Purification and Characterisation of Organic Compounds MCQs with Answers – Part 1 (NEET Chemistry)

Explanation: Crystallisation is a purification method based on differences in the solubilities of a compound and its impurities in a suitable solvent. A desirable property of the solvent is that the compound dissolves appreciably at higher temperature but remains sparingly soluble at lower temperature. Heating therefore allows a relatively large amount of the impure compound to enter the solution. When the concentrated hot solution is cooled, the solvent can no longer retain the same amount of dissolved compound. The excess compound separates from the solution as crystals. Suitable soluble impurities remain predominantly in the liquid phase, allowing the crystals to be separated by filtration. The separation does not depend on converting the compound into vapour or chemically destroying its impurities. Importantly, effective crystallisation requires consideration of impurity solubility as well as the temperature dependence of the desired compound's solubility.

Explanation: Statement I is correct because the mother liquor is the liquid remaining after crystals have been separated from the cooled solution. It generally contains impurities that remain soluble under the crystallisation conditions. It also retains some quantity of the desired organic compound because the compound usually has a finite solubility even at the lower temperature. Statement II is incorrect because saturation at a higher temperature does not imply that solubility becomes zero on cooling. Only the quantity exceeding the lower-temperature solubility can crystallise under ideal equilibrium conditions. The remaining dissolved compound contributes to incomplete recovery. Consequently, obtaining relatively pure crystals is not equivalent to recovering the entire amount of the original compound. This distinction between purity and recovery is essential when evaluating the effectiveness of crystallisation. Therefore, only Statement I is correct.

Explanation: The initial requirement is to transfer the desired organic compound into solution without dissolving the insoluble impurity. Heating with a suitable solvent accomplishes this because the desired compound has appreciable solubility at higher temperature. The insoluble impurity must then be removed while the solution is sufficiently hot to keep the desired compound dissolved. If substantial cooling occurs before this filtration, the desired compound may crystallise together with the insoluble impurity. The hot filtrate is therefore collected separately from the undissolved material. On subsequent cooling, the solubility of the desired compound decreases and crystals form. These crystals are recovered by filtration from the mother liquor. The procedure separates the insoluble impurity before crystallisation rather than allowing it to become mixed with the recovered solid. The correct sequence follows the distinct purposes of dissolution, impurity removal, crystallisation and final separation.

Explanation: \(\textbf{Solvent-selection principle:}\) An effective crystallisation solvent must dissolve the required amount of P when hot while retaining relatively little P after cooling. It should also keep the unwanted impurity Q dissolved rather than causing it to precipitate with P. For S1, \(100\,\mathrm{mL}\) of hot solvent can dissolve \(50\,\mathrm{g}\) of P, which is sufficient for the supplied \(40\,\mathrm{g}\). At lower temperature, only \(5\,\mathrm{g}\) of P remains dissolved. The potential crystal recovery is therefore: \[ m_{\mathrm{P,\ crystals}}=40-5=35\,\mathrm{g} \] The complete \(3\,\mathrm{g}\) of Q can remain dissolved because its lower-temperature solubility is \(10\,\mathrm{g}\) per \(100\,\mathrm{mL}\). For S2, the lower-temperature solubility of P is \(40\,\mathrm{g}\), so the available solution need not produce any P crystals on cooling. For S3, P crystallises readily, but the lower-temperature solubility of Q is only \(0.5\,\mathrm{g}\), creating a risk of impurity precipitation and contamination. For S4, the hot solvent cannot dissolve the entire \(40\,\mathrm{g}\) of P in the specified \(100\,\mathrm{mL}\). Thus S1 simultaneously satisfies the essential hot-solubility, cold-solubility and impurity-separation requirements. The solvent with the lowest cold solubility of P is not necessarily the best unless the behaviour of the impurity is also considered.

