Solutions MCQs with Answers – Part 4 (NEET Chemistry)

Explanation: \(\textbf{Construct the measured-pressure equation:}\) For an ideal binary liquid: \[ p_{\text{volatile}}=x_Ap_A^\circ+(1-x_A)p_B^\circ \] Let the unknown constant inert-gas partial pressure be \(c\). The measured total pressure is: \[ p_{\text{measured}}=x_Ap_A^\circ+(1-x_A)p_B^\circ+c \] Rearranging: \[ p_{\text{measured}}=(p_B^\circ+c)+x_A(p_A^\circ-p_B^\circ) \] \(\textbf{Determine the measured graph slope:}\) Using the first two observations: \[ s=\frac{115-100}{0.500-0.200} \] \[ s=\frac{15.0}{0.300} \] \[ s=50.0\,\mathrm{mmHg} \] Therefore: \[ p_A^\circ-p_B^\circ=50.0\,\mathrm{mmHg} \] \(\textbf{Recover the intercept combination:}\) At \(x_A=0.200\): \[ 100=(p_B^\circ+c)+0.200(50.0) \] \[ 100=(p_B^\circ+c)+10.0 \] \[ p_B^\circ+c=90.0\,\mathrm{mmHg} \] \(\textbf{Check the third observation:}\) At \(x_A=0.800\), the predicted measured pressure is: \[ p_{\text{measured}}=90.0+0.800(50.0) \] \[ p_{\text{measured}}=130\,\mathrm{mmHg} \] This agrees with the table. The third observation confirms the linear trend but does not supply another independent parameter. \(\textbf{Identify the remaining ambiguity:}\) The measurements establish: \[ p_A^\circ-p_B^\circ=50.0 \] \[ p_B^\circ+c=90.0 \] These are only two independent constraints involving three unknown quantities. For example, one compatible set is: \[ p_A^\circ=120,\quad p_B^\circ=70.0,\quad c=20.0 \] Another compatible set is: \[ p_A^\circ=130,\quad p_B^\circ=80.0,\quad c=10.0 \] Both produce exactly the same measured pressure-composition line. \(\textbf{Interpret the experimental limitation:}\) A constant inert-gas pressure shifts the entire measured total-pressure graph upward without changing its slope. The slope consequently reveals the difference between the pure-component vapour pressures. However, the graph intercept combines the pure-B pressure with the inert-gas contribution. Additional composition points on the same straight line cannot separate those two contributions. A separate measurement of inert-gas pressure or one pure-component vapour pressure is required. \(\textbf{Conclusion:}\) The pure-component pressure difference is \(50.0\,\mathrm{mmHg}\), but the three individual pressure quantities cannot be uniquely reconstructed.

Explanation: \(\textbf{Determine the initial equilibrium pressure:}\) Initially, the volatile components are equimolar. \[ x_{A,i}=x_{B,i}=0.500 \] Therefore: \[ p_i=0.500(120)+0.500(80.0) \] \[ p_i=60.0+40.0 \] \[ p_i=100\,\mathrm{mmHg} \] \(\textbf{Reconstruct the liquid after adding A:}\) Adding \(1.00\,\mathrm{mol}\) pure A produces: \[ n_{A,f}=3.00\,\mathrm{mol} \] \[ n_{B,f}=2.00\,\mathrm{mol} \] Let the additional nonvolatile-solute amount be \(n_X\). The total liquid molecular amount becomes: \[ n_{\text{total},f}=5.00+n_X \] \(\textbf{Express the final volatile partial pressures:}\) Under ideal behaviour: \[ p_{A,f}=\frac{3.00}{5.00+n_X}(120) \] \[ p_{B,f}=\frac{2.00}{5.00+n_X}(80.0) \] The nonvolatile solute contributes no equilibrium vapour pressure. \(\textbf{Apply the unchanged-total-pressure condition:}\) The desired final total pressure is \(100\,\mathrm{mmHg}\). \[ 100=\frac{3.00(120)+2.00(80.0)}{5.00+n_X} \] \[ 100=\frac{360+160}{5.00+n_X} \] \[ 100=\frac{520}{5.00+n_X} \] Therefore: \[ 5.00+n_X=5.20 \] \[ n_X=0.200\,\mathrm{mol} \] \(\textbf{Calculate the final partial pressure of A:}\) \[ p_{A,f}=\frac{3.00}{5.20}(120) \] \[ p_{A,f}=\frac{900}{13}\,\mathrm{mmHg} \] \(\textbf{Recover the final vapour composition:}\) Because the final total pressure is \(100\,\mathrm{mmHg}\): \[ y_{A,f}=\frac{p_{A,f}}{100} \] \[ y_{A,f}=\frac{900/13}{100} \] \[ y_{A,f}=\frac9{13} \] \(\textbf{Check the two competing effects:}\) Adding more of the more volatile A tends to increase the total equilibrium vapour pressure. Adding nonvolatile X dilutes both volatile components and tends to decrease their combined pressure. For the calculated amount of X, these two effects exactly cancel in the total pressure. However, their partial-pressure ratio changes because the A-to-B molecular ratio changes. \(\textbf{Interpret the compositional result:}\) Initially, the vapour mole fraction of A is \(60.0/100=0.600\). Finally, it is \(\frac9{13}\), which is greater. Thus, maintaining the same total vapour pressure does not maintain the same equilibrium vapour composition. The pressure constraint fixes the amount of X, while the changed volatile-component mole ratio fixes the new vapour composition. \(\textbf{Final answer:}\) \[ \boxed{n_X=0.200\,\mathrm{mol},\qquad y_{A,f}=\frac9{13}} \]

