Solutions MCQs with Answers – Part 2 (NEET Chemistry)

Explanation: \(\textbf{Interpret mixture P:}\) Positive deviation means that the observed vapour pressure exceeds the corresponding ideal Raoult-law prediction. The relatively weaker unlike-molecule interactions allow molecules to escape more readily into the vapour phase. When the deviation is sufficiently pronounced, a pressure maximum can occur at an intermediate composition. The composition corresponding to that behaviour can form a minimum-boiling azeotrope. Therefore, P is associated with minimum-boiling azeotropic behaviour. \(\textbf{Interpret mixture Q:}\) Negative deviation means that the observed vapour pressure is lower than the corresponding ideal prediction. Relatively stronger unlike-molecule attractions reduce the escaping tendency of the components. When this reduction is sufficiently pronounced, a pressure minimum can occur at an intermediate composition. That behaviour can produce a maximum-boiling azeotrope. Therefore, Q is associated with maximum-boiling azeotropic behaviour. \(\textbf{Explain the phase-composition condition:}\) An azeotrope is characterised by equality of the equilibrium liquid and vapour compositions. This condition holds for both minimum-boiling and maximum-boiling azeotropes. It is not restricted to either positive or negative deviation. \(\textbf{Explain the distillation limitation:}\) Ordinary fractional distillation relies on a difference between liquid and equilibrium vapour compositions. At an azeotropic composition, that difference disappears. Consequently, repeated ordinary fractional distillation cannot change the component ratio beyond that azeotropic limitation by simple equilibrium vapour enrichment. The vapour does not become progressively richer in one component when the mixture is exactly at the azeotropic composition. \(\textbf{Conclusion:}\) P forms a minimum-boiling azeotrope, Q forms a maximum-boiling azeotrope, and both have identical equilibrium liquid and vapour compositions at their azeotropic points.

Explanation: \(\textbf{Construct the common ideal-pressure relation:}\) For an ideal binary solution: \[ p_{\text{ideal}}=x_Ap_A^\circ+(1-x_A)p_B^\circ \] Substitute the pure-component pressures: \[ p_{\text{ideal}}=100x_A+60(1-x_A) \] \[ p_{\text{ideal}}=60+40x_A \] \(\textbf{Use the correct percentage reference:}\) For each mixture: \[ \%\text{ deviation}=\frac{p_{\text{observed}}-p_{\text{ideal}}}{p_{\text{ideal}}}\times100 \] A positive result indicates positive deviation. A negative result indicates negative deviation. The largest magnitude is obtained by comparing the absolute values. \(\textbf{Evaluate mixture P:}\) \[ p_{\text{ideal,P}}=60+40(0.250)=70.0\,\mathrm{mmHg} \] \[ \%\text{ deviation}_P=\frac{74.0-70.0}{70.0}\times100 \] \[ \%\text{ deviation}_P=+5.71\% \] \(\textbf{Evaluate mixture Q:}\) \[ p_{\text{ideal,Q}}=60+40(0.500)=80.0\,\mathrm{mmHg} \] \[ \%\text{ deviation}_Q=\frac{70.0-80.0}{80.0}\times100 \] \[ \%\text{ deviation}_Q=-12.50\% \] \(\textbf{Evaluate mixture R:}\) \[ p_{\text{ideal,R}}=60+40(0.750)=90.0\,\mathrm{mmHg} \] \[ \%\text{ deviation}_R=\frac{99.0-90.0}{90.0}\times100 \] \[ \%\text{ deviation}_R=+10.00\% \] \(\textbf{Evaluate mixture S:}\) \[ p_{\text{ideal,S}}=60+40(0.600)=84.0\,\mathrm{mmHg} \] \[ \%\text{ deviation}_S=\frac{90.0-84.0}{84.0}\times100 \] \[ \%\text{ deviation}_S=+7.14\% \] \(\textbf{Compare the magnitudes:}\) The absolute percentage deviations are \(5.71\%\), \(12.50\%\), \(10.00\%\) and \(7.14\%\). The largest belongs to Q. Its observed pressure is below its ideal prediction, so the deviation is negative. \(\textbf{Interpret the comparison:}\) The largest absolute difference in pressure and the largest percentage difference need not always occur in the same mixture. Percentage deviation requires normalising each observed-minus-ideal difference by the ideal pressure at that specific composition. Using one common reference pressure for all four mixtures would be incorrect. Thus, Q has the greatest percentage-deviation magnitude. \(\textbf{Final answer:}\) Mixture Q exhibits the greatest deviation magnitude, equal to \(12.50\%\), and its deviation is negative.

Explanation: \(\textbf{Evaluate Statement I:}\) Statement I is incorrect. The reported experimental vapour pressure was measured at \(30^\circ\mathrm C\). The reference Raoult-law pressure was calculated using pure-component vapour pressures at \(25^\circ\mathrm C\). These quantities refer to different temperatures and therefore do not constitute a valid direct test of positive or negative deviation. Heating itself increases the equilibrium vapour pressures of volatile liquids. A measured pressure greater than a lower-temperature ideal reference may arise partly or entirely from this temperature difference. \(\textbf{Evaluate Statement II:}\) Statement II is correct. The solution must be brought back to the temperature at which the pure-component reference pressures are known. It must then be allowed to establish liquid-vapour equilibrium. Only after this temperature-controlled equilibration can the observed total pressure be compared meaningfully with the ideal Raoult-law prediction at \(25^\circ\mathrm C\). \(\textbf{Evaluate Statement III:}\) Statement III is correct. Vapour pressure depends strongly on temperature. Changing temperature modifies the equilibrium distribution of molecules between the liquid and vapour phases. This effect is distinct from the pressure deviation caused by unlike-molecule interactions at a fixed composition and temperature. Therefore, temperature differences can confound the interpretation of pressure-deviation experiments. \(\textbf{Evaluate Statement IV:}\) Statement IV is incorrect. Equal liquid and vapour compositions are an azeotropic condition. Heat evolution on mixing does not establish that condition. A temporary rise in measured pressure also does not establish the equality of equilibrium phase compositions. Vapour-composition evidence would be needed to identify azeotropic behaviour. \(\textbf{Conclusion:}\) Statements II and III are correct. The question distinguishes a temperature-controlled equilibrium comparison from an uncontrolled observation made immediately after mixing.

