Solutions MCQs with Answers – Part 5 (NEET Chemistry)

Explanation: \(\textbf{Determine NaCl's effective factor:}\) Complete dissociation of NaCl produces two ions. Under partial dissociation: \[ i_P=1+\alpha_P \] \[ i_P=1+0.800 \] \[ i_P=1.80 \] \(\textbf{Determine Na₂SO₄'s effective factor:}\) Complete dissociation produces three ions per formula unit. Therefore: \[ i_Q=1+2\alpha_Q \] \[ i_Q=1+2(0.400) \] \[ i_Q=1.80 \] \(\textbf{Compare the freezing-point depressions:}\) Both analytical molalities are: \[ m=0.200\,\mathrm{mol\,kg^{-1}} \] Their effective particle molalities are: \[ m_{\text{effective},P}=1.80(0.200) \] \[ m_{\text{effective},P}=0.360\,\mathrm{mol\,kg^{-1}} \] \[ m_{\text{effective},Q}=1.80(0.200) \] \[ m_{\text{effective},Q}=0.360\,\mathrm{mol\,kg^{-1}} \] Both use the same solvent constant. Therefore, their ideal freezing-point depressions are equal. \(\textbf{Compare the osmotic concentrations:}\) Both preparations contain \(0.100\,\mathrm{mol}\) analytical electrolyte in \(0.500\,\mathrm L\). Their analytical molarities are equal. The effective molarity of each is: \[ C_{\text{effective}}=1.80\left(\frac{0.100}{0.500}\right) \] \[ C_{\text{effective}}=0.360\,\mathrm M \] At the same temperature: \[ \pi_P=\pi_Q \] \(\textbf{Interpret the equal particle factors:}\) The two electrolytes have different dissociation degrees. However, the number of product ions per fully dissociated formula unit also differs. A higher dissociation degree for a two-ion electrolyte can therefore produce the same effective particle factor as a lower degree for a three-ion electrolyte. The charge magnitude of an ion does not itself count as an additional particle. The equality of their colligative properties follows from equal effective particle concentrations, not identical ionic species. \(\textbf{Conclusion:}\) P and Q have equal freezing-point depressions and osmotic pressures under the stipulated conditions.

Explanation: \(\textbf{Recover the original particle factor:}\) For dimerisation: \[ i=1-\frac{\alpha}{2} \] Initially: \[ i_1=1-\frac{0.500}{2} \] \[ i_1=0.750 \] \(\textbf{Determine the analytical molality:}\) The corrected cryoscopic relation is: \[ \Delta T_f=iK_fm \] Therefore: \[ m_1=\frac{\Delta T_f}{i_1K_f} \] \[ m_1=\frac{0.750}{0.750(5.00)} \] \[ m_1=0.200\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Recover the original amount and mass:}\) \[ n_{X,1}=m_1w_S \] \[ n_{X,1}=0.200(0.400) \] \[ n_{X,1}=0.0800\,\mathrm{mol} \] Therefore: \[ w_{X,1}=0.0800(200) \] \[ w_{X,1}=16.0\,\mathrm g \] \(\textbf{Apply homogeneous withdrawal:}\) Half the original homogeneous solution contains half the analytical X amount and half the solvent mass. \[ n_{X,\text{withdrawn}}=0.0400\,\mathrm{mol} \] \[ w_{S,\text{withdrawn}}=0.200\,\mathrm{kg} \] \(\textbf{Apply the additional solvent dilution:}\) Exactly \(200\,\mathrm g\) pure S is added. The new solvent mass is: \[ w_{S,2}=0.200+0.200 \] \[ w_{S,2}=0.400\,\mathrm{kg} \] The analytical X amount remains \(0.0400\,\mathrm{mol}\). Thus: \[ m_2=\frac{0.0400}{0.400} \] \[ m_2=0.100\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Apply the changed association degree:}\) The final degree of dimerisation is \(20.0\%\). \[ i_2=1-\frac{0.200}{2} \] \[ i_2=0.900 \] \(\textbf{Calculate the final depression:}\) \[ \Delta T_{f,2}=i_2K_fm_2 \] \[ \Delta T_{f,2}=0.900(5.00)(0.100) \] \[ \Delta T_{f,2}=0.450\,\mathrm K \] \(\textbf{Check both preparation stages:}\) The initial mass of X was \(16.0\,\mathrm g\). Homogeneous withdrawal halves both solute and solvent amounts without changing the initial analytical molality of the withdrawn fraction. Subsequently adding \(200\,\mathrm g\) solvent halves the withdrawn fraction's molality, while the changed dimerisation degree raises its particle factor from \(0.750\) to \(0.900\). Using the final particle factor yields a depression of \(0.450\,\mathrm K\), not \(0.225\,\mathrm K\). Thus, the initial mass and final depression jointly match option B. \(\textbf{Final answer:}\) \[ \boxed{w_{X,\text{original}}=16.0\,\mathrm g,\qquad\Delta T_{f,\text{final}}=0.450\,\mathrm K} \]

