Purification and Characterisation of Organic Compounds MCQs with Answers – Part 2 (NEET Chemistry)

Explanation: \(\textbf{Apply the immiscible-liquid boiling principle:}\) For the stated system containing water and P: \[ P_{\mathrm{total}}=P_{\mathrm{water}}+P_P \] The mixture boils when: \[ P_{\mathrm{total}}=P_{\mathrm{external}} \] \(\textbf{Evaluate }85^\circ\mathrm{C}\textbf{:}\) The combined vapour pressure is: \[ P_{\mathrm{total}}=440+120 \] \[ P_{\mathrm{total}}=560\,\mathrm{mmHg} \] This equals the external pressure in Trial I. Therefore: \[ T_{\mathrm{b,I}}=85^\circ\mathrm{C} \] \(\textbf{Evaluate }92^\circ\mathrm{C}\textbf{:}\) The combined vapour pressure is: \[ P_{\mathrm{total}}=570+160 \] \[ P_{\mathrm{total}}=730\,\mathrm{mmHg} \] This equals the external pressure in Trial II. Consequently: \[ T_{\mathrm{b,II}}=92^\circ\mathrm{C} \] \(\textbf{Check the higher-temperature observation:}\) At \(96^\circ\mathrm{C}\): \[ P_{\mathrm{total}}=650+185 \] \[ P_{\mathrm{total}}=835\,\mathrm{mmHg} \] This exceeds the external pressures used in both trials. Because the combined vapour pressure increases continuously with temperature, the required boiling conditions are reached at lower temperatures. \(\textbf{Interpret the pressure dependence:}\) Trial II has a higher external pressure than Trial I. The immiscible mixture must therefore develop a higher combined vapour pressure before boiling. This requires a higher temperature. The result is consistent with the general pressure dependence of boiling, even though two immiscible liquids jointly contribute to the vapour pressure. \(\textbf{Final answer:}\) \[ \boxed{\left(85^\circ\mathrm{C},\ 92^\circ\mathrm{C}\right)} \]

Explanation: \(\textbf{Examine pure P at atmospheric pressure:}\) The vapour pressure of pure P at \(95^\circ\mathrm{C}\) is: \[ P_P=150\,\mathrm{mmHg} \] For pure P to boil under Condition I: \[ P_P=P_{\mathrm{external}} \] However: \[ 150\,\mathrm{mmHg}\lt 760\,\mathrm{mmHg} \] Therefore, pure P has not reached its boiling condition at \(95^\circ\mathrm{C}\) under ordinary atmospheric pressure. \(\textbf{Examine pure P under reduced pressure:}\) In Condition II: \[ P_{\mathrm{external}}=150\,\mathrm{mmHg} \] The vapour pressure of P at the specified temperature is also \(150\,\mathrm{mmHg}\). Thus: \[ P_P=P_{\mathrm{external}} \] Pure P satisfies its boiling condition under Condition II. \(\textbf{Examine the immiscible mixture:}\) For Condition III, the total vapour pressure is the sum of the separate vapour pressures of the two immiscible liquids. \[ P_{\mathrm{total}}=P_{\mathrm{water}}+P_P \] Substituting: \[ P_{\mathrm{total}}=610+150 \] \[ P_{\mathrm{total}}=760\,\mathrm{mmHg} \] This equals the external pressure. Therefore, the immiscible mixture also satisfies its boiling condition at \(95^\circ\mathrm{C}\). \(\textbf{Compare the two successful methods:}\) Condition II achieves boiling by reducing the external pressure until it equals the vapour pressure of pure P. Condition III achieves boiling at atmospheric pressure because water contributes to the total vapour pressure of the immiscible mixture. The same organic substance therefore reaches a boiling condition at the same temperature through two different physical arrangements. Condition I does not satisfy the boiling criterion. This comparison distinguishes the operating principles of reduced-pressure distillation and steam distillation without treating them as identical processes.