Explanation: \(\textbf{Separate liquid and vapour compositions:}\) The liquid contains \(25.0\%\) A and \(75.0\%\) B by mole. \[ x_A=0.250,\qquad x_B=0.750 \] The equilibrium vapour contains equal mole fractions of A and B. \[ y_A=0.500,\qquad y_B=0.500 \] \(\textbf{Recover actual partial pressures:}\) Dalton's law connects vapour mole fraction with partial pressure. \[ p_A=y_Ap_{\text{total}} \] \[ p_A=0.500(90.0) \] \[ p_A=45.0\,\mathrm{mmHg} \] Similarly: \[ p_B=0.500(90.0) \] \[ p_B=45.0\,\mathrm{mmHg} \] \(\textbf{Use Raoult's law to infer the pure-A pressure:}\) Because the liquid mixture is ideal: \[ p_A=x_Ap_A^\circ \] Therefore: \[ p_A^\circ=\frac{p_A}{x_A} \] \[ p_A^\circ=\frac{45.0}{0.250} \] \[ p_A^\circ=180\,\mathrm{mmHg} \] \(\textbf{Infer the pure-B pressure:}\) \[ p_B=x_Bp_B^\circ \] \[ p_B^\circ=\frac{45.0}{0.750} \] \[ p_B^\circ=60.0\,\mathrm{mmHg} \] \(\textbf{Verify the reconstructed ideal mixture:}\) The ideal total pressure predicted from these pure-component values is: \[ p_{\text{ideal}}=0.250(180)+0.750(60.0) \] \[ p_{\text{ideal}}=45.0+45.0 \] \[ p_{\text{ideal}}=90.0\,\mathrm{mmHg} \] This matches the measured pressure. The vapour contains equal component amounts even though the liquid contains three times as much B as A. \(\textbf{Interpret the volatility difference:}\) A has three times the pure-component vapour pressure of B. Therefore, A is sufficiently more volatile to contribute the same vapour partial pressure despite its lower liquid mole fraction. The distinction between liquid and vapour compositions is essential to reconstructing the pure-component pressures. \(\textbf{Final answer:}\) \[ \boxed{p_A^\circ=180\,\mathrm{mmHg},\qquad p_B^\circ=60.0\,\mathrm{mmHg}} \]

Explanation: \(\textbf{Express the ideal pressure as a composition function:}\) Let the unknown liquid mole fraction of A be \(x_A\). Then: \[ x_B=1-x_A \] The Raoult-law total pressure at the same temperature and composition would be: \[ p_{\text{ideal}}=x_Ap_A^\circ+x_Bp_B^\circ \] Substituting: \[ p_{\text{ideal}}=100x_A+60(1-x_A) \] \[ p_{\text{ideal}}=60+40x_A \] \(\textbf{Translate positive deviation into a quantitative condition:}\) Positive deviation means that the observed total pressure exceeds the ideal reference at the same composition. Therefore: \[ p_{\text{observed}}\gt p_{\text{ideal}} \] The observed pressure is \(82.0\,\mathrm{mmHg}\). \[ 82.0\gt 60+40x_A \] \[ 22.0\gt 40x_A \] \[ x_A\lt 0.550 \] \(\textbf{Interpret the inequality:}\) The value \(x_A=0.550\) would give an ideal pressure of exactly \(82.0\,\mathrm{mmHg}\). At that composition, the measured pressure would equal the ideal value rather than exceed it. Any greater mole fraction of A would give an ideal reference pressure above \(82.0\,\mathrm{mmHg}\). The stated positive deviation would then be contradicted. \(\textbf{Determine whether an exact composition is available:}\) The condition does not specify the magnitude of the positive deviation. It gives only its direction. Therefore, the exact mole fraction cannot be recovered from the total pressure alone. Several liquid compositions below \(0.550\) could satisfy the supplied positive-deviation condition. The stronger restriction \(x_A\lt 0.400\) is not established by the information given. \(\textbf{Chemical interpretation:}\) Applying the ideal total-pressure equation as an equality to the observed non-ideal pressure would incorrectly return \(x_A=0.550\). The experimentally established deviation must instead be applied as an inequality. This converts an otherwise underdetermined composition problem into a valid bound. \(\textbf{Final answer:}\) \[ \boxed{x_A\lt 0.550} \]