Explanation: \(\textbf{Establish the ternary composition constraint:}\) The liquid contains three molecular components. Its mole fractions must satisfy: \[ x_A+x_B+x_X=1 \] All mole fractions must be nonnegative. Therefore: \[ x_A+x_B\leq1 \] \(\textbf{Recover liquid composition from vapour data:}\) Dalton's law gives: \[ p_A=y_Ap_{\text{total}} \] \[ p_B=(1-y_A)p_{\text{total}} \] Because the liquid is ideal: \[ x_A=\frac{p_A}{p_A^\circ} \] \[ x_B=\frac{p_B}{p_B^\circ} \] \(\textbf{Evaluate preparation P:}\) The total volatile pressure is \(60.0\,\mathrm{mmHg}\), and \(y_A=0.500\). \[ p_A=0.500(60.0)=30.0\,\mathrm{mmHg} \] \[ p_B=30.0\,\mathrm{mmHg} \] Thus: \[ x_A=\frac{30.0}{100}=0.300 \] \[ x_B=\frac{30.0}{50.0}=0.600 \] \[ x_A+x_B=0.900 \] The remaining mole fraction of X is \(0.100\), which is physically possible. \(\textbf{Evaluate preparation Q:}\) \[ p_A=\frac47(70.0)=40.0\,\mathrm{mmHg} \] \[ p_B=30.0\,\mathrm{mmHg} \] Therefore: \[ x_A=\frac{40.0}{100}=0.400 \] \[ x_B=\frac{30.0}{50.0}=0.600 \] \[ x_A+x_B=1.000 \] This represents the limiting composition with \(x_X=0\), which is physically possible. \(\textbf{Evaluate preparation R:}\) \[ p_A=0.500(70.0)=35.0\,\mathrm{mmHg} \] \[ p_B=35.0\,\mathrm{mmHg} \] Therefore: \[ x_A=\frac{35.0}{100}=0.350 \] \[ x_B=\frac{35.0}{50.0}=0.700 \] The sum is: \[ x_A+x_B=1.050 \] This would require: \[ x_X=1-1.050=-0.050 \] A negative liquid mole fraction is impossible. Thus, R is inconsistent with the stipulated ideal ternary model. \(\textbf{Evaluate preparation S:}\) \[ p_A=0.750(80.0)=60.0\,\mathrm{mmHg} \] \[ p_B=20.0\,\mathrm{mmHg} \] Therefore: \[ x_A=0.600 \] \[ x_B=0.400 \] Their sum is unity, so S is physically possible with no X present. \(\textbf{Interpret the consistency check:}\) The measured volatile vapour pressure and vapour composition independently determine both volatile liquid mole fractions under Raoult's law. The nonvolatile solute occupies the remaining fraction. A negative remainder reveals an impossible molecular composition rather than merely an unusually concentrated solution. The component-wise calculation is essential because total vapour pressure alone does not reveal the liquid inventory. \(\textbf{Conclusion:}\) Preparation R cannot satisfy the stipulated ideal ternary composition constraint.

Explanation: \(\textbf{Establish the common ideal-pressure reference:}\) Every sample is measured at the same temperature and liquid composition. Therefore, all four have the same Raoult-law total-pressure reference. \[ p_{\text{ideal}}=x_Ap_A^\circ+x_Bp_B^\circ \] Substituting: \[ p_{\text{ideal}}=0.400(100)+0.600(60.0) \] \[ p_{\text{ideal}}=40.0+36.0 \] \[ p_{\text{ideal}}=76.0\,\mathrm{mmHg} \] \(\textbf{Interpret sample P:}\) P has observed pressure \(73.0\,\mathrm{mmHg}\), lower than the ideal total pressure. The negative deviation is: \[ \Delta p_P=73.0-76.0 \] \[ \Delta p_P=-3.00\,\mathrm{mmHg} \] Heat release and volume contraction are consistent with the usual NCERT picture of stronger unlike-molecule attractions and a negative deviation. P therefore displays a mutually consistent set of observations. \(\textbf{Interpret sample Q:}\) Q has pressure \(82.0\,\mathrm{mmHg}\), above the ideal value: \[ \Delta p_Q=82.0-76.0 \] \[ \Delta p_Q=+6.00\,\mathrm{mmHg} \] Heat absorption and expansion are commonly associated with comparatively weaker unlike-molecule attractions and positive deviation. Q therefore displays the usual internally consistent positive-deviation pattern. \(\textbf{Interpret sample R:}\) R has observed total pressure \(76.0\,\mathrm{mmHg}\), equal to the ideal reference. Its heat absorption and contraction still indicate nonzero mixing changes. A total-pressure match at one composition does not by itself establish the full ideal-solution conditions, so R does not demonstrate the stipulated standard pattern. \(\textbf{Interpret sample S:}\) S has pressure \(82.0\,\mathrm{mmHg}\), indicating positive total-pressure deviation. However, heat release is generally associated with stronger unlike-molecule attractions and negative deviation in the standard NCERT comparison. S's expansion also makes its combined observations unlike the standard consistent positive- or negative-deviation pattern. \(\textbf{Select the consistent pair:}\) Sample P combines negative pressure deviation with exothermic mixing and contraction. Sample Q combines positive pressure deviation with endothermic mixing and expansion. Among the listed alternatives, only P and Q both satisfy the requested NCERT-aligned interpretation. Thus, option A is correct.

