
ⓐ. Also sublimable
ⓑ. Insoluble in solvent
ⓒ. Non-volatile and do not vaporize
ⓓ. Soluble only in water
Correct Answer: Non-volatile and do not vaporize
Explanation: Sublimation relies on the difference in volatility. The solid that sublimes turns into vapor, while non-volatile impurities remain behind. This difference allows clean separation without needing solvents or melting.
204. In sublimation, the solid is collected after cooling on a surface known as the:
ⓐ. Condenser plate
ⓑ. Receiver flask
ⓒ. Filtering funnel
ⓓ. Fractionation column
Correct Answer: Condenser plate
Explanation: The vapors formed during sublimation solidify onto a cold surface called the condenser plate. This allows collection of purified crystals. The rest of the apparatus contains the impure solid that remains behind after heating.
205. Which of the following conditions helps in efficient sublimation?
ⓐ. High pressure
ⓑ. Very high temperature
ⓒ. Low pressure or vacuum
ⓓ. High moisture atmosphere
Correct Answer: Low pressure or vacuum
Explanation: Lowering pressure reduces the temperature required for sublimation. Vacuum sublimation is often used for heat-sensitive compounds, as it prevents decomposition and speeds up the sublimation process.
206. Why is sublimation preferred over crystallization for some compounds?
ⓐ. Because it purifies non-volatile solids better
ⓑ. Because it works even when a suitable solvent is not available
ⓒ. Because it requires high temperatures to melt the solid
ⓓ. Because it removes both volatile and non-volatile impurities
Correct Answer: Because it works even when a suitable solvent is not available
Explanation: Some compounds do not dissolve well or may react with solvents. Sublimation does not require solvents at all. It is ideal for volatile organic solids that can vaporize without decomposing, making purification simple and solvent-free.
207. Which safety precaution is essential while performing sublimation?
ⓐ. Using strong acids during heating
ⓑ. Heating the sublimation mixture very rapidly
ⓒ. Ensuring good ventilation to avoid inhaling vapors
ⓓ. Avoiding any cooling source
Correct Answer: Ensuring good ventilation to avoid inhaling vapors
Explanation: Sublimed vapors from organic solids (like naphthalene) can be harmful if inhaled. Proper ventilation, fume hoods, or ventilation chambers prevent inhalation. Slow, controlled heating also avoids decomposition or fire hazard.
208. Sublimation will NOT work for which of the following?
ⓐ. Benzoic acid
ⓑ. Naphthalene
ⓒ. Ammonium chloride
ⓓ. Sodium chloride
Correct Answer: Sodium chloride
Explanation: Sodium chloride (NaCl) is ionic and has a very high melting/boiling point. It does not sublime under normal laboratory conditions. Sublimation works only for molecular solids like naphthalene, NH₄Cl, camphor, etc.
209. What happens to non-sublimable impurities during sublimation?
ⓐ. They remain in the heating flask
ⓑ. They decompose into gas
ⓒ. They vaporize with the solid
ⓓ. They convert into crystals
Correct Answer: They remain in the heating flask
Explanation: Only substances capable of sublimation turn into vapor. Impurities that do not sublime simply stay behind in the flask or dish. This selective vaporization allows very clean separation with minimal contamination.
210. Which best describes the principle behind sublimation?
ⓐ. Separation based on solubility differences
ⓑ. Separation based on boiling point differences
ⓒ. Separation based on vapor pressure differences
ⓓ. Separation based on density differences
Correct Answer: Separation based on vapor pressure differences
Explanation: Sublimation works because sublimable solids have high vapor pressure at relatively low temperatures. This allows them to vaporize readily, while impurities with low vapor pressure remain solid. The vapor then re-solidifies as pure crystals on a cold surface.
211. Simple distillation is most suitable for separating liquids that:
ⓐ. Have very close boiling points
ⓑ. Decompose before boiling
ⓒ. Are insoluble in each other
ⓓ. Differ significantly in boiling points (≥ 25–30°C)
Correct Answer: Differ significantly in boiling points (≥ 25–30°C)
Explanation: Simple distillation works best when the boiling points of the two liquids differ by at least 25–30°C. With a large boiling point difference, one liquid vaporizes easily while the other remains in the flask, making separation efficient. Close-boiling mixtures require fractional distillation instead.
212. Fractional distillation is used when two liquids:
ⓐ. Have a boiling point difference less than 25°C
ⓑ. Are solids at room temperature
ⓒ. React with each other during heating
ⓓ. Cannot form vapors
Correct Answer: Have a boiling point difference less than 25°C
Explanation: Fractional distillation is ideal for liquids whose boiling points differ by less than 25°C. The fractionating column provides repeated vaporization–condensation cycles, improving separation efficiency. It is commonly used in petroleum refining and separating ethanol–water mixtures.
213. Which of the following statements correctly describes distillation under reduced pressure (vacuum distillation)?
ⓐ. It increases the boiling point of liquids
ⓑ. It decreases the boiling point of liquids, allowing them to distill at lower temperatures
ⓒ. It removes impurities based on solubility differences
ⓓ. It is used for separating solids from liquids
Correct Answer: It decreases the boiling point of liquids, allowing them to distill at lower temperatures
Explanation: Vacuum distillation reduces external pressure, which lowers the boiling point of heat-sensitive liquids. This prevents decomposition and allows safe distillation of substances like glycerol or oils that form char at high temperatures.
214. Which apparatus component is essential for fractional distillation but not for simple distillation?
ⓐ. Condenser
ⓑ. Round-bottom flask
ⓒ. Receiver flask
ⓓ. Fractionating column
Correct Answer: Fractionating column
Explanation: A fractionating column is required for fractional distillation because it creates multiple vapor–condensation cycles. This allows better separation of liquids with close boiling points. Simple distillation does not require such a column.
215. What is the main purpose of using a fractionating column?
ⓐ. To cool the vapors
ⓑ. To increase surface area for repeated condensation and vaporization
ⓒ. To heat the mixture more effectively
ⓓ. To collect the distillate
Correct Answer: To increase surface area for repeated condensation and vaporization
Explanation: The fractionating column contains glass beads or plates that increase the surface area, promoting multiple vapor–liquid equilibria. This helps separate liquids with very close boiling points, improving distillation efficiency.
216. Simple distillation cannot effectively separate ethanol and water because:
ⓐ. Their boiling points are too close
ⓑ. Water reacts with ethanol
ⓒ. Ethanol does not vaporize
ⓓ. Ethanol decomposes at 78°C
Correct Answer: Their boiling points are too close
Explanation: Ethanol (78°C) and water (100°C) have a boiling point difference of only 22°C, which is too small for effective separation by simple distillation. Fractional distillation is required to separate such mixtures due to repeated vaporization cycles.