Explanation: \(\textbf{Initial material balance:}\) The initial mass of P is: \[ m_{\mathrm{P,initial}}=24.0\,\mathrm{g} \] The initial mass of Q is: \[ m_{\mathrm{Q,initial}}=36.0\,\mathrm{g} \] The total mass is: \[ m_{\mathrm{initial}}=24.0+36.0 \] \[ m_{\mathrm{initial}}=60.0\,\mathrm{g} \] \(\textbf{P collected in F1:}\) \[ m_{\mathrm{P,F1}}=15.0\times\frac{80.0}{100} \] \[ m_{\mathrm{P,F1}}=12.0\,\mathrm{g} \] \(\textbf{P collected in F2:}\) \[ m_{\mathrm{P,F2}}=20.0\times\frac{45.0}{100} \] \[ m_{\mathrm{P,F2}}=9.0\,\mathrm{g} \] \(\textbf{P collected in F3:}\) \[ m_{\mathrm{P,F3}}=15.0\times\frac{15.0}{100} \] \[ m_{\mathrm{P,F3}}=2.25\,\mathrm{g} \] \(\textbf{Total P recovered in distillate:}\) \[ m_{\mathrm{P,distillate}}=12.0+9.0+2.25 \] \[ m_{\mathrm{P,distillate}}=23.25\,\mathrm{g} \] \(\textbf{P remaining in the flask:}\) By conservation of the mass of P: \[ m_{\mathrm{P,residue}}=24.0-23.25 \] \[ m_{\mathrm{P,residue}}=0.75\,\mathrm{g} \] The residue has a total mass of \(10.0\,\mathrm{g}\). Therefore: \[ \%\mathrm{\ P\ in\ residue}=\frac{0.75}{10.0}\times100 \] \[ \%\mathrm{\ P\ in\ residue}=7.5\% \] \(\textbf{Independent consistency check:}\) The total masses of the collected fractions and residue satisfy: \[ 15.0+20.0+15.0+10.0=60.0\,\mathrm{g} \] The P balance also satisfies: \[ 23.25+0.75=24.0\,\mathrm{g} \] The residue therefore contains: \[ m_{\mathrm{Q,residue}}=10.0-0.75 \] \[ m_{\mathrm{Q,residue}}=9.25\,\mathrm{g} \] This is consistent with preferential removal of the lower-boiling component P during fractionation. The residue becomes enriched in Q as the more volatile P is removed. The problem illustrates how composition data and conservation of individual components can be combined to infer an unmeasured fraction composition. \(\textbf{Final answer:}\) \[ \boxed{7.5\%} \]

Explanation: \(\textbf{Relate distribution to phase amounts:}\) The distribution coefficient is: \[ K=\frac{C_{\mathrm{organic}}}{C_{\mathrm{aqueous}}} \] Since concentration equals mass divided by volume: \[ K=\frac{m_{\mathrm{organic}}/V_{\mathrm{organic}}}{m_{\mathrm{aqueous}}/V_{\mathrm{aqueous}}} \] Rearranging: \[ \frac{m_{\mathrm{organic}}}{m_{\mathrm{aqueous}}} =K\frac{V_{\mathrm{organic}}}{V_{\mathrm{aqueous}}} \] For any extraction stage: \[ m_{\mathrm{initial}}=m_{\mathrm{organic}}+m_{\mathrm{aqueous}} \] Combining these relations: \[ m_{\mathrm{aqueous}}= \frac{m_{\mathrm{initial}}}{1+K(V_{\mathrm{organic}}/V_{\mathrm{aqueous}})} \] \(\textbf{Procedure I — single extraction:}\) Here: \[ V_{\mathrm{organic}}=100\,\mathrm{mL} \] \[ V_{\mathrm{aqueous}}=100\,\mathrm{mL} \] Therefore: \[ m_{\mathrm{aqueous,I}}= \frac{12.0}{1+3.0(100/100)} \] \[ m_{\mathrm{aqueous,I}}=\frac{12.0}{4.0} \] \[ m_{\mathrm{aqueous,I}}=3.00\,\mathrm{g} \] The recovered mass is: \[ m_{\mathrm{recovered,I}}=12.0-3.00 \] \[ m_{\mathrm{recovered,I}}=9.00\,\mathrm{g} \] \(\textbf{Procedure II — first extraction:}\) Each solvent portion has volume \(50\,\mathrm{mL}\). The mass remaining after the first contact is: \[ m_{\mathrm{aqueous,1}}= \frac{12.0}{1+3.0(50/100)} \] \[ m_{\mathrm{aqueous,1}}=\frac{12.0}{2.5} \] \[ m_{\mathrm{aqueous,1}}=4.80\,\mathrm{g} \] \(\textbf{Procedure II — second extraction:}\) The second fresh solvent portion acts on the remaining \(4.80\,\mathrm{g}\) of P. Thus: \[ m_{\mathrm{aqueous,2}}= \frac{4.80}{1+3.0(50/100)} \] \[ m_{\mathrm{aqueous,2}}=\frac{4.80}{2.5} \] \[ m_{\mathrm{aqueous,2}}=1.92\,\mathrm{g} \] The total P recovered through both organic extracts is: \[ m_{\mathrm{recovered,II}}=12.0-1.92 \] \[ m_{\mathrm{recovered,II}}=10.08\,\mathrm{g} \] \(\textbf{Compare recovery:}\) \[ \Delta m=10.08-9.00 \] \[ \Delta m=1.08\,\mathrm{g} \] \(\textbf{Interpretation:}\) Both procedures use \(100\,\mathrm{mL}\) of the same extraction solvent in total. However, the second procedure exposes the remaining aqueous solute to fresh solvent after the first extract has been removed. The repeated equilibrium distribution therefore leaves less P in the aqueous phase. The improvement follows from the distribution relation and not from any chemical conversion of P. \(\textbf{Final answer:}\) \[ \boxed{1.08\,\mathrm{g}} \]