Explanation: \(\textbf{Calculate the original liquid composition:}\) The initial amounts are: \[ n_A=2.00\,\mathrm{mol} \] \[ n_B=3.00\,\mathrm{mol} \] Therefore: \[ x_{A,1}=\frac25 \] \[ x_{B,1}=\frac35 \] \(\textbf{Determine the original equilibrium pressure:}\) At \(T_1\): \[ p_1=x_{A,1}p_{A,1}^\circ+x_{B,1}p_{B,1}^\circ \] \[ p_1=\frac25(120)+\frac35(60.0) \] \[ p_1=48.0+36.0 \] \[ p_1=84.0\,\mathrm{mmHg} \] \(\textbf{Represent the changed-temperature composition:}\) At \(T_2\), the amounts of volatile A and B remain \(2.00\) and \(3.00\,\mathrm{mol}\). Let the added nonvolatile amount be \(n_X\). The total liquid molecular amount is: \[ n_{\text{total},2}=5.00+n_X \] \(\textbf{Construct the final pressure equation:}\) The new pure-component pressures are: \[ p_{A,2}^\circ=90.0\,\mathrm{mmHg} \] \[ p_{B,2}^\circ=150\,\mathrm{mmHg} \] The final total pressure is: \[ p_2=\frac{2.00(90.0)+3.00(150)}{5.00+n_X} \] \[ p_2=\frac{180+450}{5.00+n_X} \] \[ p_2=\frac{630}{5.00+n_X} \] \(\textbf{Apply the equal-total-pressure condition:}\) The desired final pressure is the original \(84.0\,\mathrm{mmHg}\). \[ 84.0=\frac{630}{5.00+n_X} \] \[ 5.00+n_X=\frac{630}{84.0} \] \[ 5.00+n_X=7.50 \] \[ n_X=2.50\,\mathrm{mol} \] \(\textbf{Calculate the final component pressures:}\) At the new total molecular amount of \(7.50\,\mathrm{mol}\): \[ p_{A,2}=\frac{2.00}{7.50}(90.0) \] \[ p_{A,2}=24.0\,\mathrm{mmHg} \] For B: \[ p_{B,2}=\frac{3.00}{7.50}(150) \] \[ p_{B,2}=60.0\,\mathrm{mmHg} \] Their sum is the required \(84.0\,\mathrm{mmHg}\). \(\textbf{Determine the changed vapour composition:}\) \[ y_{A,2}=\frac{p_{A,2}}{p_2} \] \[ y_{A,2}=\frac{24.0}{84.0} \] \[ y_{A,2}=\frac27 \] \(\textbf{Interpret the combined temperature and dilution effects:}\) At \(T_1\), A has the greater pure-component vapour pressure. At \(T_2\), B has the greater pure-component vapour pressure. Heating therefore changes the relative volatility ordering of the two components. Adding nonvolatile X reduces both volatile partial pressures at \(T_2\), allowing the original total pressure to be restored. However, X does not alter the A-to-B molecular ratio, so it cannot prevent the temperature-induced reversal in vapour enrichment. Equal total pressures at two temperatures do not imply equal vapour compositions. \(\textbf{Final answer:}\) \[ \boxed{n_X=2.50\,\mathrm{mol},\qquad y_{A,2}=\frac27} \]

Explanation: \(\textbf{Recover the solvent molecular amount:}\) The solution contains \(180\,\mathrm g\) water. \[ n_w=\frac{180}{18.0} \] \[ n_w=10.0\,\mathrm{mol} \] \(\textbf{Apply the exact relative-lowering relation:}\) For an ideal solution of a nonvolatile molecular solute: \[ L=\frac{p_w^\circ-p_w}{p_w^\circ} \] \[ L=x_X=\frac{n_X}{n_w+n_X} \] Substituting: \[ \frac16=\frac{n_X}{10.0+n_X} \] \[ 10.0+n_X=6n_X \] \[ 5n_X=10.0 \] \[ n_X=2.00\,\mathrm{mol} \] \(\textbf{Determine the actual molar mass:}\) \[ M_X=\frac{w_X}{n_X} \] \[ M_X=\frac{60.0}{2.00} \] \[ M_X=30.0\,\mathrm{g\,mol^{-1}} \] \(\textbf{Reconstruct the student's approximation:}\) The student assumes: \[ L\approx\frac{n_X}{n_w} \] Consequently: \[ n_{X,\text{approx}}=\frac16(10.0) \] \[ n_{X,\text{approx}}=\frac53\,\mathrm{mol} \] The apparent molar mass obtained from this approximate amount is: \[ M_{\text{approx}}=\frac{60.0}{5/3} \] \[ M_{\text{approx}}=36.0\,\mathrm{g\,mol^{-1}} \] \(\textbf{Calculate the percentage overestimation:}\) The correct molar mass is the reference. \[ \%\text{ overestimation}=\frac{36.0-30.0}{30.0}\times100 \] \[ \%\text{ overestimation}=20.0\% \] \(\textbf{Explain the direction of error:}\) The dilute approximation neglects solute molecules in the denominator of the exact mole-fraction expression. At the stated concentration, this omission is appreciable. It causes the inferred solute amount to be smaller than the true amount. For a fixed measured solute mass, underestimating molecular amount leads to overestimating molar mass. The approximation is appropriate only when solute moles are sufficiently small compared with solvent moles. \(\textbf{Final answer:}\) \[ \boxed{M_X=30.0\,\mathrm{g\,mol^{-1}},\qquad\text{overestimation}=20.0\%} \]

Explanation: \(\textbf{Establish the conserved solvent inventory:}\) The amount of water is unchanged throughout the replacement. \[ n_w=\frac{180}{18.0} \] \[ n_w=10.0\,\mathrm{mol} \] \(\textbf{Calculate the initial glucose amount:}\) \[ n_{\text{glucose},i}=\frac{18.0}{180} \] \[ n_{\text{glucose},i}=0.100\,\mathrm{mol} \] \(\textbf{Determine the initial relative lowering:}\) For a nonvolatile molecular solute in an ideal solution: \[ L=\frac{n_{\text{solute}}}{n_w+n_{\text{solute}}} \] Therefore: \[ L_i=\frac{0.100}{10.0+0.100} \] \[ L_i=\frac1{101} \] \(\textbf{Reconstruct the remaining glucose:}\) The mass of glucose remaining after removal is: \[ w_{\text{glucose},f}=18.0-9.00 \] \[ w_{\text{glucose},f}=9.00\,\mathrm g \] Its molecular amount is: \[ n_{\text{glucose},f}=\frac{9.00}{180} \] \[ n_{\text{glucose},f}=0.0500\,\mathrm{mol} \] \(\textbf{Calculate the introduced urea amount:}\) \[ n_{\text{urea}}=\frac{9.00}{60.0} \] \[ n_{\text{urea}}=0.150\,\mathrm{mol} \] \(\textbf{Determine the final dissolved-particle amount:}\) Both solutes remain independently dissolved as molecules. \[ n_{\text{solute},f}=0.0500+0.150 \] \[ n_{\text{solute},f}=0.200\,\mathrm{mol} \] \(\textbf{Calculate the final relative lowering:}\) \[ L_f=\frac{0.200}{10.0+0.200} \] \[ L_f=\frac1{51} \] \(\textbf{Compare the two states:}\) \[ \frac{L_f}{L_i}=\frac{1/51}{1/101} \] \[ \frac{L_f}{L_i}=\frac{101}{51} \] \(\textbf{Interpret the selective replacement:}\) The total mass of dissolved solute remains \(18.0\,\mathrm g\). However, replacing glucose with an equal mass of lower-molar-mass urea increases the number of solute molecules. The total number of dissolved particles doubles, but the exact relative lowering increases by slightly less than a factor of two. This difference arises because the solute molecules contribute to the denominator of the mole-fraction relation. Equal solute mass does not imply equal relative vapour-pressure lowering. \(\textbf{Final answer:}\) \[ \boxed{\frac{L_f}{L_i}=\frac{101}{51}} \]