Explanation: \(\textbf{Determine the complementary liquid mole fraction:}\) \[ x_A=0.400 \] \[ x_B=1-0.400 \] \[ x_B=0.600 \] \(\textbf{Construct the Raoult-law prediction:}\) For an ideal binary solution: \[ p_{\text{ideal}}=x_Ap_A^\circ+x_Bp_B^\circ \] Substituting the given values: \[ p_{\text{ideal}}=0.400(100)+0.600(80.0) \] \[ p_{\text{ideal}}=40.0+48.0 \] \[ p_{\text{ideal}}=88.0\,\mathrm{mmHg} \] \(\textbf{Compare the experimental pressure:}\) The observed pressure is: \[ p_{\text{observed}}=75.0\,\mathrm{mmHg} \] The pressure difference relative to the ideal prediction is: \[ \Delta p=p_{\text{observed}}-p_{\text{ideal}} \] \[ \Delta p=75.0-88.0 \] \[ \Delta p=-13.0\,\mathrm{mmHg} \] Thus, the actual pressure is \(13.0\,\mathrm{mmHg}\) lower than the ideal value. \(\textbf{Interpret the deviation:}\) A negative observed-minus-ideal pressure difference establishes negative deviation from Raoult's law. Such behaviour is generally associated with relatively strong unlike-molecule attractions. These interactions reduce molecular escaping tendency compared with the ideal reference. \(\textbf{Use the endpoint pressures as a further check:}\) The observed mixture pressure of \(75.0\,\mathrm{mmHg}\) is even lower than the smaller pure-component pressure of \(80.0\,\mathrm{mmHg}\). An ideal binary solution cannot have such a total pressure at an interior composition because its pressure is a weighted average of the two pure-component pressures. Therefore, the measured value provides an especially clear indication of non-ideal behaviour. \(\textbf{Limit the conclusion appropriately:}\) The low pressure at one composition does not, by itself, establish the existence or exact composition of a maximum-boiling azeotrope. Identifying an azeotrope requires appropriate information about phase composition or the full equilibrium behaviour. The certain conclusion from the supplied data is negative deviation of \(13.0\,\mathrm{mmHg}\). \(\textbf{Final answer:}\) \[ \boxed{p_{\text{observed}}-p_{\text{ideal}}=-13.0\,\mathrm{mmHg}} \]

Explanation: \(\textbf{Identify the distinguishing azeotropic condition:}\) Azeotropic behaviour is characterised by equality of the equilibrium liquid and vapour compositions at the azeotropic point. For a binary mixture, the relevant condition is: \[ x_A=y_A \] Equality for A automatically implies equality for B because both phases have mole fractions summing to unity. \(\textbf{Evaluate Record P:}\) Record P has \(x_A=0.250\) and \(y_A=0.400\). The equilibrium vapour is richer in A than the liquid. This difference can provide a driving force for enrichment during distillation. Record P therefore does not satisfy the azeotropic equality condition. \(\textbf{Evaluate Record Q:}\) Record Q has \(x_A=0.450\) and \(y_A=0.450\). The liquid and vapour have identical compositions. The stem additionally specifies constant-boiling behaviour for this record. Together, these observations identify the azeotropic composition. \(\textbf{Evaluate Record R:}\) Record R has \(x_A=0.600\) and \(y_A=0.540\). The vapour contains a smaller mole fraction of A than the liquid. However, the phase compositions are still unequal. Record R is therefore not the azeotropic point. \(\textbf{Evaluate Record S:}\) Record S also has unequal liquid and vapour compositions. The fact that its liquid contains the highest mole fraction of A among the listed samples does not establish azeotropy. Azeotropy is determined by phase-composition equality, not by maximising the concentration of a selected component. \(\textbf{Explain the distillation consequence:}\) Ordinary fractional distillation relies on differences between the compositions of the liquid and the vapour produced from it. At Record Q, the equilibrium vapour has the same composition as the liquid. Therefore, simple repeated vapourisation and condensation at that azeotropic composition cannot progressively enrich either component. \(\textbf{Conclusion:}\) Record Q is the azeotropic composition. The relevant evidence is equality of liquid and vapour compositions together with the stated constant-boiling behaviour.

Explanation: \(\textbf{Evaluate the Assertion:}\) The Assertion is incorrect because positive deviation does not automatically guarantee azeotrope formation. Positive deviation means that the observed equilibrium vapour pressure lies above the corresponding Raoult-law prediction. The amount and composition dependence of this excess pressure can vary among liquid mixtures. Azeotrope formation requires additional phase-equilibrium conditions. In particular, the liquid and equilibrium vapour must have identical compositions at the azeotropic point. A sufficiently large positive deviation may produce a minimum-boiling azeotrope, but a smaller positive deviation need not do so. Therefore, the word "every" makes the Assertion false. \(\textbf{Evaluate the Reason:}\) The Reason is correct. When the unlike A-B interactions are relatively weaker than the interactions that they replace, molecules are retained less strongly in the mixed liquid. Their escaping tendency increases. The resulting equilibrium vapour pressure exceeds the ideal Raoult-law reference at the corresponding composition and temperature. This is the usual molecular explanation for positive deviation. \(\textbf{Distinguish deviation from azeotropy:}\) Pressure deviation and azeotrope formation are related but different concepts. A pressure curve may lie above the ideal straight line without developing the phase-composition equality required for an azeotrope. When positive deviation is sufficiently pronounced, a pressure maximum and minimum-boiling azeotropic behaviour may occur. Without evidence of that stronger condition, azeotrope formation cannot be concluded. The Reason correctly explains positive deviation but does not support the Assertion's universal claim. \(\textbf{Conclusion:}\) The Assertion is false, while the Reason is true.