217. Distillation under reduced pressure is MOST useful for:
ⓐ. Liquids with very low boiling points
ⓑ. Liquids that are highly volatile
ⓒ. Liquids that decompose at high temperatures
ⓓ. Liquids that are solid at room temperature
Correct Answer: Liquids that decompose at high temperatures
Explanation: Some organic liquids (e.g., glycerol, essential oils) decompose before they reach their boiling points at atmospheric pressure. By lowering the pressure, they boil at much lower temperatures, preventing decomposition and allowing safe distillation.
218. Why is the receiving flask often cooled in fractional distillation?
ⓐ. To speed up vaporization
ⓑ. To avoid loss of product by evaporation
ⓒ. To help impurities crystallize
ⓓ. To separate solids in the mixture
Correct Answer: To avoid loss of product by evaporation
Explanation: The distillate may still be warm and volatile. Cooling the receiving flask prevents vapors from escaping, ensuring accurate collection and reducing product loss. This is especially important for low-boiling compounds.
219. Which principle does simple and fractional distillation primarily depend on?
ⓐ. Differences in boiling points
ⓑ. Differences in melting points
ⓒ. Differences in solubility
ⓓ. Differences in density
Correct Answer: Differences in boiling points
Explanation: Distillation separates components based on boiling point differences. When a mixture is heated, the component with the lower boiling point vaporizes first and is collected after condensation. Greater boiling point differences give better separation.
220. Which statement is correct regarding vacuum distillation?
ⓐ. It is used only for inorganic liquids
ⓑ. It raises the boiling point of the liquid
ⓒ. It prevents decomposition of heat-sensitive compounds
ⓓ. It requires a fractionating column always
Correct Answer: It prevents decomposition of heat-sensitive compounds
Explanation: Many organic liquids decompose or char at their normal boiling points. Vacuum distillation lowers the boiling point by reducing external pressure, allowing distillation at safer, lower temperatures. This makes it ideal for purifying sensitive compounds like oils, glycerol, and steroids.
221. Steam distillation is mainly used for separating substances that are:
ⓐ. Insoluble in water and volatile with steam
ⓑ. Soluble in water and non-volatile
ⓒ. Non-volatile solids
ⓓ. High-boiling ionic compounds
Correct Answer: Insoluble in water and volatile with steam
Explanation: Steam distillation is ideal for compounds that are immiscible with water but have significant vapor pressure when co-distilled with steam. These include essential oils, aromatic compounds, and some organic liquids. High-boiling substances can distill at lower temperatures because the total vapor pressure is the sum of vapor pressures of water + the organic compound.
222. Steam distillation allows high-boiling liquids to distill at:
ⓐ. Temperatures below their normal boiling point
ⓑ. Exactly 100°C
ⓒ. Higher temperatures than their boiling point
ⓓ. Their melting point
Correct Answer: Temperatures below their normal boiling point
Explanation: In steam distillation, the combined vapor pressure of water and the organic compound reaches atmospheric pressure at a temperature below the boiling point of either component. This protects heat-sensitive molecules (like natural oils) from decomposition.
223. Steam distillation is most commonly used for extracting:
ⓐ. Inorganic salts
ⓑ. Water-soluble solids
ⓒ. Essential oils from plant materials
ⓓ. Alkali metal halides
Correct Answer: Essential oils from plant materials
Explanation: Essential oils (clove oil, eucalyptus oil, rose oil) are volatile, water-insoluble compounds. Steam helps vaporize them at low temperatures, preventing thermal decomposition. They condense with water and can be separated easily.
224. Why do two immiscible liquids boil together in steam distillation?
ⓐ. Because they become miscible at high temperature
ⓑ. Because their combined vapor pressure equals atmospheric pressure
ⓒ. Because they react together
ⓓ. Because they form hydrogen bonds
Correct Answer: Because their combined vapor pressure equals atmospheric pressure
Explanation: Immiscible liquids obey Dalton’s Law of Partial Pressures. The mixture boils when:
P₍water₎ + P₍organic compound₎ = 1 atm
This lowers the boiling point, allowing co-distillation at safer temperatures.
225. Which apparatus component is essential for steam distillation?
ⓐ. Fractionating column
ⓑ. Steam generator
ⓒ. Rotary evaporator
ⓓ. Desiccator
Correct Answer: Steam generator
Explanation: A steam generator supplies continuous steam that passes through the organic mixture. The steam carries the volatile compound into the condenser. This is the primary feature distinguishing steam distillation from simple distillation.
226. What is collected in the condenser during steam distillation?
ⓐ. A mixture of water and organic distillate
ⓑ. Only water
ⓒ. Only the organic compound
ⓓ. Only impurities
Correct Answer: A mixture of water and organic distillate
Explanation: Both water vapor and organic vapor condense together. After condensation, the two phases typically separate because the organic compound is insoluble in water—making separation easy using a separating funnel.
227. Steam distillation is ineffective for which type of compound?
ⓐ. Aromatic oils
ⓑ. High-boiling water-insoluble compounds
ⓒ. Organic solids that volatilize with steam
ⓓ. Water-soluble liquids
Correct Answer: Water-soluble liquids
Explanation: If a compound dissolves in water, it will not form an immiscible layer and cannot benefit from combined vapor pressures. Water-soluble substances distill with water only at very high temperatures, making steam distillation ineffective.
228. Which principle allows heat-sensitive compounds to be distilled by steam distillation?
ⓐ. Lowering vapor pressure
ⓑ. Raising the boiling point
ⓒ. Co-distillation with water at reduced temperature
ⓓ. Lowering freezing point
Correct Answer: Co-distillation with water at reduced temperature
Explanation: Steam distillation lowers the effective boiling point by combining vapor pressures. This protects sensitive organic compounds that would otherwise char or decompose at normal boiling points.
229. In steam distillation, the ratio of masses of water and organic compound collected is determined by:
ⓐ. Their densities
ⓑ. Their melting points
ⓒ. Their solubility in steam
ⓓ. The ratio of their vapor pressures
Correct Answer: The ratio of their vapor pressures
Explanation: According to Dalton’s Law, the mass ratio distilled depends on:
$\frac{m_{organic}}{m_{water}} = \frac{P_{organic} \cdot M_{organic}}{P_{water} \cdot M_{water}}$
Thus, vapor pressure directly influences how much of each component co-distills.