Explanation: \(\textbf{Separate vapour pressure from total gas pressure:}\) The measured gas pressure includes water vapour and inert gas. For the pure-water vessel: \[ p_{\text{measured,pure}}=p_w^\circ+p_{\text{inert}} \] Therefore: \[ p_w^\circ=130-20.0 \] \[ p_w^\circ=110\,\mathrm{mmHg} \] \(\textbf{Recover the solution's water-vapour pressure:}\) \[ p_{\text{measured,solution}}=p_w+p_{\text{inert}} \] \[ p_w=120-20.0 \] \[ p_w=100\,\mathrm{mmHg} \] \(\textbf{Calculate the relative lowering:}\) \[ L=\frac{p_w^\circ-p_w}{p_w^\circ} \] \[ L=\frac{110-100}{110} \] \[ L=\frac1{11} \] \(\textbf{Determine the solvent molecular amount:}\) The solution contains \(180\,\mathrm g\) water. \[ n_w=\frac{180}{18.0} \] \[ n_w=10.0\,\mathrm{mol} \] \(\textbf{Apply the exact mole-fraction relation:}\) For the ideal solution: \[ L=x_X=\frac{n_X}{n_w+n_X} \] Therefore: \[ \frac1{11}=\frac{n_X}{10.0+n_X} \] \[ 10.0+n_X=11n_X \] \[ n_X=1.00\,\mathrm{mol} \] \(\textbf{Recover the molar mass:}\) \[ M_X=\frac{60.0}{1.00} \] \[ M_X=60.0\,\mathrm{g\,mol^{-1}} \] \(\textbf{Interpret the experimental correction:}\) Relative lowering is defined using the equilibrium vapour pressure of the solvent, not the total gas pressure of the chamber. The inert gas contributes the same pressure in both measurements, but its contribution does not disappear from the denominator of an uncorrected pressure ratio. Subtracting it first is essential. The corrected solvent pressures yield the solute mole fraction, which determines the solute amount and molar mass. \(\textbf{Final answer:}\) \[ \boxed{L=\frac1{11},\qquad M_X=60.0\,\mathrm{g\,mol^{-1}}} \]

Explanation: \(\textbf{Translate relative lowering into mole fraction:}\) For an ideal aqueous solution containing nonvolatile molecular solutes: \[ L=x_{\text{solute,total}} \] Therefore, the original samples have total solute mole fractions: \[ x_{s,P}=0.100 \] \[ x_{s,Q}=0.200 \] \(\textbf{Interpret the equal-sample condition:}\) The samples contain equal total numbers of liquid-phase molecules. This does not mean they contain equal numbers of water molecules. It means the sum of solvent and solute molecules is equal for the two selected samples. Let each contain \(N\) total moles of liquid components. \(\textbf{Reconstruct sample P:}\) \[ n_{s,P}=0.100N \] \[ n_{w,P}=0.900N \] \(\textbf{Reconstruct sample Q:}\) \[ n_{s,Q}=0.200N \] \[ n_{w,Q}=0.800N \] \(\textbf{Combine the molecular inventories:}\) The mixed solution contains total dissolved solute amount: \[ n_{s,\text{mixed}}=0.100N+0.200N \] \[ n_{s,\text{mixed}}=0.300N \] Its water amount is: \[ n_{w,\text{mixed}}=0.900N+0.800N \] \[ n_{w,\text{mixed}}=1.700N \] The total amount is: \[ n_{\text{mixed,total}}=2.000N \] \(\textbf{Determine the resulting mole fractions:}\) \[ x_{s,\text{mixed}}=\frac{0.300N}{2.000N} \] \[ x_{s,\text{mixed}}=0.150 \] Therefore: \[ x_{w,\text{mixed}}=1-0.150 \] \[ x_{w,\text{mixed}}=0.850 \] \(\textbf{Connect the composition with vapour-pressure lowering:}\) For the ideal mixed solution: \[ L_{\text{mixed}}=x_{s,\text{mixed}} \] \[ L_{\text{mixed}}=0.150 \] \(\textbf{Explain why averaging works here:}\) Mole fraction is a ratio based on total molecular amount. The two samples are specifically chosen to have equal total molecular amounts. Their combined solute mole fraction is therefore the arithmetic mean of the original solute mole fractions. An arithmetic average would not generally be justified if the samples instead contained equal masses, equal volumes or equal amounts of water. The stated sampling condition determines the correct weighting. \(\textbf{Final answer:}\) \[ \boxed{L_{\text{mixed}}=0.150,\qquad x_{w,\text{mixed}}=0.850} \]