Explanation: \(\textbf{Construct the ideal pressure-composition relation:}\) For an ideal binary mixture: \[ p_{\text{ideal}}=x_Ap_A^\circ+(1-x_A)p_B^\circ \] Substituting the pure-component vapour pressures: \[ p_{\text{ideal}}=120x_A+80(1-x_A) \] \[ p_{\text{ideal}}=80+40x_A \] \(\textbf{Compare Record P with the ideal reference:}\) \[ p_{\text{ideal,P}}=80+40(0.200) \] \[ p_{\text{ideal,P}}=88.0\,\mathrm{mmHg} \] The observed pressure is \(86.0\,\mathrm{mmHg}\). \[ \Delta p_P=86.0-88.0=-2.00\,\mathrm{mmHg} \] \(\textbf{Compare Record Q with the ideal reference:}\) \[ p_{\text{ideal,Q}}=80+40(0.400) \] \[ p_{\text{ideal,Q}}=96.0\,\mathrm{mmHg} \] The observed pressure is \(90.0\,\mathrm{mmHg}\). \[ \Delta p_Q=90.0-96.0=-6.00\,\mathrm{mmHg} \] \(\textbf{Compare Record R with the ideal reference:}\) \[ p_{\text{ideal,R}}=80+40(0.600) \] \[ p_{\text{ideal,R}}=104\,\mathrm{mmHg} \] The observed pressure is \(102\,\mathrm{mmHg}\). \[ \Delta p_R=102-104=-2.00\,\mathrm{mmHg} \] \(\textbf{Evaluate Statement I:}\) All three observed-minus-ideal differences are negative. Therefore, every listed record shows negative deviation. Statement I is correct. \(\textbf{Evaluate Statement II:}\) The absolute deviations are \(2.00\,\mathrm{mmHg}\), \(6.00\,\mathrm{mmHg}\) and \(2.00\,\mathrm{mmHg}\). Record Q has the greatest deviation magnitude among the three measurements. Statement II is correct. \(\textbf{Evaluate Statement III:}\) A maximum-boiling azeotrope can occur in a system showing sufficiently strong negative deviation. However, the three total-pressure measurements do not establish the equality of liquid and equilibrium vapour compositions at \(x_A=0.400\). Nor do they establish that a pressure minimum occurs at that exact composition. The data therefore cannot prove the asserted azeotropic point. Statement III is incorrect. \(\textbf{Evaluate Statement IV:}\) The actual vapour mole fraction would be: \[ y_A=\frac{p_A}{p_A+p_B} \] The table reports only the sum of the actual partial pressures. Because the mixtures are non-ideal, the individual actual partial pressures cannot simply be replaced by their Raoult-law values. Additional component partial-pressure or vapour-composition measurements would be needed. Statement IV is correct. \(\textbf{Conclusion:}\) Statements I, II and IV are correct. The table establishes the sign and sampled magnitude of total-pressure deviations, but it does not uniquely determine phase compositions or prove azeotrope formation.

Explanation: \(\textbf{Interpret the total-pressure line:}\) For an ideal binary liquid solution: \[ p_{\text{total}}=x_Ap_A^\circ+(1-x_A)p_B^\circ \] Rearranging: \[ p_{\text{total}}=p_B^\circ+x_A(p_A^\circ-p_B^\circ) \] The vertical intercept is therefore the pure vapour pressure of B. The slope is the difference between the pure vapour pressures of A and B. \(\textbf{Recover the slope from two measured points:}\) \[ \text{slope}=\frac{110-70.0}{0.750-0.250} \] \[ \text{slope}=\frac{40.0}{0.500} \] \[ \text{slope}=80.0\,\mathrm{mmHg} \] Hence: \[ p_A^\circ-p_B^\circ=80.0\,\mathrm{mmHg} \] \(\textbf{Recover the pure-B pressure:}\) Using the point at \(x_A=0.250\): \[ 70.0=p_B^\circ+0.250(80.0) \] \[ 70.0=p_B^\circ+20.0 \] \[ p_B^\circ=50.0\,\mathrm{mmHg} \] \(\textbf{Recover the pure-A pressure:}\) \[ p_A^\circ=50.0+80.0 \] \[ p_A^\circ=130\,\mathrm{mmHg} \] \(\textbf{Calculate partial pressures at the requested composition:}\) At \(x_A=0.500\), the liquid mole fraction of B is also \(0.500\). \[ p_A=0.500(130)=65.0\,\mathrm{mmHg} \] \[ p_B=0.500(50.0)=25.0\,\mathrm{mmHg} \] Therefore: \[ p_{\text{total}}=65.0+25.0 \] \[ p_{\text{total}}=90.0\,\mathrm{mmHg} \] \(\textbf{Determine the vapour composition:}\) \[ y_A=\frac{p_A}{p_{\text{total}}} \] \[ y_A=\frac{65.0}{90.0} \] \[ y_A=\frac{13}{18} \] \(\textbf{Check the graph reconstruction:}\) At \(x_A=0.250\), the reconstructed equation gives \(50+0.250(80)=70\,\mathrm{mmHg}\). At \(x_A=0.750\), it gives \(50+0.750(80)=110\,\mathrm{mmHg}\). Both experimental points are reproduced. The vapour is richer in A than the equimolar liquid because A has the higher pure-component vapour pressure. \(\textbf{Final answer:}\) \[ \boxed{p_A^\circ=130\,\mathrm{mmHg},\quad p_B^\circ=50.0\,\mathrm{mmHg},\quad y_A=\frac{13}{18}} \]