230. Steam distillation is preferred for clove oil because:
ⓐ. Clove oil is water-soluble
ⓑ. Clove oil has a very low boiling point
ⓒ. Clove oil decomposes at its normal boiling point
ⓓ. Steam reacts with eugenol
Correct Answer: Clove oil decomposes at its normal boiling point
Explanation: Clove oil contains eugenol, which decomposes at temperatures close to its boiling point. Steam distillation allows it to distill at ~100°C, preventing decomposition and preserving the aroma and chemical structure.
231. Chromatography works on the basic principle of:
ⓐ. Difference in melting points
ⓑ. Difference in densities of components
ⓒ. Difference in boiling points
ⓓ. Difference in adsorption or partition of components between phases
Correct Answer: Difference in adsorption or partition of components between phases
Explanation: Chromatography separates mixtures because different components have different affinities toward the stationary phase and mobile phase. Components strongly adsorbed on the stationary phase travel slowly, while those with weaker interactions travel faster, creating separation. This principle applies to paper, TLC, column, and gas chromatography.
232. In paper chromatography, the stationary phase is:
ⓐ. Silica gel
ⓑ. Alumina
ⓒ. Water molecules adsorbed on cellulose fibers
ⓓ. Paraffin wax
Correct Answer: Water molecules adsorbed on cellulose fibers
Explanation: The paper used in chromatography contains cellulose, which holds water molecules on its surface. This water acts as the stationary phase. The mobile phase (a solvent) rises through capillary action. Compounds separate based on their partition between these two phases.
233. In thin-layer chromatography (TLC), the stationary phase is usually:
ⓐ. Water
ⓑ. Silica gel or alumina coated on a plate
ⓒ. Graphite
ⓓ. Alcohol
Correct Answer: Silica gel or alumina coated on a plate
Explanation: TLC plates are coated with silica gel (SiO₂) or alumina (Al₂O₃), which act as polar stationary phases. Different components adsorb to these surfaces differently. More strongly adsorbed components travel a shorter distance, giving distinct spots.
234. The movement of solvent in chromatography is mainly due to:
ⓐ. Capillary action
ⓑ. Evaporation
ⓒ. Osmosis
ⓓ. Electrolysis
Correct Answer: Capillary action
Explanation: Both paper chromatography and TLC rely on capillary action to pull the solvent upward. The solvent carries dissolved components along the stationary phase. Differences in solubility and adsorption cause separation and distinct travel distances.
235. The $R_f$ value of a component is calculated by:
ⓐ. $\dfrac{\text{Total paper length}}{\text{Distance of solvent front}}$
ⓑ. $\dfrac{\text{Distance traveled by solvent}}{\text{Distance traveled by component}}$
ⓒ. $\dfrac{\text{Distance traveled by component}}{\text{Distance traveled by solvent front}}$
ⓓ. $\dfrac{\text{Distance between two spots}}{\text{Total distance}}$
Correct Answer: $\dfrac{\text{Distance traveled by component}}{\text{Distance traveled by solvent front}}$
Explanation: The Retention factor ($R_f$) is a ratio.
$R_f = \frac{\text{Distance traveled by the solute}}{\text{Distance traveled by the solvent front}}$
It indicates how strongly a compound is attracted to the stationary phase. Higher $R_f$ means weaker adsorption.
236. A compound with a very high $R_f$ value in TLC indicates:
ⓐ. Strong interaction with stationary phase
ⓑ. Weak interaction with stationary phase
ⓒ. Stationary phase is too polar
ⓓ. Solvent is not appropriate
Correct Answer: Weak interaction with stationary phase
Explanation: If the compound travels far—almost with the solvent front—its adsorption on the stationary phase is weak. This means it is more soluble in the mobile phase. Such compounds have high $R_f$ values and are often non-polar in polar stationary phases like silica gel.
237. Why is the starting line drawn in pencil in paper chromatography?
ⓐ. Pencil is decorative
ⓑ. Pencil lines dissolve in solvent
ⓒ. Ink dissolves in solvent and interferes with separation
ⓓ. Pencil reacts with samples
Correct Answer: Ink dissolves in solvent and interferes with separation
Explanation: Pens contain dyes that travel with the solvent, producing extra spots and ruining results. Pencil contains graphite, which is insoluble and does not move. This ensures pure results without contamination.
238. What is the main advantage of TLC over paper chromatography?
ⓐ. TLC is slower
ⓑ. TLC requires expensive equipment
ⓒ. TLC cannot identify compounds
ⓓ. TLC gives better separation and sharper spots
Correct Answer: TLC gives better separation and sharper spots
Explanation: TLC provides faster, sharper, and clearer separations because the stationary phase (silica or alumina) is more efficient. It allows separation of complex mixtures and works well for both polar and non-polar compounds. Paper chromatography is less efficient.
239. Visualization of colorless spots in TLC is often done using:
ⓐ. UV light or iodine chamber
ⓑ. Candle flame
ⓒ. Water immersion
ⓓ. Simple sunlight
Correct Answer: UV light or iodine chamber
Explanation: Many organic compounds are colorless. UV light causes fluorescent background with dark spots. Alternatively, an iodine chamber temporarily stains compounds brown. These techniques help detect spots that are otherwise invisible.
240. Steam-volatile compounds like essential oils are better analyzed by paper/TLC chromatography because:
ⓐ. They dissolve in graphite
ⓑ. They must be melted before analysis
ⓒ. They cannot be separated by solvents
ⓓ. They often contain multiple organic components separable by polarity
Correct Answer: They often contain multiple organic components separable by polarity
Explanation: Essential oils consist of terpenes, phenols, alcohols, and esters. These compounds differ in polarity and affinity for stationary phases. Chromatography separates them into distinct spots, helping in purity checking, identification, and quality control.
241. In the CuO test for detecting carbon and hydrogen, the organic compound is heated strongly with:
ⓐ. Copper(II) oxide
ⓑ. Zinc dust
ⓒ. Sodium carbonate
ⓓ. Calcium chloride
Correct Answer: Copper(II) oxide
Explanation: In the CuO test, the organic compound is heated with CuO, which acts as an oxidizing agent. Carbon in the compound is oxidized to CO₂, and hydrogen is oxidized to H₂O. CuO itself is reduced to Cu, confirming oxidation. This test helps identify the presence of C and H in any organic sample.
242. During the CuO test, the carbon in the organic compound is converted into:
ⓐ. Carbon monoxide
ⓑ. Methane
ⓒ. Charcoal
ⓓ. Carbon dioxide
Correct Answer: Carbon dioxide
Explanation: When heated with CuO, carbon present in the organic molecule is fully oxidized to CO₂ gas. This gas is later passed through limewater for detection. The appearance of milkiness confirms CO₂ and thus confirms carbon in the original sample.