Explanation: \(\textbf{Given:}\) Mass of crude sample: \[ m_{\mathrm{sample}}=50.0\,\mathrm{g} \] Mass percentage of P: \[ w_{\mathrm P}=90.0\% \] Lower-temperature solubility of P: \[ S_{\mathrm{cold}}=9.0\,\mathrm{g}/100\,\mathrm{mL} \] \(\textbf{Initial quantity of the desired compound:}\) The starting material is not pure P. Therefore, the mass of P must first be calculated from its percentage composition. \[ m_{\mathrm{P,\ initial}}=\frac{90.0}{100}\times50.0 \] \[ m_{\mathrm{P,\ initial}}=45.0\,\mathrm{g} \] \(\textbf{Quantity remaining in the mother liquor:}\) At equilibrium after cooling, the \(100\,\mathrm{mL}\) solution retains \(9.0\,\mathrm{g}\) of P. \[ m_{\mathrm{P,\ dissolved}}=9.0\,\mathrm{g} \] \(\textbf{Crystal recovery:}\) The remaining P is recovered as crystals. \[ m_{\mathrm{P,\ crystals}}=45.0-9.0 \] \[ m_{\mathrm{P,\ crystals}}=36.0\,\mathrm{g} \] The relevant percentage recovery must be calculated relative to the original quantity of P, not the entire crude sample. \[ \%\mathrm{\ recovery}=\frac{36.0}{45.0}\times100 \] \[ \%\mathrm{\ recovery}=80.0\% \] \(\textbf{Consistency check:}\) The unrecovered \(9.0\,\mathrm{g}\) of P represents \(20\%\) of the original P. The impurity contributes to the crude sample mass but does not increase the quantity of P available for crystallisation. \(\textbf{Final answer:}\) \[ \boxed{80\%} \]

Explanation: The Assertion is true because removal of coloured impurities and removal of other soluble impurities are not necessarily the same process. Activated charcoal possesses a surface capable of adsorbing suitable coloured substances present in a solution. These impurities become associated with the charcoal rather than remaining dissolved in the liquid phase. Removal of the charcoal carrying the adsorbed impurities can therefore reduce or eliminate the colour of the solution. However, other impurities may not be appreciably adsorbed under the same conditions. Such impurities may remain dissolved even after the solution becomes colourless. Subsequent crystallisation separates the desired compound according to solubility differences, while appropriately soluble impurities remain in the mother liquor. The Reason is consequently true and directly explains why colour removal does not establish the complete removal of dissolved impurities. Activated charcoal should not be confused with the agent responsible for the actual crystallisation of the desired compound.

Explanation: Statement I is correct because R remains undissolved even when the mixture is heated with the selected solvent. Removing it by hot filtration prevents this insoluble material from remaining with the desired crystals. The filtrate contains dissolved P and Q. Statement II is correct because cooling decreases the solubility of P but does not necessarily reduce it to zero. Consequently, some P remains in the mother liquor together with Q. Statement III is incorrect because evaporating the mother liquor completely can recover both its residual P and the soluble impurity Q. The resulting residue cannot automatically be regarded as purer than crystals separated from the mother liquor. Statement IV is correct because the equilibrium quantity of dissolved P depends on temperature. If cooling stops at a temperature where P is still appreciably soluble, less P will crystallise than at a suitable lower temperature. The overall purification therefore relies on selective dissolution, removal of insoluble material, controlled crystallisation and separation from dissolved impurities. These operations have different functions and cannot be replaced indiscriminately by complete evaporation.

Explanation: \(\textbf{Verify dissolution:}\) The hot solubility of P is sufficient to dissolve the original \(54\,\mathrm{g}\) in either solvent volume. \[ S_{\mathrm{hot}}=70\,\mathrm{g}/100\,\mathrm{mL} \] The complete initial quantity of P is therefore present in solution before cooling in both trials. \(\textbf{First trial:}\) The lower-temperature solubility is: \[ S_{\mathrm{cold}}=10\,\mathrm{g}/100\,\mathrm{mL} \] In \(100\,\mathrm{mL}\) of solvent: \[ m_{\mathrm{dissolved,1}}=\frac{10}{100}\times100=10\,\mathrm{g} \] The mass recovered as crystals is: \[ m_{\mathrm{crystals,1}}=54-10=44\,\mathrm{g} \] \(\textbf{Second trial:}\) The final solvent volume is increased to \(150\,\mathrm{mL}\). \[ m_{\mathrm{dissolved,2}}=\frac{10}{100}\times150 \] \[ m_{\mathrm{dissolved,2}}=15\,\mathrm{g} \] Hence: \[ m_{\mathrm{crystals,2}}=54-15=39\,\mathrm{g} \] \(\textbf{Compare recoveries:}\) \[ \Delta m=44-39 \] \[ \Delta m=5\,\mathrm{g} \] Both trials started with the same amount of P, but the second mother liquor can retain more P because of its larger volume. The additional solvent reduces recovery without changing the intrinsic lower-temperature solubility of P. \(\textbf{Final answer:}\) \[ \boxed{5\,\mathrm{g}} \]