Explanation: \(\textbf{First extraction — transfer of P:}\) Initially: \[ m_{\mathrm{P,0}}=20.0\,\mathrm{g} \] The first solvent portion extracts \(60\%\) of P. \[ m_{\mathrm{P,1}}=0.60\times20.0 \] \[ m_{\mathrm{P,1}}=12.0\,\mathrm{g} \] \(\textbf{First extraction — transfer of Q:}\) Initially: \[ m_{\mathrm{Q,0}}=5.0\,\mathrm{g} \] The first portion extracts \(20\%\) of Q. \[ m_{\mathrm{Q,1}}=0.20\times5.0 \] \[ m_{\mathrm{Q,1}}=1.0\,\mathrm{g} \] The first dried residue has mass: \[ m_{\mathrm{residue,1}}=12.0+1.0 \] \[ m_{\mathrm{residue,1}}=13.0\,\mathrm{g} \] Its percentage purity is: \[ \%\mathrm{\ purity_1}=\frac{12.0}{13.0}\times100 \] \[ \%\mathrm{\ purity_1}\approx92.3\% \] \(\textbf{Determine the remaining aqueous composition:}\) After the first extraction: \[ m_{\mathrm{P,remaining}}=20.0-12.0 \] \[ m_{\mathrm{P,remaining}}=8.0\,\mathrm{g} \] For Q: \[ m_{\mathrm{Q,remaining}}=5.0-1.0 \] \[ m_{\mathrm{Q,remaining}}=4.0\,\mathrm{g} \] \(\textbf{Second extraction:}\) The second fresh solvent portion removes \(60\%\) of the remaining P. \[ m_{\mathrm{P,2}}=0.60\times8.0 \] \[ m_{\mathrm{P,2}}=4.80\,\mathrm{g} \] It removes \(20\%\) of the remaining Q. \[ m_{\mathrm{Q,2}}=0.20\times4.0 \] \[ m_{\mathrm{Q,2}}=0.80\,\mathrm{g} \] The second residue has mass: \[ m_{\mathrm{residue,2}}=4.80+0.80 \] \[ m_{\mathrm{residue,2}}=5.60\,\mathrm{g} \] Therefore: \[ \%\mathrm{\ purity_2}=\frac{4.80}{5.60}\times100 \] \[ \%\mathrm{\ purity_2}\approx85.7\% \] \(\textbf{Explain the purity difference:}\) The first extraction preferentially removes P, leaving the aqueous solution relatively enriched in Q. Although the same fractional extraction behaviour applies in the second contact, the composition of the material entering that contact has changed. The second extract consequently contains a greater relative proportion of Q. Thus successive extraction can increase total recovery of P while later collected extracts need not have the same purity as the first. \(\textbf{Final answer:}\) \[ \boxed{\left(92.3\%,\ 85.7\%\right)} \]