243. During the CuO test, hydrogen present in the organic compound forms:
ⓐ. Hydrogen peroxide
ⓑ. Water
ⓒ. Hydrogen gas
ⓓ. Methanol
Correct Answer: Water
Explanation: Hydrogen atoms in the organic compound are oxidized to water (H₂O) when heated with copper oxide. This water is absorbed by anhydrous copper sulfate or calcium chloride placed in the tube. Blue-colored CuSO₄ indicates presence of hydrogen.
244. In the CuO test, the presence of carbon is confirmed by:
ⓐ. Turning moist blue litmus red
ⓑ. Formation of black residue
ⓒ. Formation of brown fumes
ⓓ. Formation of white precipitate in limewater
Correct Answer: Formation of white precipitate in limewater
Explanation: The CO₂ produced during the test is bubbled through limewater (Ca(OH)₂).
$Ca(OH)_2 + CO_2 \rightarrow CaCO_3 \downarrow + H_2O$
The white precipitate of CaCO₃ confirms the presence of carbon in the organic sample.
245. The water produced in the CuO test is detected using:
ⓐ. Blue litmus paper
ⓑ. Sodium chloride
ⓒ. Anhydrous copper sulfate
ⓓ. Copper metal
Correct Answer: Anhydrous copper sulfate
Explanation: Anhydrous CuSO₄ is white. When it absorbs water, it becomes blue hydrated copper sulfate.
$CuSO_4 + 5H_2O \rightarrow CuSO_4\cdot5H_2O$
A color change from white to blue confirms the presence of hydrogen in the organic compound.
246. Copper(II) oxide acts as what during the CuO test?
ⓐ. Reducing agent
ⓑ. Oxidizing agent
ⓒ. Catalyst
ⓓ. Precipitating agent
Correct Answer: Oxidizing agent
Explanation: CuO oxidizes carbon to CO₂ and hydrogen to H₂O while itself getting reduced to Cu metal (reddish-brown). This ability to oxidize organic elements makes CuO essential for detecting C and H.
247. In the CuO test, which change in copper oxide confirms the reaction has occurred?
ⓐ. CuO turns into metallic copper
ⓑ. CuO turns white
ⓒ. CuO dissolves completely
ⓓ. CuO becomes green
Correct Answer: CuO turns into metallic copper
Explanation: Black CuO is reduced by carbon/hydrogen to reddish-brown copper metal during oxidation:
$CuO + C \rightarrow Cu + CO_2$
This color change visually confirms oxidation of the organic sample.
248. Limewater turns milky during the CuO test due to the formation of:
ⓐ. Calcium bicarbonate
ⓑ. Calcium carbonate
ⓒ. Calcium sulfate
ⓓ. Calcium oxide
Correct Answer: Calcium carbonate
Explanation: Passing CO₂ into limewater forms insoluble CaCO₃, which appears as milkiness.
$Ca(OH)_2 + CO_2 \rightarrow CaCO_3 \downarrow$
This is a standard confirmatory test for carbon dioxide, thus confirming carbon.
249. What is the primary purpose of the CuO test?
ⓐ. To detect carbon and hydrogen in organic compounds
ⓑ. To identify nitrogen
ⓒ. To test for sulfur
ⓓ. To detect halogens
Correct Answer: To detect carbon and hydrogen in organic compounds
Explanation: The CuO test is designed specifically to determine whether the sample contains carbon and hydrogen. Carbon forms CO₂, and hydrogen forms H₂O, each confirmed by separate tests (limewater and anhydrous CuSO₄).
250. Which observation confirms the presence of hydrogen in an organic compound during the CuO test?
ⓐ. Limewater turns milky
ⓑ. Pungent gas is evolved
ⓒ. Formation of black precipitate
ⓓ. White CuSO₄ turns blue
Correct Answer: White CuSO₄ turns blue
Explanation: Water vapors produced during oxidation are absorbed by anhydrous CuSO₄, causing a color change to blue. This is a clear indication that hydrogen was present in the original organic molecule.
251. In Lassaigne’s test for nitrogen, the organic compound is fused with:
ⓐ. Copper metal
ⓑ. Sodium metal
ⓒ. Zinc dust
ⓓ. Potassium carbonate
Correct Answer: Sodium metal
Explanation: In Lassaigne’s test, the organic compound is fused with fresh sodium metal. Sodium breaks the covalent bonds of organic molecules, converting covalently bound nitrogen into ionic sodium cyanide (NaCN). This water-soluble form allows further testing for nitrogen. Without sodium fusion, nitrogen remains in covalent form and cannot be detected by classical wet tests.
252. During sodium fusion, nitrogen present in the compound is converted into:
ⓐ. Sodium nitrate ($NaNO_3$)
ⓑ. Sodium nitrite ($NaNO_2$)
ⓒ. Sodium bicarbonate ($NaHCO_3$)
ⓓ. Sodium cyanide ($NaCN$)
Correct Answer: Sodium cyanide ($NaCN$)
Explanation: When the organic compound is fused with sodium, nitrogen reacts with sodium and carbon to form water-soluble sodium cyanide (NaCN). This cyanide ion reacts later with FeSO₄ to form complexes that ultimately give the Prussian blue color. Formation of NaCN is essential to detect nitrogen in organic compounds.
253. In the Prussian blue test, the sodium extract is first treated with:
ⓐ. Lead acetate
ⓑ. Silver nitrate
ⓒ. Ferrous sulfate (FeSO₄)
ⓓ. Barium chloride
Correct Answer: Ferrous sulfate (FeSO₄)
Explanation: FeSO₄ is added to the sodium cyanide extract. Cyanide ions react with Fe²⁺ to form sodium ferrocyanide ($Na_4$Fe(CN)_6$$). On acidification with dilute HCl, some Fe²⁺ is oxidized to Fe³⁺, which then reacts with ferrocyanide to produce the characteristic Prussian blue precipitate. This confirms the presence of nitrogen.
254. The appearance of Prussian blue color confirms the presence of:
ⓐ. Sulfur
ⓑ. Halogens
ⓒ. Nitrogen
ⓓ. Phosphorus
Correct Answer: Nitrogen
Explanation: The Prussian blue test depends on forming ferric ferrocyanide ($Fe_4$Fe(CN)_6$_3$), a deep blue colored compound. This color appears only if nitrogen in the organic compound has formed cyanide during sodium fusion. No other elemental impurity produces this exact blue coloration.