Explanation: \(\textbf{Determine the analytical polymer molarity:}\) The mass concentration is \(8.00\,\mathrm{g\,L^{-1}}\). \[ C_0=\frac{8.00}{4.00\times10^4} \] \[ C_0=2.00\times10^{-4}\,\mathrm M \] \(\textbf{Calculate the unassociated pressure reference:}\) If every polymer molecule remained independent: \[ \pi_0=C_0RT \] \[ \pi_0=(2.00\times10^{-4})(0.0820)(300) \] \[ \pi_0=0.00492\,\mathrm{atm} \] \(\textbf{Recover the particle factor:}\) The observed osmotic pressure is \(0.00369\,\mathrm{atm}\). \[ i=\frac{\pi_{\text{observed}}}{\pi_0} \] \[ i=\frac{0.00369}{0.00492} \] \[ i=0.750 \] \(\textbf{Determine the dimerisation degree:}\) For: \[ 2X\rightleftharpoons X_2 \] the particle factor is: \[ i=1-\frac{\alpha}{2} \] Substitution gives: \[ 0.750=1-\frac{\alpha}{2} \] \[ \alpha=0.500 \] \[ \alpha=50.0\% \] \(\textbf{Calculate the apparent molar mass:}\) \[ M_{\text{app}}=\frac{M_{\text{normal}}}{i} \] \[ M_{\text{app}}=\frac{4.00\times10^4}{0.750} \] \[ M_{\text{app}}\approx5.33\times10^4\,\mathrm{g\,mol^{-1}} \] \(\textbf{Verify the association interpretation:}\) For one analytical mole of original polymer molecules, \(0.500\,\mathrm{mol}\) remain monomeric. The other \(0.500\,\mathrm{mol}\) original molecules form \(0.250\,\mathrm{mol}\) dimers. The solution consequently contains \(0.750\,\mathrm{mol}\) independent solute species per analytical mole. That particle reduction explains the lower measured osmotic pressure. \(\textbf{Interpret the method:}\) Osmotic pressure responds to the number of independently moving dissolved solute species. The large normal molar mass makes the analytical molarity small even at a measurable mass concentration. Dimerisation reduces the effective molarity further. Using the uncorrected osmotic expression therefore gives an apparently larger molecular mass. \(\textbf{Final answer:}\) \[ \boxed{\alpha=50.0\%,\qquad M_{\text{app}}\approx5.33\times10^4\,\mathrm{g\,mol^{-1}}} \]

Explanation: \(\textbf{Evaluate the Assertion:}\) The normal molar mass is established independently of the colligative measurements. Both measurements give a larger apparent molar mass. This is consistent with fewer effective solute particles than would exist if all original molecules remained separate. Agreement between the two methods therefore does not prove that association is absent. Observation R also independently establishes that associated species are present. The Assertion is true. \(\textbf{Evaluate the Reason:}\) Freezing-point depression depends on effective dissolved-particle molality. Osmotic pressure depends on effective dissolved-particle molarity and absolute temperature. Despite the different concentration denominators, both methods count independently moving solute species under their respective ideal dilute approximations. If the association behaviour is effectively the same under both measurement conditions, both can indicate the same abnormal molecular mass. The Reason is true. \(\textbf{Establish the cause-and-effect connection:}\) Association reduces the number of effective solute particles. Both observed colligative effects consequently become smaller than their unassociated predictions. Applying the ordinary molecular-mass expressions without correcting the particle count produces a larger apparent molar mass in both methods. Their agreement is therefore compatible with association rather than evidence against it. The Reason directly explains the Assertion. \(\textbf{Interpret the experimental limitation:}\) A colligative measurement primarily supplies information about effective particle concentration. Identifying aggregate size or molecular structure usually requires additional chemical information or an explicitly justified association model. Different experimental methods may agree on particle counting while remaining unable to identify the detailed association mechanism. Both A and R are true, and R correctly explains A.