Explanation: \(\textbf{Determine the original liquid composition:}\) The initial liquid contains two volatile components. \[ n_{\text{initial,total}}=2.00+3.00=5.00\,\mathrm{mol} \] \[ x_{A,i}=\frac25 \] \[ x_{B,i}=\frac35 \] \(\textbf{Calculate the initial equilibrium pressures:}\) Raoult's law gives: \[ p_{A,i}=\frac25(120)=48.0\,\mathrm{mmHg} \] \[ p_{B,i}=\frac35(80.0)=48.0\,\mathrm{mmHg} \] Therefore: \[ p_{\text{total},i}=48.0+48.0 \] \[ p_{\text{total},i}=96.0\,\mathrm{mmHg} \] The original vapour mole fraction of A is: \[ y_{A,i}=\frac{48.0}{96.0}=\frac12 \] \(\textbf{Reconstruct the changed liquid composition:}\) The addition of X increases the total liquid molecular amount without changing the individual amounts of A and B. \[ n_{\text{final,total}}=2.00+3.00+5.00 \] \[ n_{\text{final,total}}=10.00\,\mathrm{mol} \] The final liquid mole fractions are: \[ x_{A,f}=\frac2{10}=0.200 \] \[ x_{B,f}=\frac3{10}=0.300 \] \[ x_{X,f}=\frac5{10}=0.500 \] \(\textbf{Calculate the new volatile-component pressures:}\) Because X is nonvolatile, it contributes no partial vapour pressure. \[ p_{A,f}=0.200(120)=24.0\,\mathrm{mmHg} \] \[ p_{B,f}=0.300(80.0)=24.0\,\mathrm{mmHg} \] Therefore: \[ p_{\text{total},f}=24.0+24.0 \] \[ p_{\text{total},f}=48.0\,\mathrm{mmHg} \] \(\textbf{Determine the final vapour composition:}\) \[ y_{A,f}=\frac{24.0}{48.0} \] \[ y_{A,f}=\frac12 \] \(\textbf{Explain the invariant:}\) Adding the nonvolatile solute decreases the mole fractions of A and B by the same factor. Their partial vapour pressures consequently decrease by the same factor. The ratio of those partial pressures is unchanged. Hence, the vapour composition remains unchanged even though the total equilibrium pressure is halved. \(\textbf{Final answer:}\) \[ \boxed{p_{\text{total}}:96.0\rightarrow48.0\,\mathrm{mmHg},\qquad y_A=\frac12\ \text{unchanged}} \]

Explanation: \(\textbf{Identify the saturation condition:}\) A pure liquid in equilibrium with its saturated vapour exerts a characteristic vapour pressure at fixed temperature. However, that pressure can be established only if sufficient substance is present to supply the required amount of vapour. If all the liquid evaporates before saturation is reached, the vapour pressure is determined by the amount of gas actually present. The two vessels must therefore be checked separately for saturation. \(\textbf{Calculate the moles needed for saturated vapour:}\) For the available vapour volume: \[ pV=nRT \] At saturation: \[ n_{\text{sat}}=\frac{p^\circ V}{RT} \] \[ n_{\text{sat}}=\frac{0.100(1.00)}{0.0820(300)} \] \[ n_{\text{sat}}=\frac{0.100}{24.6} \] \[ n_{\text{sat}}=0.004065\,\mathrm{mol} \] \(\textbf{Convert the saturation requirement into mass:}\) \[ w_{\text{sat}}=n_{\text{sat}}M_X \] \[ w_{\text{sat}}=0.004065(123) \] \[ w_{\text{sat}}=0.500\,\mathrm g \] Thus, approximately \(0.500\,\mathrm g\) of X must be present in the vapour phase to produce the saturated pressure in the specified vessel. \(\textbf{Evaluate vessel P:}\) Vessel P contains only \(0.250\,\mathrm g\) of X, which is below the required saturation mass. Therefore, all of X evaporates, and no liquid remains. \[ n_P=\frac{0.250}{123} \] \[ n_P=0.002033\,\mathrm{mol} \] Its final pressure is: \[ p_P=\frac{n_PRT}{V} \] \[ p_P=\frac{0.002033(0.0820)(300)}{1.00} \] \[ p_P=0.0500\,\mathrm{atm} \] \(\textbf{Evaluate vessel Q:}\) Vessel Q initially contains \(0.750\,\mathrm g\) X. This exceeds the \(0.500\,\mathrm g\) needed to saturate the vapour space. \[ w_{\text{liquid remaining}}=0.750-0.500 \] \[ w_{\text{liquid remaining}}=0.250\,\mathrm g \] Liquid and saturated vapour can therefore coexist at equilibrium. The final vapour pressure equals the saturated vapour pressure: \[ p_Q=0.100\,\mathrm{atm} \] \(\textbf{Physical check:}\) The pressure in P must be below saturation because its entire sample is insufficient to supply the saturated vapour quantity. The pressure in Q cannot exceed the saturated value while liquid and vapour coexist at the specified temperature. This establishes two different pressure regimes in vessels with identical temperature and volume. \(\textbf{Final answer:}\) \[ \boxed{p_P=0.0500\,\mathrm{atm},\qquad p_Q=0.100\,\mathrm{atm}} \]