Explanation: The original extraction has transferred most of the desired compound P from the aqueous phase into solvent X. However, the collected organic extract also contains a small amount of impurity Q, so extraction has not achieved complete purification. The supplied solubility information indicates that Q is more favourably distributed into water than into X. P, in contrast, strongly prefers the organic phase. Washing the organic extract with a suitable amount of water can therefore promote transfer of Q from X into the aqueous wash layer. Much of P can remain in X because its solubility behaviour favours that phase. After separation of the wash water, the organic solvent can be removed to recover a residue enriched in P. This sequence uses a difference in solubility while the compounds are still dissolved in separable liquid phases. Evaporating X before washing removes the organic liquid phase required for that ordinary liquid-liquid extraction step. A later wash of the resulting solid may still affect the mixture, but its outcome depends on the solubility and physical behaviour of the solids and is not guaranteed to equal that of washing the organic solution. Neither washing procedure implies chemical conversion of Q into an inactive substance. Also, preferential retention of P in X does not mean that absolutely no P can be lost into the aqueous wash. The correct inference is that washing before solvent removal provides a suitable opportunity for selective impurity transfer under the stated solubility conditions.

Explanation: Chromatographic separation requires the mixture components to migrate differently under the selected stationary-phase and mobile-phase conditions. In S1, both P and Q have relatively small retardation factors. Their values are also close, indicating that both compounds remain near the baseline and are poorly discriminated by the selected system. In S3, both compounds have retardation factors close to unity. They move almost as far as the solvent front, but their relative migration distances remain very similar. Such near-front co-migration does not provide strong evidence of effective separation. S2 produces substantially different retardation factors, with P at \(0.38\) and Q at \(0.67\). These values indicate a more useful distinction between the migration behaviours of the two compounds. P remains comparatively more strongly retained, while Q travels appreciably farther with the mobile phase. The chromatographic spots are therefore more likely to be distinguishable under S2 than under either of the other stated conditions, assuming otherwise comparable development. This conclusion does not require identifying the molecular masses of P and Q. It also does not imply that the solvent producing the greatest absolute migration is automatically the best. Appropriate chromatographic development requires useful discrimination between the components rather than simply maximising or minimising their retardation factors.

Explanation: Paper chromatography is treated as partition chromatography in the NCERT framework. The retained water in the paper acts as the stationary liquid phase, while the developing solvent moves through the paper by capillary action. A compound more strongly associated with the stationary water phase generally spends a greater proportion of its migration process in that phase. Consequently, it tends to move less readily with the mobile solvent under otherwise comparable conditions. In the first solvent system, P is preferentially retained by the stationary phase, while Q is more favourably distributed into the mobile phase. This behaviour supports Q travelling farther than P. The second experiment changes the relative distribution behaviour of both compounds. P becomes more favourable in the mobile phase, while Q becomes relatively more strongly associated with the stationary water phase. Under sufficiently different distributions, P may consequently travel farther than Q. The relative migration order therefore need not remain fixed when the developing solvent changes. The experiment does not require a chemical reaction involving either substance. Nor does a change in mobile solvent automatically change the principal partition mechanism into adsorption chromatography. The decisive property is the relative distribution of each component between the phases established by the selected chromatographic system.

Explanation: The separation of compounds by TLC depends on their different interactions with the stationary adsorbent and developing solvent. However, the physical manner in which a sample is applied can also affect the clarity of the resulting chromatogram. Applying an excessive quantity or producing a broad initial sample zone can lead to poorly defined chromatographic regions. This can make it difficult to distinguish components even when they possess different migration behaviours. In the repeated experiment, the amount of sample is reduced and the initial spot is made narrow. Under the stated conditions, P now produces a compact region that can be distinguished from Q. The improvement therefore concerns the physical resolution of the chromatographic zones rather than a necessary alteration in molecular identity. The stationary phase remains silica gel in both experiments. No reaction or chemical conversion is described, and none is required to account for the observations. The experiment also provides no basis for claiming that the molecular mass of P changed between the two trials. A narrow and appropriately sized starting zone allows the differences in component migration to be observed more clearly. The correct interpretation distinguishes chromatographic separation behaviour from the quality of sample application and zone formation.

Explanation: The supplied data establish two different physical-property distinctions relevant to purification. P and Q both sublime appreciably under the available heating conditions. Consequently, ordinary collection of the complete sublimate would not necessarily separate them from each other. R is non-sublimable, so sublimation could remove R from a deposited P-Q mixture, but the deposited fraction would still contain both P and Q. Solvent S provides a more useful combined separation route. When the mixture is heated with S, P and Q dissolve while R remains undissolved. Hot filtration can therefore remove R before the solution is cooled. The filtrate contains dissolved P and Q. On cooling, the solubility of P decreases appreciably, favouring its crystallisation. Q remains highly soluble and can therefore remain predominantly in the mother liquor. The resulting crystals are consequently enriched in P relative to the original mixture. Collecting suspended solids before removing R would risk including the insoluble impurity in the final product. The desired procedure must distinguish the role of hot filtration from the subsequent crystallisation stage. The correct sequence exploits the insolubility of R and the contrasting cold solubilities of P and Q rather than relying on their shared ability to sublime.