Explanation: \(\textbf{Recover the total effective particle concentration:}\) The ideal osmotic relation is: \[ \pi=C_{\text{effective,total}}RT \] Therefore: \[ C_{\text{effective,total}}=\frac{\pi}{RT} \] \[ C_{\text{effective,total}}=\frac{5.658}{0.0820(300)} \] \[ C_{\text{effective,total}}=\frac{5.658}{24.6} \] \[ C_{\text{effective,total}}=0.230\,\mathrm M \] \(\textbf{Separate the glucose contribution:}\) Glucose remains molecularly dissolved. Therefore: \[ C_{\text{effective,glucose}}=0.100\,\mathrm M \] \(\textbf{Recover the electrolyte contribution:}\) \[ C_{\text{effective,MX}_2}=0.230-0.100 \] \[ C_{\text{effective,MX}_2}=0.130\,\mathrm M \] \(\textbf{Determine the electrolyte particle factor:}\) Its analytical molarity is \(0.0650\,\mathrm M\). \[ i_{\mathrm{MX_2}}=\frac{0.130}{0.0650} \] \[ i_{\mathrm{MX_2}}=2.00 \] \(\textbf{Calculate its dissociation degree:}\) Complete dissociation produces three independent ions. \[ i=1+2\alpha \] Therefore: \[ 2.00=1+2\alpha \] \[ \alpha=0.500 \] \[ \alpha=50.0\% \] \(\textbf{Calculate glucose's osmotic contribution:}\) \[ \pi_{\text{glucose}}=0.100(0.0820)(300) \] \[ \pi_{\text{glucose}}=2.46\,\mathrm{atm} \] Its fraction of the total osmotic pressure is: \[ f_{\text{glucose}}=\frac{2.46}{5.658} \] \[ f_{\text{glucose}}=\frac{10}{23} \] \(\textbf{Verify the two components independently:}\) The salt contributes \(0.0650(2.00)=0.130\,\mathrm M\) effective particles. Adding glucose gives \(0.130+0.100=0.230\,\mathrm M\), consistent with the measured \(5.658\,\mathrm{atm}\) at \(300\,\mathrm K\). Glucose's fraction is determined from its own effective particle concentration divided by the total, not from the relative number of salt formula units. The resulting pair matches option A. \(\textbf{Final answer:}\) \[ \boxed{\alpha=50.0\%,\qquad f_{\text{glucose}}=\frac{10}{23}} \]

Explanation: \(\textbf{Evaluate Statement I:}\) Adding pure water increases the solvent mass without changing the solute molecular amount. Molality therefore decreases. For the same solvent and molecular nonelectrolyte, freezing-point depression is proportional to molality. The depression consequently decreases. Statement I is correct. \(\textbf{Evaluate Statement II:}\) Molality depends on dissolved solute amount divided by solvent mass. Neither changes during the stated warming experiment. Molality therefore remains constant. Molarity, however, depends on the final solution volume. If warming changes the liquid volume, its molarity may change despite unchanged material amounts. Statement II is correct. \(\textbf{Evaluate Statement III:}\) Pure ice contains solvent but excludes the nonvolatile dissolved solute under the stated model. Removing ice therefore reduces liquid solvent mass while retaining dissolved X. The molality of X in the remaining liquid increases. Its freezing-point depression increases accordingly. Statement III is correct. \(\textbf{Evaluate Statement IV:}\) A greater freezing-point depression means the equilibrium freezing temperature lies further below the pure-solvent freezing temperature. Removing pure ice concentrates the remaining solution. It therefore lowers, rather than raises, the equilibrium freezing temperature of the remaining liquid. Statement IV is incorrect. \(\textbf{Integrated interpretation:}\) Colligative behaviour is governed by the concentration measure appropriate to the property. Changing solvent mass directly changes molality. Changing solution volume can affect molarity without changing molality. Selective solvent crystallisation concentrates the remaining liquid and changes its subsequent freezing equilibrium. The correct statements are I, II and III.