255. During Lassaigne’s test, the addition of dilute HCl after FeSO₄ is necessary to:
ⓐ. Oxidize Fe²⁺ to Fe³⁺
ⓑ. Neutralize the solution
ⓒ. Remove sulfur impurities
ⓓ. Remove halogens
Correct Answer: Oxidize Fe²⁺ to Fe³⁺
Explanation: Acidification converts some Fe²⁺ to Fe³⁺, which is essential because Fe³⁺ reacts with the ferrocyanide complex to produce Prussian blue. Without acidification and oxidation, the characteristic blue color will not form, even if nitrogen is present.
256. Which of the following reactions leads to the final Prussian blue precipitate?
ⓐ. $Fe^{3+}$ reacting with chloride ions
ⓑ. $Fe^{2+}$ reacting with hydroxide ions
ⓒ. $Fe^{3+}$ reacting with ferrocyanide ions
ⓓ. $Fe^{2+}$ reacting with nitrate ions
Correct Answer: $Fe^{3+}$ reacting with ferrocyanide ions
Explanation: After sodium fusion and treatment with FeSO₄, ferrocyanide ions $$Fe(CN)_6$^{4−}$ form. When Fe³⁺ ions (produced upon acidification) react with these ions, an insoluble deep blue compound called ferric ferrocyanide forms. This gives the classical Prussian blue color.
257. An organic compound gives no Prussian blue color in Lassaigne’s test. Which is the LEAST likely reason?
ⓐ. No nitrogen is present
ⓑ. Sodium fusion was incomplete
ⓒ. Cyanide ions were destroyed due to strong heating
ⓓ. FeSO₄ was added in excess
Correct Answer: FeSO₄ was added in excess
Explanation: Excess FeSO₄ does not prevent the Prussian blue reaction. However, absence of nitrogen, improper fusion, or overheating (which decomposes cyanide to nitrogen gas) can all lead to a false negative. Thus, adding excess FeSO₄ is not a reason for test failure.
258. Freshly prepared FeSO₄ solution is necessary in the Prussian blue test because:
ⓐ. It must be odorless
ⓑ. Fe²⁺ ions easily oxidize to Fe³⁺ in air
ⓒ. It acts as a strong reducing agent
ⓓ. It neutralizes the extract
Correct Answer: Fe²⁺ ions easily oxidize to Fe³⁺ in air
Explanation: FeSO₄ solutions slowly oxidize when exposed to oxygen. For the test to work properly, Fe²⁺ must first react with NaCN to form ferrocyanide. If Fe³⁺ is already present, this step is hindered, making the test unreliable. Therefore, a fresh FeSO₄ solution is essential.
259. What is the purpose of boiling the Lassaigne’s extract before testing for nitrogen?
ⓐ. To dissolve sodium cyanide in water
ⓑ. To remove excess sodium
ⓒ. To evaporate the organic compound
ⓓ. To decompose FeSO₄
Correct Answer: To dissolve sodium cyanide in water
Explanation: After sodium fusion, the fused mass containing NaCN must be extracted into water by boiling. This produces the sodium extract (Lassaigne’s extract). Only in dissolved form can NaCN react with FeSO₄ to form the diagnostic Prussian blue color.
260. The blue precipitate formed in nitrogen detection is chemically known as:
ⓐ. Ferrous sulfate
ⓑ. Ferric hydroxide
ⓒ. Ferric ferrocyanide
ⓓ. Ferrous cyanide
Correct Answer: Ferric ferrocyanide
Explanation: The final product responsible for the Prussian blue color is ferric ferrocyanide, a complex inorganic blue pigment. It forms when Fe³⁺ ions react with the ferrocyanide ion formed from Fe²⁺ and NaCN. This strongly colored compound is a highly reliable indicator of nitrogen in organic substances.
261. In Lassaigne’s test for sulfur, the organic compound is fused with sodium metal to convert sulfur into:
ⓐ. Sodium sulfate ($Na_2SO_4$)
ⓑ. Sodium sulfide ($Na_2S$)
ⓒ. Sodium thiosulfate ($Na_2S_2O_3$)
ⓓ. Sodium sulfite ($Na_2SO_3$)
Correct Answer: Sodium sulfide ($Na_2S$)
Explanation: During sodium fusion, sulfur present in the organic compound reacts with sodium to form sodium sulfide (Na₂S). This ionic, water-soluble form allows further testing. Without sodium fusion, sulfur remains covalently bonded and cannot be detected. The formation of Na₂S is the basis for both Lead Acetate and Sodium Nitroprusside tests.
262. In the lead acetate test for sulfur, sodium sulfide reacts with lead acetate to produce:
ⓐ. White precipitate of lead sulfate
ⓑ. Brown precipitate of lead dioxide
ⓒ. Black precipitate of lead sulfide
ⓓ. Yellow precipitate of lead chromate
Correct Answer: Black precipitate of lead sulfide
Explanation: The Lassaigne’s extract containing Na₂S is treated with lead acetate solution. Sodium sulfide reacts to form black lead sulfide (PbS):
$Na_2S + Pb(CH_3COO)_2 \rightarrow PbS \downarrow + 2CH_3COONa$
The formation of black PbS is a very strong confirmation of sulfur in the organic sample.
263. What color change is observed in the lead acetate test that confirms sulfur?
ⓐ. Green solution
ⓑ. Yellow precipitate
ⓒ. White crystals
ⓓ. Black precipitate
Correct Answer: Black precipitate
Explanation: Lead acetate reacts with sulfide ions to give black PbS, which is highly insoluble and easily recognizable. No other element in Lassaigne’s extract produces this intense black color, making the test very specific for sulfur detection.
264. In the sodium nitroprusside test, sodium sulfide reacts with sodium nitroprusside to give a color that confirms sulfur. What is the color?
ⓐ. Blood red
ⓑ. Deep blue
ⓒ. Purple/violet
ⓓ. Yellow
Correct Answer: Purple/violet
Explanation: Sodium nitroprusside reacts with sulfide ions to produce a purple or violet-colored complex, confirming sulfur. The reaction is very sensitive and can detect even small amounts of sulfide ions. This test complements the lead acetate test for reliability.
265. Which ion from sodium fusion is responsible for sulfur detection in the lead acetate test?
ⓐ. $CN^-$
ⓑ. $S^{2-}$
ⓒ. $SO_4^{2-}$
ⓓ. $HSO_3^-$
Correct Answer: $S^{2-}$
Explanation: During sodium fusion, sulfur becomes sulfide ions (S²⁻). These ions react with lead acetate to form PbS. If fusion is incomplete, sulfide ions won’t form and the test may give a false negative. Thus, presence of S²⁻ is essential.