Explanation: Simple distillation ideally separates a volatile liquid from a suitable nonvolatile impurity through selective vaporisation and subsequent condensation. The volatile component P enters the vapour phase, while nonvolatile Q is expected to remain in the heated flask. However, the presence of Q in the collected distillate does not automatically establish that Q underwent equilibrium vaporisation. Vigorous boiling can mechanically carry small droplets of the original liquid mixture into the vapour outlet. These droplets contain dissolved or suspended material from the original liquid phase and may therefore transport Q without Q itself becoming appreciably volatile. If such droplets reach the receiving system, the collected distillate can become contaminated. TLC analysis can reveal the resulting contamination through a detectable spot matching reference Q. However, TLC establishes chromatographic evidence for the component rather than identifying the physical mechanism by which it reached the receiver. The absence of decomposition evidence also does not justify assuming a change in chemical identity or a sudden decrease in the boiling point of Q. The observations are therefore compatible with mechanical carry-over as an alternative to genuine co-vaporisation. A reliable interpretation must distinguish the composition of the collected material from the physical process responsible for transporting each constituent. This distinction is essential when assessing whether the distillation conditions have produced effective purification.

Explanation: \(\textbf{Initial composition:}\) The original sample contains: \[ m_{\mathrm{P,initial}}=24.0\,\mathrm{g} \] \[ m_{\mathrm{Q,initial}}=4.0\,\mathrm{g} \] \[ m_{\mathrm{R,initial}}=2.0\,\mathrm{g} \] The total crude mass is: \[ m_{\mathrm{crude}}=24.0+4.0+2.0 \] \[ m_{\mathrm{crude}}=30.0\,\mathrm{g} \] \(\textbf{Hot filtration:}\) R remains insoluble and is removed by hot filtration. The question specifies that the dissolved substances are not lost during this operation. Therefore, the filtrate still contains the original \(24.0\,\mathrm{g}\) of P and \(4.0\,\mathrm{g}\) of Q. \(\textbf{First crystallisation:}\) After cooling, \(6.0\,\mathrm{g}\) of P remains dissolved. The amount of P crystallised is: \[ m_{\mathrm{P,first}}=24.0-6.0 \] \[ m_{\mathrm{P,first}}=18.0\,\mathrm{g} \] Q remains completely dissolved in the first mother liquor. The first crystalline crop therefore contains \(18.0\,\mathrm{g}\) of P under the stated assumptions. \(\textbf{Second crystallisation:}\) The second purification stage begins with the \(18.0\,\mathrm{g}\) of P recovered from the first stage. During recrystallisation, \(20\%\) of this P remains dissolved in the new mother liquor. \[ m_{\mathrm{P,lost\ in\ second}}=\frac{20}{100}\times18.0 \] \[ m_{\mathrm{P,lost\ in\ second}}=3.60\,\mathrm{g} \] The final mass of P recovered as crystals is: \[ m_{\mathrm{P,final}}=18.0-3.60 \] \[ m_{\mathrm{P,final}}=14.40\,\mathrm{g} \] \(\textbf{Calculate overall recovery:}\) The denominator must be the amount of P originally present in the crude sample. \[ \%\mathrm{\ recovery}=\frac{14.40}{24.0}\times100 \] \[ \%\mathrm{\ recovery}=60.0\% \] \(\textbf{Check the two-stage recovery:}\) The first crystallisation recovers: \[ \frac{18.0}{24.0}\times100=75.0\% \] The second crystallisation retains \(80.0\%\) of the P entering that stage. Therefore: \[ \%\mathrm{\ overall\ recovery}=0.75\times0.80\times100 \] \[ \%\mathrm{\ overall\ recovery}=60.0\% \] The two calculations agree. The insoluble impurity R is removed by filtration and does not enter the recovery calculation for P. Similarly, Q remains in the mother liquor and does not contribute to the final crystalline product under the stated assumptions. The decrease in recovered mass arises from the finite solubility of P during the two crystallisation stages. The final crystalline product may be pure P while representing only \(60\%\) of the desired compound originally present. \(\textbf{Final answer:}\) \[ \boxed{60.0\%} \]