Explanation: \(\textbf{Calculate the analytical solute amount:}\) The normal molar mass is \(100\,\mathrm{g\,mol^{-1}}\). \[ n_X=\frac{5.00}{100} \] \[ n_X=0.0500\,\mathrm{mol} \] \(\textbf{Determine the original analytical molality:}\) The solvent mass is \(0.500\,\mathrm{kg}\). \[ m_0=\frac{0.0500}{0.500} \] \[ m_0=0.100\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Construct the undimerised cryoscopic prediction:}\) \[ \Delta T_{f,0}=K_fm_0 \] \[ \Delta T_{f,0}=4.00(0.100) \] \[ \Delta T_{f,0}=0.400\,\mathrm K \] \(\textbf{Recover the observed particle factor:}\) \[ i=\frac{\Delta T_{f,\text{observed}}}{\Delta T_{f,0}} \] \[ i=\frac{0.240}{0.400} \] \[ i=0.600 \] \(\textbf{Determine the dimerisation degree:}\) For: \[ 2X\rightleftharpoons X_2 \] \[ i=1-\frac{\alpha}{2} \] Therefore: \[ 0.600=1-\frac{\alpha}{2} \] \[ \alpha=0.800 \] \[ \alpha=80.0\% \] \(\textbf{Recover the effective amount in the second preparation:}\) The second preparation has the same analytical X molality and the same degree of association. For the same \(0.500\,\mathrm{kg}\) solvent basis, its effective particle amount is: \[ n_{\text{effective}}=i n_X \] \[ n_{\text{effective}}=0.600(0.0500) \] \[ n_{\text{effective}}=0.0300\,\mathrm{mol} \] \(\textbf{Determine effective molarity:}\) The final solution volume is: \[ V=0.400\,\mathrm L \] Therefore: \[ C_{\text{effective}}=\frac{0.0300}{0.400} \] \[ C_{\text{effective}}=0.0750\,\mathrm M \] \(\textbf{Calculate osmotic pressure:}\) \[ \pi=C_{\text{effective}}RT \] \[ \pi=0.0750(0.0820)(300) \] \[ \pi=1.845\,\mathrm{atm} \] \(\textbf{Interpret the linked measurements:}\) The freezing-point measurement establishes the effective particle factor of X under the stated association model. The same factor may be applied to the second preparation only because its association degree is explicitly held unchanged. Osmotic pressure then requires the effective particle amount per litre of final solution, rather than per kilogram of solvent. The two measurements therefore share a particle-count correction but use different concentration denominators. \(\textbf{Final answer:}\) \[ \boxed{\alpha=80.0\%,\qquad\pi=1.845\,\mathrm{atm}} \]

Explanation: \(\textbf{Identify the chemical reaction:}\) Strong acid and strong base undergo neutralisation. The net ionic reaction is: \[ \mathrm{H^+(aq)+OH^-(aq)\rightarrow H_2O(l)} \] The spectator ions remain dissolved. Osmotic pressure must be calculated from the particles present after the reaction, rather than by counting reactant ions that have already been consumed. \(\textbf{Determine the initial ionic amounts:}\) Complete dissociation of HCl supplies: \[ n_{\mathrm{H^+}}=0.0600\,\mathrm{mol} \] \[ n_{\mathrm{Cl^-}}=0.0600\,\mathrm{mol} \] Complete dissociation of NaOH supplies: \[ n_{\mathrm{Na^+}}=0.0400\,\mathrm{mol} \] \[ n_{\mathrm{OH^-}}=0.0400\,\mathrm{mol} \] \(\textbf{Apply the neutralisation