Explanation: \(\textbf{Reconstruct the initial liquid composition:}\) The initial total amount is: \[ n_{\text{total}}=2.00+6.00=8.00\,\mathrm{mol} \] Therefore: \[ x_{A,i}=\frac28=\frac14 \] \[ x_{B,i}=\frac68=\frac34 \] \(\textbf{Connect vapour composition with volatility:}\) For ideal liquid and vapour behaviour: \[ \frac{y_A}{y_B}=\frac{p_A}{p_B} \] Using Raoult's law: \[ \frac{y_A}{y_B}=\frac{x_Ap_A^\circ}{x_Bp_B^\circ} \] Initially, the equilibrium vapour is equimolar. \[ \frac{y_A}{y_B}=1 \] Therefore: \[ 1=\frac{(1/4)p_A^\circ}{(3/4)p_B^\circ} \] \[ \frac{p_A^\circ}{p_B^\circ}=3.00 \] \(\textbf{Interpret the revised vapour composition:}\) After adding B: \[ y_{A,f}=0.300 \] \[ y_{B,f}=0.700 \] Hence: \[ \frac{y_{A,f}}{y_{B,f}}=\frac37 \] \(\textbf{Find the required final liquid composition:}\) The ratio of the pure-component vapour pressures remains \(3.00\) because temperature is unchanged. \[ \frac37=3\left(\frac{x_{A,f}}{x_{B,f}}\right) \] \[ \frac{x_{A,f}}{x_{B,f}}=\frac17 \] The liquid must therefore contain seven moles of B for each mole of A. \(\textbf{Reconstruct the final amount of B:}\) The amount of A is conserved at \(2.00\,\mathrm{mol}\). \[ \frac{n_{A,f}}{n_{B,f}}=\frac17 \] \[ \frac{2.00}{n_{B,f}}=\frac17 \] \[ n_{B,f}=14.0\,\mathrm{mol} \] The additional amount of B is: \[ n_{B,\text{added}}=14.0-6.00 \] \[ n_{B,\text{added}}=8.00\,\mathrm{mol} \] \(\textbf{Verify the revised vapour composition:}\) The final liquid contains \(2.00\,\mathrm{mol}\) A and \(14.0\,\mathrm{mol}\) B. \[ x_{A,f}=\frac2{16}=\frac18 \] \[ x_{B,f}=\frac{14}{16}=\frac78 \] Their equilibrium partial-pressure ratio is: \[ \frac{p_A}{p_B}=\frac{(1/8)(3)}{7/8} \] \[ \frac{p_A}{p_B}=\frac37 \] Thus: \[ y_{A,f}=\frac3{3+7}=0.300 \] The observed final vapour composition is reproduced. \(\textbf{Final answer:}\) \[ \boxed{\frac{p_A^\circ}{p_B^\circ}=3.00,\qquad n_{B,\text{added}}=8.00\,\mathrm{mol}} \]

Explanation: \(\textbf{Establish the ideal reference for A:}\) The liquid mole fractions are: \[ x_A=0.400,\qquad x_B=0.600 \] Raoult's law predicts: \[ p_{A,\text{ideal}}=x_Ap_A^\circ \] \[ p_{A,\text{ideal}}=0.400(150) \] \[ p_{A,\text{ideal}}=60.0\,\mathrm{mmHg} \] The observed value is \(70.0\,\mathrm{mmHg}\), which is greater than its ideal reference. A therefore has a positive individual partial-pressure difference. \(\textbf{Establish the ideal reference for B:}\) \[ p_{B,\text{ideal}}=x_Bp_B^\circ \] \[ p_{B,\text{ideal}}=0.600(100) \] \[ p_{B,\text{ideal}}=60.0\,\mathrm{mmHg} \] The observed value is \(56.0\,\mathrm{mmHg}\), below the ideal reference. B therefore has a negative individual partial-pressure difference. \(\textbf{Compare the total pressures:}\) The ideal total is: \[ p_{\text{ideal,total}}=60.0+60.0 \] \[ p_{\text{ideal,total}}=120\,\mathrm{mmHg} \] The observed total is: \[ p_{\text{observed,total}}=70.0+56.0 \] \[ p_{\text{observed,total}}=126\,\mathrm{mmHg} \] The observed total exceeds the ideal value. \(\textbf{Interpret the separate contributions:}\) The component differences are: \[ \Delta p_A=+10.0\,\mathrm{mmHg} \] \[ \Delta p_B=-4.00\,\mathrm{mmHg} \] Their sum is: \[ \Delta p_{\text{total}}=+6.00\,\mathrm{mmHg} \] The positive difference for A more than compensates for the negative difference for B. Therefore, the total mixture shows positive deviation at the tested composition. \(\textbf{Conceptual distinction:}\) The sign of the total-pressure deviation is established by the sum of both measured partial pressures relative to the sum of their ideal values. It does not follow that every component must individually have a pressure difference of the same sign. Nor do opposite signs imply exact cancellation. The supplied observations demonstrate why total-pressure behaviour and individual partial-pressure behaviour must be distinguished. Thus, option C is correct.

Explanation: \(\textbf{Cylinder P — Pure liquid equilibrium:}\) The saturated vapour pressure of a pure liquid depends on its identity and temperature. Moving the piston outward initially decreases the vapour pressure because the existing vapour occupies a greater volume. The pressure is temporarily below the saturation value. The liquid then undergoes net evaporation until saturated equilibrium is restored. Because liquid remains present and temperature is unchanged, the final equilibrium vapour pressure returns to its original value. \(\textbf{Cylinder Q — Binary liquid equilibrium:}\) The equilibrium vapour pressure of an ideal binary solution depends on temperature and liquid composition. Its total pressure is: \[ p_{\text{total}}=x_Ap_A^\circ+x_Bp_B^\circ \] Here, A is more volatile because its pure-component vapour pressure is greater. Consequently, the equilibrium vapour is richer in A than the liquid. When the enlarged vapour space is supplied by additional evaporation, the newly transferred material preferentially removes A from the liquid. The remaining liquid therefore becomes relatively poorer in A. \(\textbf{Effect on the new equilibrium pressure:}\) At fixed temperature, decreasing \(x_A\) in an ideal solution with \(p_A^\circ\gt p_B^\circ\) decreases the total equilibrium vapour pressure. Thus, unlike the pure-liquid case, the final equilibrium pressure of the binary mixture need not return to its original value. Under the described preferential evaporation, it can settle at a lower value. \(\textbf{Central distinction:}\) The pure liquid retains its chemical identity and composition during evaporation. The binary solution may change composition because the liquid and equilibrium vapour generally have different component ratios. Temperature independence of the final pressure under a fixed-temperature volume change applies directly to a saturated pure liquid, not to every multicomponent liquid. Therefore, option B correctly distinguishes the two cylinders.