Explanation: Fractional distillation improves the separation of suitable miscible liquids through repeated vaporisation and condensation within the fractionating column. The ascending vapour and descending condensed liquid exchange heat, allowing further separation according to relative volatility. Because P is more volatile than Q, the vapour moving upwards can become progressively enriched in P. In Trial I, the mole fraction of P increases substantially from 0.45 at the column entrance to 0.88 near the condenser. This observation indicates effective enrichment of the more volatile component. Trial II begins with the same incoming vapour composition but produces an upper vapour containing only 0.52 mole fraction of P. The greatly reduced descending liquid flow and poor vapour-liquid contact provide a physical explanation for this smaller improvement. Rapid heating does not automatically increase chromatographic or distillation selectivity; the intended repeated separation stages must operate effectively. The observations do not require any chemical conversion of Q into P. Nor do they establish that the intrinsic boiling-point order has reversed. The correct interpretation connects column operation with the measured change in vapour composition rather than judging performance merely from the quantity of vapour produced.

Explanation: \(\textbf{Initial two-liquid condition:}\) For the specified immiscible-liquid system, the total equilibrium vapour pressure is the sum of the separate vapour pressures when both liquid phases are present. \[ P_{\mathrm{total}}=P_{\mathrm{water}}+P_P \] Substituting: \[ P_{\mathrm{total}}=630+130 \] \[ P_{\mathrm{total}}=760\,\mathrm{mmHg} \] This equals the external pressure. Therefore, the original two-liquid mixture satisfies the boiling condition at \(95^\circ\mathrm{C}\). \(\textbf{After disappearance of liquid P:}\) The stated additive relation applies to the system containing both immiscible liquid phases. Once the liquid P phase has disappeared, water is the only remaining liquid phase under the specified conditions. The vapour pressure of water at \(95^\circ\mathrm{C}\) is: \[ P_{\mathrm{water}}=630\,\mathrm{mmHg} \] The imposed external pressure remains: \[ P_{\mathrm{external}}=760\,\mathrm{mmHg} \] Therefore: \[ P_{\mathrm{water}}\lt P_{\mathrm{external}} \] The remaining water alone does not satisfy its boiling condition at this temperature. \(\textbf{Chemical interpretation:}\) The contribution of P to the original total vapour pressure is essential to the lower-temperature boiling of the immiscible mixture. Once P is no longer present as a separate liquid phase, its former contribution cannot simply be retained in the calculation for the remaining pure water. Water would require a higher temperature to develop sufficient vapour pressure for boiling at the same external pressure. Thus the validity of the vapour-pressure summation must be checked against the actual phases present, not applied mechanically after the composition of the flask has changed.

Explanation: Glycerol is a high-boiling organic liquid, and its recovery from spent-lye is a textbook application of distillation under reduced pressure. The relevant difficulty is the high temperature required to boil glycerol under ordinary atmospheric pressure. Reducing the external pressure decreases the vapour pressure required for boiling. Because the vapour pressure of a liquid increases with temperature, glycerol can reach its boiling condition at a lower temperature under suitable reduced pressure. This allows the volatile glycerol to enter the vapour phase without requiring the same thermal exposure as ordinary atmospheric-pressure distillation. The glycerol vapour is then condensed and collected separately. The dissolved substances described as nonvolatile do not appreciably enter the vapour phase under the selected conditions. They consequently remain predominantly in the original flask. Fractional distillation is specifically valuable when multiple volatile liquid components require separation, but the present problem emphasises a high-boiling liquid and effectively nonvolatile dissolved material. Steam distillation is not automatically suitable for every high-boiling organic liquid because it has its own steam-volatility and immiscibility requirements. The preferred method must therefore address both the boiling-temperature limitation and the volatility difference between the desired substance and its impurities.

Explanation: The component that should be collected as the final product depends on the desired substance, not merely on which part of the apparatus receives condensate. In the present mixture, P is nonvolatile under the operating conditions. X and Q are volatile organic liquids that can enter the vapour phase during distillation. Their vapours can be condensed and collected as separate fractions because the question specifies that their volatility difference is sufficient for effective fractionation. P, however, does not appreciably vaporise during this operation. It therefore becomes concentrated in the original distillation flask as X and Q are removed. The distillation apparatus separates the volatile components from the nonvolatile residue even though the objective is to retain the residue rather than recover a particular distillate. This is different from purifying a volatile organic liquid from a nonvolatile impurity, where the desired substance would usually be collected after condensation. The largest starting component need not be the first recovered desired substance. Nor does the fact that P was dissolved in X force it to vaporise with X. Under the stated assumptions, the flask residue is the P-containing fraction, while the distillates contain the removed volatile materials.