Explanation: \(\textbf{Express relative lowering through solute amount:}\) For an ideal solution of a nonvolatile molecular solute: \[ L=\frac{n_X}{n_S+n_X} \] The solvent amount remains \(10.0\,\mathrm{mol}\). Therefore: \[ L=\frac{n_X}{10.0+n_X} \] \(\textbf{Evaluate the initial state:}\) Before addition: \[ n_X=0 \] Therefore: \[ L_0=0 \] \(\textbf{Evaluate the first addition:}\) After adding \(1.00\,\mathrm{mol}\) X: \[ L_1=\frac{1.00}{10.0+1.00} \] \[ L_1=\frac1{11} \] The first increase is: \[ \Delta L_1=L_1-L_0 \] \[ \Delta L_1=\frac1{11} \] \(\textbf{Evaluate the second addition:}\) After adding another mole: \[ L_2=\frac{2.00}{10.0+2.00} \] \[ L_2=\frac16 \] The second increase is: \[ \Delta L_2=L_2-L_1 \] \[ \Delta L_2=\frac16-\frac1{11} \] \[ \Delta L_2=\frac5{66} \] \(\textbf{Compare the two increments:}\) \[ \frac5{66}\lt \frac1{11} \] Therefore, the second equal-mole addition produces a smaller increase in relative lowering. \(\textbf{Compare the molality increments:}\) Molality is: \[ m=\frac{n_X}{w_S(\mathrm{kg})} \] The solvent mass remains fixed. Each addition contributes exactly \(1.00\,\mathrm{mol}\) X. Therefore, each addition increases molality by the same amount: \[ \Delta m=\frac{1.00}{w_S(\mathrm{kg})} \] \(\textbf{Interpret the contrasting graph behaviour:}\) Molality is directly proportional to solute amount when solvent mass remains constant. Relative lowering is not directly proportional to solute amount at finite concentration. Adding X increases both the numerator and denominator of its exact mole-fraction expression. Consequently, successive equal-mole additions produce progressively smaller increases in relative lowering. The relative-lowering graph rises but becomes progressively less steep, whereas a molality-versus-solute-amount graph remains linear. \(\textbf{Conclusion:}\) The second relative-lowering increment is smaller, but the two molality increments are equal.

Explanation: \(\textbf{Choose a molality-compatible basis:}\) Consider \(1.000\,\mathrm{kg}\) of solvent S. At molality \(0.500\,\mathrm{mol\,kg^{-1}}\), the solution contains: \[ n_X=0.500\,\mathrm{mol} \] Let the solvent molar mass be \(M_S\), expressed in \(\mathrm{g\,mol^{-1}}\). The solvent amount is: \[ n_S=\frac{1000}{M_S} \] \(\textbf{Use the exact relative-lowering relation:}\) For the ideal nonvolatile-solute solution: \[ L=\frac{n_X}{n_S+n_X} \] Substitution gives: \[ \frac1{101}=\frac{0.500}{1000/M_S+0.500} \] \(\textbf{Recover the solvent molecular amount:}\) Cross-multiplying: \[ \frac{1000}{M_S}+0.500=101(0.500) \] \[ \frac{1000}{M_S}+0.500=50.5 \] \[ \frac{1000}{M_S}=50.0 \] Therefore: \[ M_S=\frac{1000}{50.0} \] \[ M_S=20.0\,\mathrm{g\,mol^{-1}} \] \(\textbf{Use the boiling-point elevation measurement:}\) The dilute ebullioscopic relation for a molecular nonelectrolyte is: \[ \Delta T_b=K_bm \] Therefore: \[ K_b=\frac{\Delta T_b}{m} \] \[ K_b=\frac{0.400}{0.500} \] \[ K_b=0.800\,\mathrm{K\,kg\,mol^{-1}} \] \(\textbf{Verify the mole-fraction calculation:}\) One kilogram of S contains: \[ n_S=\frac{1000}{20.0} \] \[ n_S=50.0\,\mathrm{mol} \] Therefore: \[ x_X=\frac{0.500}{50.0+0.500} \] \[ x_X=\frac1{101} \] The observed relative lowering is reproduced. \(\textbf{Interpret the independent measurements:}\) Molality counts solute molecules per kilogram of solvent. Relative lowering counts solute molecules as a fraction of all molecules in the liquid. Combining those two concentration scales determines the number of solvent molecules per kilogram and hence the solvent molar mass. The boiling-point elevation then determines the independent solvent constant \(K_b\). Neither measurement alone supplies both unknown solvent properties. \(\textbf{Final answer:}\) \[ \boxed{M_S=20.0\,\mathrm{g\,mol^{-1}},\qquad K_b=0.800\,\mathrm{K\,kg\,mol^{-1}}} \]

Explanation: \(\textbf{Compare the molalities:}\) For P: \[ m_P=\frac{0.200}{0.500} \] \[ m_P=0.400\,\mathrm{mol\,kg^{-1}} \] For Q: \[ m_Q=\frac{0.300}{0.750} \] \[ m_Q=0.400\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Infer the freezing-point relationship:}\) Both solutions use water as the solvent, and both solutes remain molecularly dissolved. Therefore: \[ \Delta T_f=K_fm \] Their equal molalities imply: \[ \Delta T_{f,P}=\Delta T_{f,Q} \] \(\textbf{Calculate P's molarity:}\) \[ C_P=\frac{0.200}{0.550} \] \[ C_P=\frac4{11}\,\mathrm M \] \(\textbf{Calculate Q's molarity:}\) \[ C_Q=\frac{0.300}{0.900} \] \[ C_Q=\frac13\,\mathrm M \] \(\textbf{Determine the osmotic-pressure ratio:}\) For both molecular nonelectrolytes at the same temperature: \[ \pi=CRT \] Therefore: \[ \frac{\pi_P}{\pi_Q}=\frac{C_P}{C_Q} \] \[ \frac{\pi_P}{\pi_Q}=\frac{4/11}{1/3} \] \[ \frac{\pi_P}{\pi_Q}=\frac{12}{11} \] \(\textbf{Interpret the unequal volume scaling:}\) Q contains \(1.50\) times the solute amount in P. It also contains \(1.50\) times the water mass, so the two molalities are equal. However, Q's measured solution volume is not \(1.50\) times P's volume. Consequently, the two solute molarities are different. \(\textbf{Explain the experimental distinction:}\) Freezing-point depression is governed by effective solute-particle amount per kilogram of solvent. Osmotic pressure is governed by effective solute-particle amount per litre of solution. Equal molality therefore guarantees equal ideal freezing-point depression for the same solvent, but it does not guarantee equal osmotic pressure. The measured final volumes supply the additional information needed for that comparison. \(\textbf{Final answer:}\) \[ \boxed{\Delta T_{f,P}=\Delta T_{f,Q},\qquad\frac{\pi_P}{\pi_Q}=\frac{12}{11}} \]