266. Why must the Lassaigne’s extract be alkaline before performing the sodium nitroprusside test?
ⓐ. Sulfide ions (S²⁻) are stable only in alkaline medium
ⓑ. Nitroprusside decomposes in acidic medium
ⓒ. Lead acetate reacts only in alkaline medium
ⓓ. Fusion cannot occur without alkali
Correct Answer: Sulfide ions (S²⁻) are stable only in alkaline medium
Explanation: Sulfide ions get protonated in acidic medium to form H₂S gas, which escapes. In alkaline medium, S²⁻ remains dissolved and available to react with nitroprusside to form the characteristic violet color. Hence, alkalinity ensures accurate sulfur detection.
267. Which of the following reactions correctly describes the sodium nitroprusside test?
ⓐ. $S^{2-} + Pb^{2+} \rightarrow PbS$
ⓑ. $S^{2-} + $Fe(CN)_5NO$^{2-} \rightarrow$ violet complex
ⓒ. $S^{2-} + Cl^- \rightarrow SCl_2$
ⓓ. $S^{2-} + Fe^{3+} \rightarrow FeS$
Correct Answer: $S^{2-} + $Fe(CN)_5NO$^{2-} \rightarrow$ violet complex
Explanation: In this test, sulfide ions react with sodium nitroprusside to form a violet-colored complex. This is a highly sensitive reaction and detects trace sulfur amounts. Other options do not correspond to the nitroprusside reaction.
268. A student performs Lassaigne’s test for sulfur but gets no black precipitate with lead acetate. Which error is LEAST likely responsible?
ⓐ. Sodium fusion was incomplete
ⓑ. Sulfide ions were lost due to acidic conditions
ⓒ. The organic compound contains no sulfur
ⓓ. Using cold lead acetate solution
Correct Answer: Using cold lead acetate solution
Explanation: Lead acetate works perfectly even when cold. However, incomplete fusion, acidic medium (which removes S²⁻ as H₂S gas), or actual absence of sulfur can all result in negative tests. Thus, cold lead acetate is not a reason for test failure.
269. Which observation indicates incomplete sodium fusion while testing for sulfur?
ⓐ. Boiling of extract
ⓑ. Absence of CO₂ formation
ⓒ. Lack of Prussian blue color
ⓓ. No reaction with lead acetate despite sulfur being present
Correct Answer: No reaction with lead acetate despite sulfur being present
Explanation: If sodium fusion is incomplete, sulfur in the organic compound will not convert to S²⁻ ions, so adding lead acetate gives no black precipitate. This is a common student error and can lead to false negative results.
270. In Lassaigne’s test for sulfur, what is the advantage of using both lead acetate and sodium nitroprusside tests?
ⓐ. One detects nitrogen and one detects sulfur
ⓑ. One detects halogens and one detects sulfur
ⓒ. Together they confirm sulfur with higher reliability
ⓓ. They detect sulfur only if nitrogen is present
Correct Answer: Together they confirm sulfur with higher reliability
Explanation: Lead acetate gives black PbS, while nitroprusside gives a violet complex. Using both tests minimizes errors—if one test fails (due to incomplete fusion or mis-handling), the other may still detect sulfur. This cross-verification strengthens analytical accuracy in qualitative analysis.
271. In the AgNO₃ test for halogens, the Lassaigne’s extract is first acidified with:
ⓐ. Dilute H₂SO₄
ⓑ. Dilute HCl
ⓒ. Dilute CH₃COOH
ⓓ. Dilute HNO₃
Correct Answer: Dilute HNO₃
Explanation: The extract must be acidified with dilute nitric acid (HNO₃) to remove interfering ions such as sulfide and cyanide, which otherwise form precipitates with Ag⁺. HCl is never used because it already contains chloride ions that would give a false positive. Acidification ensures that only halide ions react with AgNO₃.
272. Halogens present in an organic compound form which ions during sodium fusion?
ⓐ. $SO_4^{2-}$
ⓑ. $CN^{-}$
ⓒ. $X^{-}$ (Cl⁻, Br⁻, I⁻)
ⓓ. $NO_3^{-}$
Correct Answer: $X^{-}$ (Cl⁻, Br⁻, I⁻)
Explanation: When fused with sodium, halogens in the organic compound convert into their corresponding sodium halides (NaCl, NaBr, NaI). These dissolve in water to give halide ions (X⁻) in the Lassaigne’s extract. These halide ions then react with AgNO₃ to form characteristic precipitates used for identification.
273. When AgNO₃ is added to the acidified Lassaigne’s extract containing chloride ions, the precipitate formed is:
ⓐ. White
ⓑ. Yellow
ⓒ. Pale yellow
ⓓ. Cream
Correct Answer: White
Explanation: Chloride ions react with Ag⁺ to give white silver chloride (AgCl):
$Ag^+ + Cl^- \rightarrow AgCl \downarrow$
AgCl is white, curdy, and dissolves in excess ammonia due to the formation of $Ag(NH₃)₂$⁺. This solubility distinguishes chloride from other halides.
274. Which halogen gives a pale yellow precipitate with AgNO₃?
ⓐ. Chlorine
ⓑ. Fluorine
ⓒ. Bromine
ⓓ. Iodine
Correct Answer: Bromine
Explanation: Bromide ions react with AgNO₃ to form silver bromide (AgBr), a pale yellow precipitate.
AgBr is sparingly soluble in ammonia (less soluble than AgCl). This limited solubility helps differentiate between chlorides and bromides during the test.
275. Iodide ions react with AgNO₃ to produce a precipitate that is:
ⓐ. Pale yellow
ⓑ. White
ⓒ. Cream
ⓓ. Deep yellow
Correct Answer: Deep yellow
Explanation: Iodide ions form silver iodide (AgI), a deep yellow precipitate that is insoluble in ammonia. This deep color and insolubility make AgI easy to distinguish from AgCl and AgBr.
276. Why is HNO₃ used before adding AgNO₃ in the halogen test?
ⓐ. To produce more halide ions
ⓑ. To remove interfering sulfide and cyanide ions
ⓒ. To form complexes with halogens
ⓓ. To increase solubility of silver salts
Correct Answer: To remove interfering sulfide and cyanide ions
Explanation: Sulfide and cyanide ions from incomplete sodium fusion can give black Ag₂S or white AgCN, which interfere with halogen detection. Acidification with HNO₃ decomposes these ions so that only halide ions remain to react with AgNO₃.