Explanation: The difficulty with solvent X alone is that P remains highly soluble even after cooling. A solution prepared only in X may therefore retain a large quantity of P rather than producing useful crystallisation. Solvent Y presents the opposite difficulty because P is sparingly soluble even at higher temperature. Consequently, Y alone is not suitable for dissolving a substantial amount of the original solid. A suitably chosen mixture of X and Y can overcome these limitations. P is first dissolved using the favourable dissolving ability of hot X. Gradual introduction of Y decreases the dissolving capacity of the mixed-solvent system and allows the composition to approach conditions suitable for crystallisation. Cooling can then cause P to separate because its solubility in the chosen mixed solvent decreases under the final conditions. The addition must be controlled so that excessive nonsolvent does not simply cause uncontrolled precipitation before the desired cooling stage. Mutual miscibility is important because the process requires a reasonably uniform mixed-solvent phase rather than two independently separated liquid layers. The suitable procedure therefore combines hot dissolution, controlled solvent-composition adjustment and cooling.

Explanation: \(\textbf{Original amount of pure compound:}\) Mass of crude sample: \[ m_0=20.0\,\mathrm{g} \] Initial mass percentage of P: \[ w_0=80.0\% \] Thus: \[ m_{\mathrm{P,0}}=\frac{80.0}{100}\times20.0 \] \[ m_{\mathrm{P,0}}=16.0\,\mathrm{g} \] \(\textbf{After the first crystallisation:}\) The recovered solid has a mass of \(14.0\,\mathrm{g}\), but it contains only \(95.0\%\) P. \[ m_{\mathrm{P,1}}=\frac{95.0}{100}\times14.0 \] \[ m_{\mathrm{P,1}}=13.3\,\mathrm{g} \] Therefore, the first crystallisation has already caused some P to remain outside the recovered crystalline fraction. \(\textbf{After recrystallisation:}\) The final recovered mass is \(11.0\,\mathrm{g}\), with \(99.0\%\) P. \[ m_{\mathrm{P,final}}=\frac{99.0}{100}\times11.0 \] \[ m_{\mathrm{P,final}}=10.89\,\mathrm{g} \] \(\textbf{Calculate overall recovery:}\) The required denominator is the original \(16.0\,\mathrm{g}\) of P, not the original crude mass or the mass recovered after the first purification. \[ \%\mathrm{\ recovery}=\frac{10.89}{16.0}\times100 \] \[ \%\mathrm{\ recovery}=68.0625\% \] \[ \%\mathrm{\ recovery}\approx68.1\% \] \(\textbf{Interpretation:}\) The purity has increased from \(80.0\%\) initially to \(99.0\%\) in the final product. Nevertheless, the recovery of P is substantially less than \(100\%\). This demonstrates why improved purity and increased recovery are not equivalent outcomes of repeated crystallisation. \(\textbf{Final answer:}\) \[ \boxed{68.1\%} \]

Explanation: Crystallisation is most effective when the desired compound and its impurities behave sufficiently differently in the selected solvent. If P and Q have comparable solubilities, cooling may not separate them as effectively as desired. Depending on their amounts and solubility behaviour, some impurity may separate along with the desired compound. Therefore, the formation of crystals alone is not proof of complete purification. Repeated crystallisation may improve the composition of the recovered crystalline material by redistributing the impurity between crystals and mother liquor. However, each crystallisation can leave some desired P dissolved in the separated liquid. This introduces a possible loss of the desired compound during successive purification stages. Simply increasing the volume of solvent is not a guaranteed solution because it can increase the amount of P remaining dissolved after cooling. The appropriate decision involves comparing the improvement in purity with the reduction in recovery. Where the solubility behaviour allows, a more selective solvent may also be considered rather than assuming repeated crystallisation will necessarily provide complete separation.