stoichiometry:}\) Hydroxide ions are limiting. Exactly \(0.0400\,\mathrm{mol}\) hydroxide ions react with the same amount of hydrogen ions. The remaining hydrogen-ion amount is: \[ n_{\mathrm{H^+,remaining}}=0.0600-0.0400 \] \[ n_{\mathrm{H^+,remaining}}=0.0200\,\mathrm{mol} \] \(\textbf{Count the particles after reaction:}\) The solution contains: \[ n_{\mathrm{H^+}}=0.0200\,\mathrm{mol} \] \[ n_{\mathrm{Na^+}}=0.0400\,\mathrm{mol} \] \[ n_{\mathrm{Cl^-}}=0.0600\,\mathrm{mol} \] Thus: \[ n_{\text{effective}}=0.0200+0.0400+0.0600 \] \[ n_{\text{effective}}=0.120\,\mathrm{mol} \] \(\textbf{Calculate the correct osmotic pressure:}\) \[ \pi=\frac{n_{\text{effective}}RT}{V} \] \[ \pi=\frac{0.120(0.0820)(300)}{0.500} \] \[ \pi=5.904\,\mathrm{atm} \] \(\textbf{Reconstruct the student's incorrect particle count:}\) The student counts all the initially introduced ions. \[ n_{\text{wrong}}=2(0.0600)+2(0.0400) \] \[ n_{\text{wrong}}=0.200\,\mathrm{mol} \] Consequently: \[ \pi_{\text{wrong}}=\frac{0.200(0.0820)(300)}{0.500} \] \[ \pi_{\text{wrong}}=9.840\,\mathrm{atm} \] \(\textbf{Explain the discrepancy:}\) Neutralisation removes equal amounts of hydrogen and hydroxide ions from the independent dissolved-particle inventory. The product water becomes part of the solvent. The number of dissolved ions therefore decreases, although charge and elemental material remain conserved. Colligative calculations involving reacting solutes require a chemical particle balance before the concentration calculation. \(\textbf{Final answer:}\) \[ \boxed{\pi_{\text{actual}}=5.904\,\mathrm{atm},\qquad\pi_{\text{wrong}}=9.840\,\mathrm{atm}} \]

Explanation: \(\textbf{Separate hydrate units from solvent molecules:}\) The formula mass includes anhydrous sodium carbonate and ten water molecules. When the crystalline hydrate dissolves, the water of crystallisation contributes to the liquid solvent inventory. It must not be counted as an additional dissolved ionic species. \(\textbf{Calculate the hydrate amount:}\) \[ n_{\text{hydrate}}=\frac{2.86}{286} \] \[ n_{\text{hydrate}}=0.0100\,\mathrm{mol} \] Therefore, the analytical amount of dissolved sodium carbonate is: \[ n_{\mathrm{Na_2CO_3}}=0.0100\,\mathrm{mol} \] \(\textbf{Recover the introduced crystal water:}\) Each mole of hydrate contains ten moles of water. \[ n_{\text{water,hydrate}}=10(0.0100) \] \[ n_{\text{water,hydrate}}=0.100\,\mathrm{mol} \] Its mass is: \[ w_{\text{water,hydrate}}=0.100(18.0) \] \[ w_{\text{water,hydrate}}=1.80\,\mathrm g \] \(\textbf{Determine the actual solvent mass:}\) \[ w_{\text{solvent}}=100.0+1.80 \] \[ w_{\text{solvent}}=101.8\,\mathrm g \] \[ w_{\text{solvent}}=0.1018\,\mathrm{kg} \] \(\textbf{Recover the effective dissolved-particle amount:}\) \[ \Delta T_f=K_f\frac{n_{\text{effective}}}{w_{\text{solvent}}} \] Therefore: \[ n_{\text{effective}}=\frac{0.3654(0.1018)}{1.86} \] \[ n_{\text{effective}}\approx0.0200\,\mathrm{mol} \] \(\textbf{Calculate