Explanation: \(\textbf{Interpret the pressure-graph endpoints:}\) At \(x_A=0\), the liquid is pure B. Therefore, the horizontal graph gives: \[ p_B^\circ=85.0\,\mathrm{mmHg} \] At \(x_A=1\), the liquid is pure A. Therefore: \[ p_A^\circ=85.0\,\mathrm{mmHg} \] The pure-component vapour pressures are equal at the specified temperature. \(\textbf{Explain the zero slope:}\) For an ideal binary solution: \[ p_{\text{total}}=p_B^\circ+x_A(p_A^\circ-p_B^\circ) \] The slope is: \[ p_A^\circ-p_B^\circ \] Because the pure-component pressures are equal, the slope is zero. This explains why the total-pressure graph is horizontal. \(\textbf{Determine the vapour composition:}\) Raoult's law gives: \[ p_A=85.0x_A \] \[ p_B=85.0(1-x_A) \] Their sum is \(85.0\,\mathrm{mmHg}\) at every composition. The vapour mole fraction of A is: \[ y_A=\frac{p_A}{p_A+p_B} \] \[ y_A=\frac{85.0x_A}{85.0} \] \[ y_A=x_A \] \(\textbf{Interpret the physical significance:}\) Neither component is preferentially enriched in the equilibrium vapour because their pure-component vapour pressures are equal. Nevertheless, their individual partial pressures still vary as the liquid composition changes. The constancy of total pressure does not mean that each component has a constant partial pressure. The result follows specifically from ideal behaviour and equal pure-component vapour pressures. No separate assumption about equal molecular masses or equal liquid volumes is required. Thus, option C is correct.

Explanation: \(\textbf{Calculate the equilibrium partial pressures:}\) The liquid mole fractions are: \[ x_A=0.300,\qquad x_B=0.700 \] For ideal behaviour: \[ p_A=x_Ap_A^\circ \] \[ p_A=0.300(160)=48.0\,\mathrm{mmHg} \] For B: \[ p_B=0.700(80.0)=56.0\,\mathrm{mmHg} \] \(\textbf{Determine total vapour pressure:}\) \[ p_{\text{total}}=48.0+56.0 \] \[ p_{\text{total}}=104\,\mathrm{mmHg} \] \(\textbf{Recover the vapour-phase mole fractions:}\) \[ y_A=\frac{48.0}{104} \] \[ y_A=\frac6{13} \] \[ y_B=\frac7{13} \] \(\textbf{Reconstruct the condensed vapour composition:}\) Complete condensation without component loss preserves the mole ratio present in the sampled vapour. Choose \(6.00\,\mathrm{mol}\) A and \(7.00\,\mathrm{mol}\) B as a convenient condensate basis. The component masses are: \[ w_{A,\text{condensate}}=6(60.0)=360\,\mathrm g \] \[ w_{B,\text{condensate}}=7(100)=700\,\mathrm g \] Therefore: \[ w_{\text{condensate}}=360+700=1060\,\mathrm g \] \(\textbf{Calculate the condensate mass percentage:}\) \[ w_{A,\text{condensate}}\%=\frac{360}{1060}\times100 \] \[ w_{A,\text{condensate}}\%=33.9623\% \] \(\textbf{Calculate the original liquid mass percentage:}\) Choose \(1.00\,\mathrm{mol}\) of original liquid mixture. It contains \(0.300\,\mathrm{mol}\) A and \(0.700\,\mathrm{mol}\) B. \[ w_{A,\text{liquid}}=0.300(60.0)=18.0\,\mathrm g \] \[ w_{B,\text{liquid}}=0.700(100)=70.0\,\mathrm g \] Hence: \[ w_{A,\text{liquid}}\%=\frac{18.0}{18.0+70.0}\times100 \] \[ w_{A,\text{liquid}}\%=20.4545\% \] \(\textbf{Explain the phase difference:}\) A is the more volatile component, so the equilibrium vapour is enriched in A relative to the original liquid. That enrichment survives complete condensation of a separately collected vapour sample. However, its mole fraction and mass percentage are different because the two components have different molar masses. The question therefore requires conversion from liquid mole fraction to vapour mole fraction and then from vapour mole fraction to condensate mass percentage. \(\textbf{Final answer:}\) \[ \boxed{w_{A,\text{condensate}}\%\approx33.96\%,\qquad w_{A,\text{liquid}}\%\approx20.45\%} \]

Explanation: \(\textbf{Identify the unchanged liquid composition:}\) The liquid mole fractions at both temperatures are: \[ x_A=0.400,\qquad x_B=0.600 \] Only the pure-component vapour pressures change between the two measurements. The liquid composition is explicitly held fixed. \(\textbf{Calculate partial pressures at }T_1\textbf{:}\) \[ p_{A,1}=0.400(80.0) \] \[ p_{A,1}=32.0\,\mathrm{mmHg} \] \[ p_{B,1}=0.600(40.0) \] \[ p_{B,1}=24.0\,\mathrm{mmHg} \] Therefore: \[ p_{\text{total},1}=32.0+24.0 \] \[ p_{\text{total},1}=56.0\,\mathrm{mmHg} \] \(\textbf{Determine the initial vapour composition:}\) \[ y_{A,1}=\frac{32.0}{56.0} \] \[ y_{A,1}=\frac47 \] \(\textbf{Calculate partial pressures at }T_2\textbf{:}\) At the higher temperature, B has the greater pure-component vapour pressure. \[ p_{A,2}=0.400(150) \] \[ p_{A,2}=60.0\,\mathrm{mmHg} \] \[ p_{B,2}=0.600(180) \] \[ p_{B,2}=108\,\mathrm{mmHg} \] Therefore: \[ p_{\text{total},2}=60.0+108 \] \[ p_{\text{total},2}=168\,\mathrm{mmHg} \] \(\textbf{Determine the revised vapour composition:}\) \[ y_{A,2}=\frac{60.0}{168} \] \[ y_{A,2}=\frac5{14} \] \(\textbf{Interpret the volatility reversal:}\) At \(T_1\), A has the greater pure-component vapour pressure. Consequently, its equilibrium vapour mole fraction exceeds its liquid mole fraction. \[ y_{A,1}=\frac47\gt 0.400 \] At \(T_2\), B has the greater pure-component vapour pressure. The equilibrium vapour is therefore relatively enriched in B. \[ y_{A,2}=\frac5{14}\lt 0.400 \] Thus, heating increases the total vapour pressure while reversing which component is preferentially enriched in the vapour. The relative volatility ordering must be determined at the actual temperature; it cannot be carried over automatically from another temperature. \(\textbf{Final answer:}\) \[ \boxed{y_{A,1}=\frac47,\qquad y_{A,2}=\frac5{14}} \]