Explanation: \(\textbf{Initial cooling condition:}\) The initial solution contains: \[ m_{\mathrm P}=50\,\mathrm{g} \] \[ V_1=500\,\mathrm{mL} \] The solubility at the final temperature is: \[ S_{\mathrm{cold}}=12\,\mathrm{g}/100\,\mathrm{mL} \] The maximum amount of P that can remain dissolved in \(500\,\mathrm{mL}\) is: \[ m_{\mathrm{capacity,1}}=\frac{12}{100}\times500 \] \[ m_{\mathrm{capacity,1}}=60\,\mathrm{g} \] Since the solution contains only \(50\,\mathrm{g}\) of P, it is not saturated at the final temperature. Thus the observation of no crystallisation is consistent with the supplied solubility. \(\textbf{Effect of concentrating the solution:}\) The volume is reduced to: \[ V_2=200\,\mathrm{mL} \] The mass of P remains \(50\,\mathrm{g}\), because the solvent is evaporated without loss of solute. At the same final temperature, the smaller volume can retain: \[ m_{\mathrm{capacity,2}}=\frac{12}{100}\times200 \] \[ m_{\mathrm{capacity,2}}=24\,\mathrm{g} \] \(\textbf{Crystal formation:}\) The amount exceeding the new lower-temperature dissolving capacity separates as crystals. \[ m_{\mathrm{crystals}}=50-24 \] \[ m_{\mathrm{crystals}}=26\,\mathrm{g} \] \(\textbf{Chemical interpretation:}\) Concentrating the hot solution does not require changing the intrinsic solubility of P. Instead, it reduces the total quantity of P that the remaining solvent can retain after cooling. This explains why a solution that initially failed to crystallise may yield crystals after suitable concentration. \(\textbf{Final answer:}\) \[ \boxed{26\,\mathrm{g}} \]

Explanation: Sublimation involves the direct conversion of a solid into vapour without passage through the liquid state under the stated conditions. Compound P possesses this property and can therefore enter the vapour phase during suitable heating. Impurity Q is specified as non-sublimable and consequently remains in the heated container. The vapour containing P travels away from the original solid mixture. When it encounters a suitably cool surface, it undergoes deposition to form solid P again. The deposited material is therefore the fraction intended for recovery. This separation differs from ordinary distillation, where the collected vapour is generally condensed to a liquid. The purification is effective because P and Q exhibit different phase-change behaviour under the selected conditions. It should not be assumed that every heated organic solid is suitable for sublimation. The method requires the substance of interest to sublime under conditions that provide useful discrimination from its impurities.

Explanation: \(\textbf{Behaviour of mixture M:}\) P sublimes readily under the selected heating conditions. Q is non-sublimable and remains predominantly in the heated container. This difference allows P to enter the vapour phase while Q is left behind. Cooling the vapour can therefore produce a deposit enriched in P, with effective purification under suitable operating conditions. \(\textbf{Behaviour of mixture N:}\) Both R and S are capable of subliming readily under the stated conditions. Heating can consequently transfer both constituents into the vapour phase. Since their vapours are collected together, deposition may produce a mixture of R and S rather than a pure sample of R. Sublimation is therefore not automatically an effective purification method merely because the desired substance is sublimable. The decisive requirement is a sufficiently useful difference between the behaviours of the desired compound and the impurities. These observations distinguish a favourable sublimation separation from a case where the same phase-change process can occur for both constituents. The statement that a deposited solid must be pure is consequently unjustified without considering the behaviour of every relevant component.

Explanation: \(\textbf{Identify the sublimable constituent:}\) Only impurity U is capable of subliming under the specified heating conditions. It can therefore pass into the vapour phase and subsequently deposit on the cooled collecting surface. The deposited fraction originates mainly from U rather than from the desired compound T. \(\textbf{Determine the heated residue:}\) T is explicitly non-sublimable and consequently remains in the heated container. Impurity V is also non-sublimable and remains there under the same conditions. The heated residue therefore contains both T and V. Removal of U reduces the number of impurities present but does not establish that the residue is pure T. The effectiveness of purification must be judged against all impurities rather than only the one removed by the selected technique. This situation illustrates why choosing an appropriate phase-change method and obtaining a single visibly separated deposit do not necessarily complete the purification. A further suitable separation would be required if pure T were desired. The correct inference depends on tracking the destination of each constituent rather than assuming that the residue or deposit is automatically pure.