Explanation: \(\textbf{Calculate the initial molality:}\) The original solution contains: \[ n_{X,i}=0.300\,\mathrm{mol} \] \[ w_{w,i}=0.500\,\mathrm{kg} \] Therefore: \[ m_i=\frac{0.300}{0.500} \] \[ m_i=0.600\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Apply the two separate removals:}\) Pure ice removes water but not X. Crystallisation of X removes dissolved solute but not water. The final water mass is: \[ w_{w,f}=500-100 \] \[ w_{w,f}=400\,\mathrm g=0.400\,\mathrm{kg} \] The final dissolved amount of X is: \[ n_{X,f}=0.300-0.0600 \] \[ n_{X,f}=0.240\,\mathrm{mol} \] \(\textbf{Calculate the final molality:}\) \[ m_f=\frac{0.240}{0.400} \] \[ m_f=0.600\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Compare freezing-point depressions:}\) The same solvent and dissolved molecular solute remain. \[ \Delta T_f=K_fm \] Consequently: \[ \frac{\Delta T_{f,f}}{\Delta T_{f,i}}=\frac{m_f}{m_i} \] \[ \frac{\Delta T_{f,f}}{\Delta T_{f,i}}=1 \] \(\textbf{Compare the solute-to-water mole ratios:}\) The ratio of dissolved solute amount to water mass is unchanged. Because the solvent identity is unchanged, the number of water molecules per kilogram is also unchanged. Therefore, the ratio \(n_X/n_w\) remains constant. \(\textbf{Apply the exact ideal relative-lowering relation:}\) \[ L=\frac{n_X}{n_X+n_w} \] The unchanged solute-to-water mole ratio implies the same solute mole fraction. Therefore: \[ L_f=L_i \] \(\textbf{Interpret the competing separation processes:}\) Removing water as pure ice tends to increase concentration. Removing dissolved X by crystallisation tends to decrease concentration. Here, both dissolved solute amount and water mass decrease by the same fractional amount. Their ratio remains constant. Consequently, the remaining liquid has the same molality and the same exact solute mole fraction as the original solution. Both colligative measures are therefore unchanged, even though the total mass of liquid has decreased. \(\textbf{Conclusion:}\) The freezing-point depression and relative vapour-pressure lowering both remain unchanged.

Explanation: \(\textbf{Represent the contributions of two solutes:}\) For independent molecular nonelectrolytes: \[ \pi=(C_X+C_Y)RT \] If the mass concentration of X is \(c_X\), expressed in \(\mathrm{g\,L^{-1}}\): \[ C_X=\frac{c_X}{M_X} \] Therefore: \[ \pi=\left(\frac{c_X}{M_X}+C_Y\right)RT \] \(\textbf{Identify the graph's slope and intercept:}\) Rearranging: \[ \pi=\frac{RT}{M_X}c_X+C_YRT \] This is a straight-line relation. The slope is: \[ s=\frac{RT}{M_X} \] The intercept is: \[ b=C_YRT \] \(\textbf{Recover the molar mass of X:}\) \[ M_X=\frac{RT}{s} \] Substitution gives: \[ M_X=\frac{0.0820(300)}{0.00820} \] \[ M_X=\frac{24.6}{0.00820} \] \[ M_X=3000\,\mathrm{g\,mol^{-1}} \] \(\textbf{Recover Y's concentration:}\) The pressure intercept is: \[ b=0.123\,\mathrm{atm} \] Therefore: \[ C_Y=\frac{b}{RT} \] \[ C_Y=\frac{0.123}{0.0820(300)} \] \[ C_Y=\frac{0.123}{24.6} \] \[ C_Y=0.00500\,\mathrm M \] \(\textbf{Interpret the different graph features:}\) Increasing X's mass concentration changes only X's contribution to the osmotic pressure. The slope therefore determines the molecular amount introduced per gram of X. The fixed Y concentration contributes a constant positive pressure offset. That offset appears as the nonzero intercept. \(\textbf{Distinguish the intercept from a molecular-mass effect:}\) A nonzero intercept does not require X to associate or dissociate. It can arise from an independently dissolved background solute whose amount per litre is constant. The slope and intercept encode different chemical information. Both must be interpreted before attributing the graph to a single-solute solution. \(\textbf{Final answer:}\) \[ \boxed{M_X=3000\,\mathrm{g\,mol^{-1}},\qquad C_Y=0.00500\,\mathrm M} \]