the carbonate particle factor:}\) \[ i=\frac{n_{\text{effective}}}{n_{\text{analytical}}} \] \[ i\approx\frac{0.0200}{0.0100} \] \[ i\approx2.00 \] \(\textbf{Infer the dissociation degree:}\) Complete dissociation produces three ions per formula unit. \[ i=1+2\alpha \] \[ 2.00=1+2\alpha \] \[ \alpha\approx0.500 \] \(\textbf{Interpret the hydrate correction:}\) Using only the initially supplied \(100.0\,\mathrm g\) water would slightly underestimate the solvent mass and distort the inferred particle factor. The water of crystallisation becomes part of the solvent, while the sodium carbonate component supplies the solute formula units. These two contributions must be treated separately before applying the colligative relation. \(\textbf{Final answer:}\) \[ \boxed{w_{\text{solvent}}=101.8\,\mathrm g,\qquad\alpha\approx50.0\%} \]

Explanation: \(\textbf{Construct the two-solute osmotic equation:}\) Let \(c_X\) be the mass concentration of X in \(\mathrm{g\,L^{-1}}\). Its molarity is: \[ C_X=\frac{c_X}{M_X} \] For two independent molecular solutes: \[ \pi=(C_X+C_Y)RT \] Substituting: \[ \pi=\frac{RT}{M_X}c_X+C_YRT \] \(\textbf{Identify the graphical parameters:}\) The slope is: \[ s=\frac{RT}{M_X} \] The vertical intercept is: \[ b=C_YRT \] \(\textbf{Recover X's molecular mass at }300\,\mathrm K\textbf{:}\) \[ M_X=\frac{RT}{s} \] \[ M_X=\frac{0.0820(300)}{0.01230} \] \[ M_X=\frac{24.6}{0.01230} \] \[ M_X=2000\,\mathrm{g\,mol^{-1}} \] \(\textbf{Determine the background molarity:}\) \[ C_Y=\frac{b}{RT} \] \[ C_Y=\frac{0.2460}{0.0820(300)} \] \[ C_Y=0.0100\,\mathrm M \] \(\textbf{Check the higher-temperature slope:}\) At \(330\,\mathrm K\): \[ s_{330}=\frac{0.0820(330)}{2000} \] \[ s_{330}=0.01353\,\mathrm{atm\,L\,g^{-1}} \] This agrees with the second graph. \(\textbf{Check the higher-temperature intercept:}\) \[ b_{330}=C_YRT \] \[ b_{330}=0.0100(0.0820)(330) \] \[ b_{330}=0.2706\,\mathrm{atm} \] This also agrees with the observation. \(\textbf{Interpret the paired graphs:}\) Both slope and intercept increase proportionally with absolute temperature. That behaviour is consistent with fixed molar concentrations of two ideal molecular solutes. A strictly temperature-independent additive pressure offset would instead have the same numerical intercept at both temperatures. The observations support the stipulated background-solute model without requiring a change in molecular mass or disappearance of Y. \(\textbf{Final answer:}\) \[ \boxed{M_X=2000\,\mathrm{g\,mol^{-1}},\qquad C_Y=0.0100\,\mathrm M} \]