Explanation: \(\textbf{Interpret the failed separation:}\) The trial sublimation shows that both P and Q can enter the vapour phase under the available heating conditions. Their vapours are collected together, so the resulting deposit contains both compounds. Repeating the same process under identical conditions does not introduce a new separation principle capable of selectively removing Q. \(\textbf{Use the new solubility information:}\) The alternative solvent provides a useful distinction between the two compounds. P dissolves appreciably at higher temperature but becomes sparingly soluble when the solution is cooled. Q remains highly soluble at both temperatures and is therefore more likely to remain in the mother liquor. Dissolving the mixture in the hot solvent allows both components to enter the solution initially. Subsequent cooling selectively favours crystallisation of P while Q remains dissolved. Filtration can then separate the P-rich crystals from the solution containing Q. Complete evaporation is unsuitable because it would recover the nonvolatile dissolved constituents together rather than exploiting their different solubilities. The correct purification strategy follows from identifying why sublimation failed and replacing it with a method that uses a genuinely different physical-property distinction. Successful purification depends on the relative behaviour of the components, not merely on whether the desired compound is compatible with a particular technique.

Explanation: Simple distillation separates substances by exploiting differences in volatility. Compound P is a volatile liquid that can vaporise when heated to its boiling temperature. Impurity Q is specified as nonvolatile under the distillation conditions, so it does not enter the vapour phase appreciably. The vapour leaving the heated flask is therefore enriched in P. Passing this vapour through a condenser removes heat and converts it back into liquid P. The condensed liquid is collected in a separate receiver, while Q remains predominantly in the original flask. The separation involves vaporisation followed by condensation rather than crystallisation or sublimation. The condenser is essential when the objective is to recover the volatile component rather than merely remove it from the mixture. The method is suitable because the supplied volatility difference allows the desired liquid and its impurity to follow different paths. Under properly controlled conditions, the collected liquid is therefore substantially purer than the original sample.

Explanation: \(\textbf{Examine mixture M:}\) The boiling-point difference between P and Q is: \[ \Delta T_M=145-65=80^\circ\mathrm{C} \] This large difference provides favourable conditions for separation by simple distillation. P is the lower-boiling component and is generally enriched in the vapour produced during the earlier stages. Condensation of this vapour therefore gives a P-rich initial distillate. \(\textbf{Examine mixture N:}\) The boiling-point difference between R and S is: \[ \Delta T_N=83-78=5^\circ\mathrm{C} \] These boiling points are relatively close. Consequently, both liquids may contribute appreciably to the vapour over a similar temperature range. Simple distillation is therefore less effective for obtaining well-separated fractions. A fractionating column improves separation through repeated vaporisation and condensation. R, having the lower boiling point, is the more volatile component under the intended conditions and becomes enriched in the ascending vapour. Hence the earlier collected fraction is expected to be R-rich. The decisive distinction is not merely that both mixtures contain two liquids, but how effectively their volatility differences can be exploited by the selected apparatus. The preferred technique must therefore be chosen separately for the two mixtures.

Explanation: The intended separation in simple distillation depends on the formation of vapour enriched in the volatile component P. Since Q is specified as nonvolatile, it should remain predominantly in the heated liquid rather than entering the vapour phase. However, the question describes a second transport mechanism: mechanical carry-over of liquid droplets. These droplets originate from the original solution and can therefore contain dissolved Q. If they reach the condenser or receiver, Q can appear in the collected liquid even though it has not vaporised. The presence of Q in the distillate consequently does not prove that the normal volatility-based separation principle has failed. Instead, it indicates that the collection process has allowed material to bypass the intended vapour-phase discrimination. Vigorous boiling and mechanical entrainment must therefore be distinguished from genuine co-vaporisation. Simply interpreting every substance found in a distillate as a component of the equilibrium vapour would give an incorrect conclusion. The observation is consistent with contamination by solution droplets, not with an automatic change in the boiling point or chemical identity of Q.

Explanation: The normal boiling points of P and Q differ by only \(4^\circ\mathrm{C}\), so their vapours may be formed over substantially overlapping temperature ranges. In simple distillation, vapour entering the condenser can consequently contain appreciable amounts of both components. Introducing a fractionating column creates conditions for repeated condensation and vaporisation before the vapour reaches the final condenser. The higher-boiling component Q tends to condense more readily within the column. The liquid and rising vapour repeatedly exchange material under the conditions established along the column. Through these repeated steps, the vapour moving upwards becomes progressively enriched in the more volatile component P. The condensate collected from the top can therefore contain a substantially greater proportion of P than the early condensate from simple distillation. This improvement does not require chemical conversion of either liquid. Nor does it mean that Q suddenly loses its ability to exert vapour pressure. Fractionation improves the effectiveness of separation by exploiting an existing difference in volatility repeatedly rather than depending on a single vaporisation-condensation stage.