Explanation: \(\textbf{Verify the observed freezing-point equality:}\) For P: \[ m_P=\frac{0.100}{0.500} \] \[ m_P=0.200\,\mathrm{mol\,kg^{-1}} \] For Q: \[ m_Q=\frac{0.150}{0.750} \] \[ m_Q=0.200\,\mathrm{mol\,kg^{-1}} \] Both use water and contain independent molecular solutes. Therefore, their freezing-point depressions are equal under the ideal dilute relation. \(\textbf{Identify the osmotic-pressure requirement:}\) Osmotic pressure depends on molarity: \[ \pi=CRT \] The two solutions are measured at the same temperature. Therefore, their osmotic pressures are proportional to their solute molarities, not directly to molalities. \(\textbf{Calculate P's molarity:}\) \[ C_P=\frac{0.100}{0.525} \] \[ C_P\approx0.190476\,\mathrm M \] \(\textbf{Calculate Q's molarity:}\) \[ C_Q=\frac{0.150}{0.800} \] \[ C_Q=0.187500\,\mathrm M \] \(\textbf{Compare the osmotic pressures:}\) \[ \frac{\pi_P}{\pi_Q}=\frac{C_P}{C_Q} \] \[ \frac{\pi_P}{\pi_Q}=\frac{0.190476}{0.187500} \] \[ \frac{\pi_P}{\pi_Q}=\frac{64}{63} \] Thus: \[ \pi_P\gt \pi_Q \] \(\textbf{Identify the conceptual error:}\) The experimenter treats equal molality as if it necessarily implied equal molarity. However, the solvent-mass and solution-volume denominators are different. The samples contain solute and solvent amounts in the same ratios, but their measured volumes do not scale in precisely the same ratio. Consequently, their molarities and osmotic pressures differ slightly. \(\textbf{Interpret the experimental significance:}\) Freezing-point depression and osmotic pressure are both colligative properties, but they need not be numerically equivalent indicators of concentration. They depend on different concentration measures. Equal freezing-point depressions establish equal effective particle molality for the same solvent. An independent solution-volume measurement is still necessary to compare osmotic pressures. \(\textbf{Final answer:}\) \[ \boxed{\frac{\pi_P}{\pi_Q}=\frac{64}{63}\gt 1} \]

Explanation: \(\textbf{Establish the initial molality:}\) The initial solution contains: \[ n_{X,i}=0.200\,\mathrm{mol} \] \[ w_{w,i}=0.400\,\mathrm{kg} \] Therefore: \[ m_i=\frac{0.200}{0.400} \] \[ m_i=0.500\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Translate the final depression measurement:}\) For a dilute molecular nonelectrolyte: \[ \Delta T_f=K_fm \] Both measurements use the same solvent. Therefore: \[ \frac{\Delta T_{f,f}}{\Delta T_{f,i}}=\frac{m_f}{m_i} \] The measured ratio is \(0.750\). Thus: \[ m_f=0.750(0.500) \] \[ m_f=0.375\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Reconstruct the final solvent mass:}\) Only water evaporates in the first operation. \[ w_{w,f}=400-100 \] \[ w_{w,f}=300\,\mathrm g \] \[ w_{w,f}=0.300\,\mathrm{kg} \] No additional water is lost during X crystallisation. \(\textbf{Determine the final dissolved amount:}\) \[ n_{X,f}=m_fw_{w,f} \] \[ n_{X,f}=0.375(0.300) \] \[ n_{X,f}=0.1125\,\mathrm{mol} \] \(\textbf{Calculate the crystallised amount:}\) \[ n_{X,\text{crystallised}}=n_{X,i}-n_{X,f} \] \[ n_{X,\text{crystallised}}=0.200-0.1125 \] \[ n_{X,\text{crystallised}}=0.0875\,\mathrm{mol} \] \(\textbf{Determine the dissolved fraction remaining:}\) \[ f_{\text{remaining}}=\frac{0.1125}{0.200} \] \[ f_{\text{remaining}}=0.5625 \] \[ f_{\text{remaining}}=\frac9{16} \] \(\textbf{Interpret the two successive processes:}\) Evaporation alone would increase molality by reducing solvent mass. Crystallisation subsequently reduces the dissolved solute amount. The observed final depression is lower than the original, showing that solute removal more than compensates for solvent evaporation. It would be incorrect to use the \(75.0\%\) depression ratio directly as the remaining fraction of X because the solvent mass also changes. Both material balances must be included. \(\textbf{Final answer:}\) \[ \boxed{n_{X,\text{crystallised}}=0.0875\,\mathrm{mol},\qquad f_{\text{remaining}}=\frac9{16}} \]

Explanation: \(\textbf{Establish the initial osmotic equality:}\) The two original solutions have the same molarity. Both glucose and urea remain molecularly dissolved. Therefore, at the same temperature: \[ \pi_P=C_PRT \] \[ \pi_Q=C_QRT \] Since: \[ C_P=C_Q=0.100\,\mathrm M \] the two solutions are initially isotonic. \(\textbf{Determine the added glucose amount:}\) \[ n_{\text{glucose,added}}=\frac{9.00}{180} \] \[ n_{\text{glucose,added}}=0.0500\,\mathrm{mol} \] The volume of P is restored to \(1.000\,\mathrm L\). Therefore, the added glucose raises its molarity by \(0.0500\,\mathrm M\). \(\textbf{Calculate P's final molarity:}\) \[ C_{P,f}=0.100+0.0500 \] \[ C_{P,f}=0.150\,\mathrm M \] \(\textbf{Determine the added urea amount:}\) \[ n_{\text{urea,added}}=\frac{9.00}{60.0} \] \[ n_{\text{urea,added}}=0.150\,\mathrm{mol} \] This increases Q's molarity by \(0.150\,\mathrm M\). \(\textbf{Calculate Q's final molarity:}\) \[ C_{Q,f}=0.100+0.150 \] \[ C_{Q,f}=0.250\,\mathrm M \] \(\textbf{Determine the osmotic direction:}\) At the common temperature: \[ \frac{\pi_{Q,f}}{\pi_{P,f}}=\frac{0.250}{0.150} \] \[ \frac{\pi_{Q,f}}{\pi_{P,f}}=\frac53 \] Q has the greater osmotic pressure. At equal mechanical pressures, net water movement is therefore from P to Q. \(\textbf{Interpret the effect of equal added masses:}\) The initial solutions have equal numbers of dissolved particles per litre. However, adding equal masses of different molecular solutes does not add equal numbers of molecules. Urea has one-third the molar mass of glucose. The same added mass of urea consequently contributes three times as many new molecules. The final solutions are no longer isotonic. \(\textbf{Conclusion:}\) The final molarities are \(0.150\,\mathrm M\) for P and \(0.250\,\mathrm M\) for Q. Because Q has the greater effective solute-particle concentration, its osmotic pressure is higher. Consequently, water moves from P to Q through the semipermeable membrane.