277. Which of the following halides does NOT react with AgNO₃?
ⓐ. Fluoride
ⓑ. Chloride
ⓒ. Bromide
ⓓ. Iodide
Correct Answer: Fluoride
Explanation: Fluoride ions do not form a precipitate with AgNO₃ because silver fluoride (AgF) is soluble in water. Thus, fluorine cannot be detected by the standard AgNO₃ test, and special tests are required to detect fluoride.
278. Which silver halide precipitate dissolves completely in excess ammonia?
ⓐ. AgBr
ⓑ. AgCl
ⓒ. AgI
ⓓ. None dissolve
Correct Answer: AgCl
Explanation: AgCl dissolves in excess ammonia due to formation of the soluble complex:
$AgCl + 2NH_3 \rightarrow $Ag(NH_3)_2$^+ + Cl^-$
AgBr dissolves only partially, while AgI is entirely insoluble, allowing clear differentiation among halides.
279. A cream-colored precipitate that dissolves partially in ammonia indicates the presence of:
ⓐ. Fluoride
ⓑ. Chloride
ⓒ. Bromide
ⓓ. Iodide
Correct Answer: Bromide
Explanation: Bromide ions form AgBr, a cream-colored precipitate. Its partial solubility in ammonia distinguishes it from AgCl (fully soluble) and AgI (insoluble). This behavior is a key identification step for bromine.
280. A student obtains a yellow precipitate that remains unchanged even after adding excess ammonia. This confirms the presence of:
ⓐ. Chlorine
ⓑ. Bromine
ⓒ. Fluorine
ⓓ. Iodine
Correct Answer: Iodine
Explanation: Silver iodide (AgI) is deep yellow and completely insoluble in ammonia. Failure to dissolve in ammonia is an important confirmatory characteristic of iodide ions. Neither AgCl nor AgBr behaves this way, ensuring accurate identification of iodine.
281. In Liebig’s method for estimation of carbon and hydrogen, the organic compound is oxidized using:
ⓐ. Copper metal
ⓑ. Potassium dichromate
ⓒ. Copper(II) oxide (CuO)
ⓓ. Sodium hydroxide
Correct Answer: Copper(II) oxide (CuO)
Explanation: In Liebig’s method, the organic compound is heated strongly with excess CuO, which oxidizes carbon to CO₂ and hydrogen to H₂O. CuO acts as a powerful oxidizing agent while being reduced to Cu. This complete oxidation is essential for accurate quantitative estimation of carbon and hydrogen in organic molecules.
282. During Liebig’s combustion method, carbon present in the organic compound is converted into:
ⓐ. Carbon monoxide
ⓑ. Carbon dioxide
ⓒ. Carbon disulfide
ⓓ. Carbon tetrachloride
Correct Answer: Carbon dioxide
Explanation: CuO oxidizes carbon completely to CO₂, which is then absorbed in KOH solution. The increase in mass of the KOH absorption bulb corresponds directly to the amount of carbon in the original sample. This quantitative conversion ensures precise carbon estimation.
283. In Liebig’s method, hydrogen present in the compound is finally collected as:
ⓐ. Water
ⓑ. Hydrogen gas
ⓒ. Hydrogen peroxide
ⓓ. Hydronium ions
Correct Answer: Water
Explanation: Hydrogen atoms in the organic sample are oxidized to water molecules during combustion. The water formed is absorbed by anhydrous CaCl₂, which increases in mass. The gain in mass directly represents the hydrogen content of the compound.
284. In Liebig’s method, CO₂ produced is absorbed by:
ⓐ. CaCl₂ tube
ⓑ. Concentrated HCl
ⓒ. KOH solution
ⓓ. Silver nitrate
Correct Answer: KOH solution
Explanation: CO₂ reacts with potassium hydroxide to form potassium carbonate:
$2KOH + CO_2 \rightarrow K_2CO_3 + H_2O$
The mass increase in the KOH absorption apparatus indicates the total CO₂ produced, which is directly used to calculate carbon percentage.
285. The water formed during combustion in Liebig’s method is absorbed by:
ⓐ. Silver nitrate solution
ⓑ. Copper sulfate solution
ⓒ. Lead acetate paper
ⓓ. Anhydrous CaCl₂
Correct Answer: Anhydrous CaCl₂
Explanation: Anhydrous calcium chloride absorbs water very effectively, forming hydrated CaCl₂. The increase in mass of the CaCl₂ tube gives the amount of water formed, which allows calculation of the hydrogen percentage in the compound.
286. After combustion, the percentage of carbon is calculated from the mass of:
ⓐ. Water produced
ⓑ. CO₂ absorbed
ⓒ. Residue remaining
ⓓ. CaCl₂ tube
Correct Answer: CO₂ absorbed
Explanation: Since all carbon in the organic compound is converted into CO₂ during combustion, the mass of CO₂ absorbed by KOH is used to calculate carbon percentage:
$\\C = \dfrac{12 \times {mass of } CO_2}{44 \times {mass of organic compound}} \times 100$
This formula directly relates the absorbed CO₂ mass to the carbon present in the original sample.
287. Hydrogen percentage in Liebig’s method is calculated using:
ⓐ. Mass of CO₂
ⓑ. Mass of unreacted CuO
ⓒ. Mass of H₂O
ⓓ. Mass of CaCO₃ formed
Correct Answer: Mass of H₂O
Explanation: Hydrogen present in the organic compound is oxidized to water. The increase in mass of the CaCl₂ tube gives the mass of water formed:
$\\H = \dfrac{2 \times {mass of } H_2O}{18 \times {mass of organic compound}} \times 100$
This formula accurately determines hydrogen content.
288. What is the main role of oxygen supply and excess CuO in Liebig’s method?
ⓐ. Speed up heating
ⓑ. Reduce CaCl₂
ⓒ. Increase solubility of CO₂
ⓓ. Ensure complete oxidation of C and H
Correct Answer: Ensure complete oxidation of C and H
Explanation: Excess CuO and continuous oxygen flow guarantee complete combustion of the organic compound. Incomplete oxidation would produce CO and unburnt carbon, leading to incorrect carbon and hydrogen percentages.
289. Which of the following can cause errors in Liebig’s estimation of carbon and hydrogen?
ⓐ. Using freshly dried CaCl₂
ⓑ. Incomplete combustion leading to formation of CO
ⓒ. Using excess CuO
ⓓ. Using pure oxygen gas
Correct Answer: Incomplete combustion leading to formation of CO
Explanation: If combustion is incomplete, carbon monoxide (CO) may form instead of CO₂. Since CO is not absorbed by KOH, the carbon percentage will be underestimated. Ensuring complete oxidation is critical for accurate results.