Explanation: \(\textbf{Identify the original saturated-vapour condition:}\) The pure volatile liquid has a saturated vapour pressure of \(0.100\,\mathrm{atm}\) at the specified temperature. Because liquid remains present, the vapour partial pressure of X is fixed by its liquid-vapour equilibrium. This partial pressure is distinct from the total pressure of a gas mixture. \(\textbf{Interpret the inert-gas addition:}\) The added gas does not react with X. It also does not change the liquid composition. At fixed temperature and vessel volume, the inert gas contributes its own partial pressure. Under the stated idealisation, its presence does not require an increase in the saturated partial pressure of X. Therefore: \[ p_X=0.100\,\mathrm{atm} \] \(\textbf{Apply Dalton's law:}\) The final gas phase contains X vapour and the inert gas. \[ p_{\text{total}}=p_X+p_{\text{inert}} \] Substituting: \[ p_{\text{total}}=0.100+0.400 \] \[ p_{\text{total}}=0.500\,\mathrm{atm} \] \(\textbf{Determine the gas-phase mole fraction:}\) For an ideal gas mixture: \[ y_X=\frac{p_X}{p_{\text{total}}} \] \[ y_X=\frac{0.100}{0.500} \] \[ y_X=0.200 \] \(\textbf{Check the remaining gas composition:}\) \[ y_{\text{inert}}=\frac{0.400}{0.500} \] \[ y_{\text{inert}}=0.800 \] The two gas-phase mole fractions sum to unity. \(\textbf{Interpret the physical result:}\) The total pressure increases because an additional gas is present. However, the partial pressure of saturated X vapour remains fixed at the original temperature. The mole fraction of X in the gas phase decreases because the gas mixture now contains many inert-gas molecules. The saturated vapour pressure of X must not be confused with the total vessel pressure. \(\textbf{Final answer:}\) \[ \boxed{p_X=0.100\,\mathrm{atm},\qquad p_{\text{total}}=0.500\,\mathrm{atm},\qquad y_X=0.200} \]

Explanation: \(\textbf{Establish the ideal-pressure bounds:}\) For an ideal binary solution: \[ p_{\text{ideal}}=x_Ap_A^\circ+(1-x_A)p_B^\circ \] Using the supplied pure-component vapour pressures: \[ p_{\text{ideal}}=100x_A+60(1-x_A) \] \[ p_{\text{ideal}}=60+40x_A \] For any binary composition, this ideal total pressure lies between \(60.0\) and \(100\,\mathrm{mmHg}\). The exact ideal reference within this interval depends on the unknown liquid composition. \(\textbf{Evaluate Statement I:}\) Mixture P has an observed pressure of \(110\,\mathrm{mmHg}\). This exceeds even the greater pure-component vapour pressure. Therefore, its observed pressure must also exceed the ideal pressure at any interior composition. P necessarily exhibits positive deviation. Statement I is correct. \(\textbf{Evaluate Statement II:}\) Mixture Q has an observed pressure of \(50.0\,\mathrm{mmHg}\). This lies below the smaller pure-component vapour pressure. Consequently, its observed pressure must be below the ideal pressure at every binary composition. Q necessarily exhibits negative deviation. Statement II is correct. \(\textbf{Evaluate Statement III:}\) Mixture R has an observed pressure of \(85.0\,\mathrm{mmHg}\), within the pure-component pressure interval. However, this does not establish ideality. For example, at \(x_A=0.250\): \[ p_{\text{ideal}}=60+40(0.250)=70.0\,\mathrm{mmHg} \] An observed pressure of \(85.0\,\mathrm{mmHg}\) would represent positive deviation. At \(x_A=0.750\), the ideal pressure would instead be \(90.0\,\mathrm{mmHg}\), making the same observed pressure a negative deviation. Therefore, Statement III is incorrect. \(\textbf{Evaluate Statement IV:}\) Mixture S has an observed pressure of \(100\,\mathrm{mmHg}\). At an interior composition such as \(x_A=0.500\), the ideal prediction is: \[ p_{\text{ideal}}=60+40(0.500)=80.0\,\mathrm{mmHg} \] An observed pressure of \(100\,\mathrm{mmHg}\) would then represent positive deviation. Equality with one pure-component pressure does not establish ideal behaviour at an unknown liquid composition. Statement IV is correct. \(\textbf{Conclusion:}\) Statements I, II and IV are correct. The pure-component pressure interval permits certain deviation conclusions when the observed pressure lies outside it, but pressures inside the interval require composition-specific comparison.