Explanation: \(\textbf{Identify the conserved material quantities:}\) The expansion occurs without loss of solute or solvent. Therefore, the number of solute molecules remains fixed. The solvent mass also remains fixed. Consequently, molality does not change. \(\textbf{Interpret the freezing-point relation:}\) For a dilute molecular nonelectrolyte: \[ \Delta T_f=K_fm \] The problem specifies that changes in \(K_f\) are neglected. Since molality is unchanged: \[ \Delta T_{f,2}=\Delta T_{f,1} \] \(\textbf{Examine the osmotic-pressure concentration:}\) For ideal dilute behaviour: \[ \pi=\frac{nRT}{V} \] The solute amount \(n\) remains fixed. The final volume is: \[ V_2=1.10V_1 \] The final absolute temperature is: \[ T_2=1.10T_1 \] \(\textbf{Calculate the osmotic-pressure ratio:}\) \[ \frac{\pi_2}{\pi_1}=\frac{nRT_2/V_2}{nRT_1/V_1} \] \[ \frac{\pi_2}{\pi_1}=\frac{T_2}{T_1}\frac{V_1}{V_2} \] \[ \frac{\pi_2}{\pi_1}=1.10\left(\frac1{1.10}\right) \] \[ \frac{\pi_2}{\pi_1}=1 \] Thus, osmotic pressure is unchanged. \(\textbf{Explain the cancellation:}\) Increasing absolute temperature tends to raise osmotic pressure. Increasing solution volume lowers the number of solute particles per litre. The two changes are proportionally equal. Their effects cancel in the ideal osmotic-pressure equation. \(\textbf{Explain why the density data are unnecessary:}\) The original density and solute molar mass would permit conversion between molality and molarity. However, the requested comparison concerns ratios between two states containing identical material amounts. The unknown solute amount cancels from the osmotic-pressure ratio. The initial density therefore need not be used. Recognising unnecessary information is part of the required inference. \(\textbf{Final answer:}\) \[ \boxed{\frac{\pi_2}{\pi_1}=1,\qquad\frac{\Delta T_{f,2}}{\Delta T_{f,1}}=1} \]

Explanation: \(\textbf{Interpret the osmotic-pressure result:}\) The problem states that osmotic-pressure analysis gives: \[ C=0.100\,\mathrm{mol\,L^{-1}} \] This means the solution contains \(0.100\,\mathrm{mol}\) dissolved X per litre of final solution. It does not specify the solvent mass in that litre. \(\textbf{Recover molality from freezing-point depression:}\) For a molecular nonelectrolyte: \[ \Delta T_f=K_fm \] Therefore: \[ m=\frac{\Delta T_f}{K_f} \] \[ m=\frac{0.186}{1.86} \] \[ m=0.100\,\mathrm{mol\,kg^{-1}} \] This means the solution contains \(0.100\,\mathrm{mol}\) dissolved X per kilogram of water. \(\textbf{Distinguish the two concentration denominators:}\) Molarity refers to litres of final solution. Molality refers to kilograms of solvent. The numerical values can coincide for a particular solution, but their definitions remain different. There is no general identity requiring molarity and molality to have the same numerical value. \(\textbf{Assess the student's statements:}\) The first numerical statement follows from the freezing-point depression. The second follows from the independently established molarity. Both are supported by the specified measurements. However, their equality is a property of this preparation, not a universal rule for aqueous solutions. \(\textbf{Interpret the combined measurement:}\) If exactly one litre of this solution is considered, it contains \(0.100\,\mathrm{mol}\) X. Because the molality is also \(0.100\,\mathrm{mol\,kg^{-1}}\), that litre corresponds to exactly one kilogram of solvent under the idealised measurements. This additional relation is inferred from combining the two concentrations. The measurements are therefore consistent, while the student's apparent assumption of universal equivalence would be incorrect. \(\textbf{Conclusion:}\) Both reported numerical concentrations are supported, but they measure different concentration quantities.

Explanation: \(\textbf{Identify the conserved solute amount:}\) The solute mass and molar mass are the same in both preparations. Therefore, the molecular amount of X is identical. Let: \[ n_X=n \] for both P and Q. \(\textbf{Compare the molalities:}\) The solvent mass in P is \(m_S\). In Q, it is \(2m_S\). Therefore: \[ m_P=\frac{n}{m_S} \] \[ m_Q=\frac{n}{2m_S} \] Consequently: \[ \frac{m_P}{m_Q}=2 \] \(\textbf{Compare freezing-point depressions:}\) Both preparations use the same solvent. For the stated molecular-nonelectrolyte model: \[ \Delta T_f=K_fm \] Therefore: \[ \frac{\Delta T_{f,P}}{\Delta T_{f,Q}}=\frac{m_P}{m_Q} \] \[ \frac{\Delta T_{f,P}}{\Delta T_{f,Q}}=2 \] \(\textbf{Compare osmotic pressures:}\) At the common absolute temperature: \[ \pi=\frac{nRT}{V} \] The dissolved amount \(n\) is identical. Therefore, osmotic pressure varies inversely with final solution volume. \[ \frac{\pi_P}{\pi_Q}=\frac{V_Q}{V_P} \] The measured volume relation is: \[ V_Q=1.50V_P \] Thus: \[ \frac{\pi_P}{\pi_Q}=1.50 \] \(\textbf{Interpret the differing ratios:}\) Doubling the solvent mass halves the molality, so the freezing-point depression in Q is half that in P. However, the final solution volume of Q is only \(1.50\) times that of P. Its solute molarity is consequently two-thirds that of P, not one-half. The two colligative ratios therefore differ. \(\textbf{Identify the controlling measurements:}\) The freezing-point comparison requires the solvent-mass ratio. The osmotic-pressure comparison requires the final solution-volume ratio. Neither ratio can be substituted for the other merely because the solute and solvent identities are the same. This distinction is essential whenever the relationship between solution volume and solvent mass is not proportional. \(\textbf{Final answer:}\) \[ \boxed{\frac{\Delta T_{f,P}}{\Delta T_{f,Q}}=2,\qquad\frac{\pi_P}{\pi_Q}=1.50} \]