Explanation: Statement I is correct because the deposited solid gives two separated TLC spots corresponding to the supplied reference positions. Under the stated detection conditions, this is evidence that the deposit contains a detectable component consistent with Q in addition to P. Therefore, the deposit cannot be regarded as chromatographically pure P. Statement II is correct because the heated residue still gives a spot matching reference P. The observation indicates that the heating and collection process has not removed all detectable P from the starting material. Thus deposition of P does not automatically imply complete recovery of the original desired compound. Statement III is correct because Q is described as having limited but nonzero sublimation under the selected conditions. Some Q can therefore enter the vapour phase and subsequently deposit with P. This provides a physical explanation for its appearance in the collected material. Statement IV is incorrect because the observations are consistent with ordinary physical transfer of the original substances. The TLC results do not establish the formation of any new decomposition product. A sublimation procedure may therefore be incomplete in two distinct ways: some desired material can remain in the residue, and some impurity can enter the collected deposit. Both aspects must be considered when evaluating purification performance.

Explanation: Column chromatography separates mixture components according to their relative interactions with the stationary adsorbent and the moving eluant. Silica gel remains the stationary phase throughout the described experiment. With the first eluant S1, P moves sufficiently readily to leave the column, whereas Q and R remain retained within it. Their retention does not imply that they are permanently attached to the stationary phase or insoluble in every possible solvent. Instead, it describes their behaviour under the specific stationary-phase and mobile-phase conditions established by S1. Introducing S2 changes the mobile phase and therefore changes the relative migration behaviour of the remaining components. The question explicitly states that Q becomes more mobile than R under the new conditions. Q can consequently be collected before R as the second eluant passes through the column. No chemical transformation is required to explain this altered movement. Silica gel remains a solid stationary adsorbent, so merely changing the eluant does not establish a change from adsorption to partition chromatography. The initial and subsequent elution sequences are therefore condition-dependent observations rather than fixed classifications based on molecular mass. The experiment illustrates how an appropriate mobile phase can be selected to obtain useful separation of components that were strongly retained under an earlier solvent condition.

Explanation: Chromatographic separation and chromatographic detection are distinct operations. The movement of P during TLC development is governed by its relative interactions with the stationary adsorbent and the moving solvent. After development, the compound may remain colourless and therefore be difficult to observe directly. Exposure to iodine vapours is a useful method for revealing suitable organic substances through interaction with iodine. The resulting visible staining need not be permanent. When the plate is removed from the iodine atmosphere, the intensity of the staining may decrease as iodine is lost from the treated region. This fading does not require the original organic compound to have moved from its chromatographic position. The question also states that another suitable revealing method still detects material corresponding to P at that position. This supports the interpretation that the disappearance of visible iodine staining should not be confused with disappearance of the analyte. The iodine observation alone cannot establish that P has decomposed. Nor does revealing a spot necessarily involve permanent chemical conversion of every detectable substance. The appropriate conclusion distinguishes the presence of chromatographic material from the temporary visibility produced by a particular detecting method.

Explanation: Chromatographic separation occurs because components move at different relative rates through or across a stationary phase. However, the physical direction of movement depends on the arrangement of the chromatographic system. In the silica-gel column described, the eluant flows downwards. A component that moves more readily with this downward-flowing mobile phase travels farther towards the lower outlet. P is therefore found below Q when P is the more mobile component. In the paper chromatographic arrangement, the solvent moves upwards by capillary action. A more mobile component now travels farther upwards from the baseline rather than downwards towards an outlet. Thus P appears above Q on the developed paper. The different directions of movement create different spatial arrangements even though P is the more mobile substance in both experiments. A band located nearer the top of a downward-flowing column is not automatically the more mobile band. Similarly, the compound appearing highest on an ascending paper chromatogram is not necessarily the most strongly retained substance. The correct interpretation requires identifying the mobile-phase direction before comparing component positions. The observations therefore agree with the same relative mobility order and do not imply a contradiction between the two chromatographic experiments.