290. Why is the organic compound mixed with CuO in Liebig’s method before combustion?
ⓐ. To prevent formation of soot
ⓑ. To prevent water evaporation
ⓒ. To slow the reaction
ⓓ. To decrease CO₂ production
Correct Answer: To prevent formation of soot
Explanation: Organic compounds may char or form soot (pure carbon) when heated alone. Mixing with CuO ensures uniform oxidation, preventing soot formation and guaranteeing complete combustion. This enhances accuracy in carbon and hydrogen estimation.
291. In Dumas method for nitrogen estimation, the organic compound is heated with excess copper oxide to produce nitrogen gas and:
ⓐ. CO only
ⓑ. CO₂ and H₂O
ⓒ. HCN
ⓓ. NO₂
Correct Answer: CO₂ and H₂O
Explanation: In Dumas method, the compound is combusted strongly in the presence of CuO, converting carbon to CO₂ and hydrogen to H₂O. Nitrogen in the sample gets converted to N₂ gas. The CO₂ and H₂O are absorbed by KOH and CaCl₂, while N₂ is collected and measured. This volume of nitrogen is used to calculate %N in the sample.
292. In Dumas method, the nitrogen gas evolved is collected over:
ⓐ. Water
ⓑ. Mercury
ⓒ. KOH solution
ⓓ. HCl solution
Correct Answer: KOH solution
Explanation: Nitrogen is collected over KOH because KOH absorbs the CO₂ formed during combustion but does not absorb N₂. Collecting over water would cause nitrogen to dissolve slightly, giving errors. Using KOH ensures pure nitrogen collection for accurate nitrogen estimation.
293. In Dumas method, nitrogen gas is produced by reducing:
ⓐ. NO₂
ⓑ. HCN
ⓒ. Oxides of nitrogen
ⓓ. All nitrogen-containing products to N₂
Correct Answer: All nitrogen-containing products to N₂
Explanation: During combustion, nitrogen in the compound may temporarily form nitrogen oxides (NO, NO₂). These are later reduced back to N₂ by passing the gases over heated copper. This ensures all nitrogen finally appears as molecular nitrogen for quantitative estimation.
294. In Kjeldahl’s method, the organic compound is digested with concentrated sulfuric acid in the presence of:
ⓐ. Copper turning
ⓑ. Sodium metal
ⓒ. Potassium sulfate + catalyst
ⓓ. Potassium nitrate
Correct Answer: Potassium sulfate + catalyst
Explanation: The mixture of conc. H₂SO₄ + K₂SO₄ + catalyst (Cu, Se, Hg) digests the organic compound. Sulfuric acid converts nitrogen to ammonium sulfate. K₂SO₄ raises the boiling point of the acid ensuring faster digestion. Catalysts accelerate the oxidative breakdown of organic matter.
295. In Kjeldahl’s method, nitrogen is converted during digestion into:
ⓐ. Nitrite
ⓑ. Ammonium sulfate
ⓒ. Cyanide
ⓓ. Nitrate
Correct Answer: Ammonium sulfate
Explanation: In the digestion step:
${Organic N} \rightarrow (NH_4)_2SO_4$
This is then released as ammonia in the distillation step. The resulting NH₃ is absorbed in standard acid and later titrated to determine nitrogen content.
296. After digestion, ammonia is liberated by adding:
ⓐ. NaOH
ⓑ. HCl
ⓒ. KOH
ⓓ. Na₂CO₃
Correct Answer: NaOH
Explanation: Adding strong alkali like NaOH converts ammonium sulfate into ammonia gas:
$(NH_4)_2SO_4 + 2NaOH \rightarrow 2NH_3 + Na_2SO_4 + 2H_2O$
The liberated NH₃ is distilled off and absorbed in standard HCl or H₂SO₄ to quantify nitrogen.
297. In Kjeldahl’s method, the ammonia produced is absorbed in:
ⓐ. Silver nitrate
ⓑ. Calcium hydroxide
ⓒ. Sodium hydroxide
ⓓ. Standard acid like HCl or H₂SO₄
Correct Answer: Standard acid like HCl or H₂SO₄
Explanation: Ammonia reacts with a known concentration of the absorbing acid:
$NH_3 + HCl \rightarrow NH_4Cl$
The remaining acid is then back-titrated with NaOH. The difference gives the amount of NH₃ absorbed, and thus the nitrogen present.
298. Kjeldahl’s method CANNOT estimate nitrogen in organic compounds containing:
ⓐ. Amines
ⓑ. Amides
ⓒ. Nitro groups
ⓓ. Proteins
Correct Answer: Nitro groups
Explanation: Nitro compounds (–NO₂), azo compounds (–N=N–), and heterocyclic nitrogen do not get converted into ammonium sulfate during digestion. Therefore, Kjeldahl method is unsuitable for these compounds. Dumas method is preferred instead.
299. In the estimation of oxygen, the organic compound is treated with:
ⓐ. Carbon monoxide
ⓑ. Hydrated copper sulfate
ⓒ. Pyrolysis in a stream of nitrogen
ⓓ. Carbon to convert oxygen into CO
Correct Answer: Carbon to convert oxygen into CO
Explanation: Oxygen estimation is performed by reacting the compound in a carbon-filled apparatus. Oxygen forms CO, whose amount is measured as CO₂ after oxidation. By knowing the CO₂ produced, the oxygen present in the organic compound is calculated.
300. In estimation of halogens by Carius method, the organic compound is heated with:
ⓐ. Concentrated HNO₃
ⓑ. Fuming H₂SO₄
ⓒ. Strong nitric acid in sealed tube
ⓓ. Silver nitrate solution
Correct Answer: Strong nitric acid in sealed tube
Explanation: The Carius method requires heating the organic compound with fuming nitric acid ($HNO_3$) AND silver nitrate ($AgNO_3$) in a sealed tube54. The $HNO_3$ oxidizes the compound, and the $AgNO_3$ (which is added before heating) precipitates the halogen as a silver halide (AgX). The answer and explanation omit the essential reagent, $AgNO_3$, from the heating step.
Welcome to Class 11 Chemistry MCQs – Chapter 12: Organic Chemistry (Part 3). This final section includes the last 100 MCQs focusing on purification methods, elemental analysis, and quantitative techniques such as Liebig’s, Dumas, and Kjeldahl methods. It also covers important questions on chromatography, solubility, and detection of elements in organic compounds — highly relevant for Board and NEET exams.
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