Explanation: \(\textbf{Recover the measured effective particle amount:}\) The freezing-point depression depends on the combined effective particle molality. \[ \Delta T_f=K_f\frac{n_{\text{effective}}}{w_w} \] The solvent mass is \(0.500\,\mathrm{kg}\). Therefore: \[ n_{\text{effective}}=\frac{1.0416(0.500)}{1.86} \] \[ n_{\text{effective}}=0.280\,\mathrm{mol} \] \(\textbf{Represent the dimerising solute:}\) For: \[ 2X\rightleftharpoons X_2 \] the particle factor is: \[ i_X=1-\frac{\alpha}{2} \] Its effective contribution is: \[ n_{X,\text{effective}}=0.100\left(1-\frac{\alpha}{2}\right) \] \[ n_{X,\text{effective}}=0.100-0.0500\alpha \] \(\textbf{Represent the dissociating electrolyte:}\) For: \[ \mathrm{MX_2\rightleftharpoons M^{2+}+2X^-} \] the factor is: \[ i_{\mathrm{MX_2}}=1+2\beta \] Therefore: \[ n_{\mathrm{MX_2},\text{effective}}=0.100(1+2\beta) \] \[ n_{\mathrm{MX_2},\text{effective}}=0.100+0.200\beta \] \(\textbf{Apply the combined particle balance:}\) \[ 0.280=0.100-0.0500\alpha+0.100+0.200\beta \] \[ 0.0800=-0.0500\alpha+0.200\beta \] Rearranging: \[ 0.200\beta=0.0800+0.0500\alpha \] \[ \beta=0.400+0.250\alpha \] \(\textbf{Apply the physical limits:}\) The degree of dimerisation satisfies: \[ 0\leq\alpha\leq1 \] Therefore: \[ 0.400\leq\beta\leq0.650 \] \(\textbf{Interpret the underdetermination:}\) If X remains entirely monomeric, the electrolyte must be \(40.0\%\) dissociated. If X becomes completely dimerised, the electrolyte must be \(65.0\%\) dissociated. Intermediate pairs also satisfy the measured freezing-point depression. One colligative measurement supplies the combined particle count but does not separately identify two independent transformation degrees. \(\textbf{Final answer:}\) \[ \boxed{0.400\leq\beta\leq0.650,\qquad\beta=0.400+0.250\alpha} \]

Explanation: \(\textbf{Establish the undimerised reference values:}\) For the common analytical molality: \[ \Delta T_{f,0}=K_fm \] \[ \Delta T_{f,0}=5.00(0.200)=1.00\,\mathrm K \] Similarly: \[ \Delta T_{b,0}=K_bm \] \[ \Delta T_{b,0}=2.50(0.200)=0.500\,\mathrm K \] \(\textbf{Establish the physical particle-factor range:}\) For dimerisation: \[ i=1-\frac{\alpha}{2} \] Because \(0\leq\alpha\leq1\): \[ 0.500\leq i\leq1.000 \] A factor outside this interval cannot arise from exclusive monomer–dimer association. \(\textbf{Evaluate record P:}\) From freezing: \[ i_{f,P}=\frac{0.750}{1.00}=0.750 \] From boiling: \[ i_{b,P}=\frac{0.450}{0.500}=0.900 \] Both lie within the permitted range. P can be explained by different permissible association degrees at the two measurement temperatures. \(\textbf{Evaluate record Q:}\) \[ i_{f,Q}=\frac{0.550}{1.00}=0.550 \] \[ i_{b,Q}=\frac{0.350}{0.500}=0.700 \] Both are permissible. Q is therefore compatible with the model. \(\textbf{Evaluate record R:}\) \[ i_{f,R}=\frac{0.800}{1.00}=0.800 \] \[ i_{b,R}=\frac{0.550}{0.500}=1.100 \] The boiling measurement requires a factor greater than unity. Exclusive dimerisation cannot increase the number of solute particles above the original monomeric count. R is therefore incompatible with the stipulated model. \(\textbf{Evaluate record S:}\) \[ i_{f,S}=\frac{0.950}{1.00}=0.950 \] \[ i_{b,S}=\frac{0.300}{0.500}=0.600 \] Both factors lie within the permitted range. S is compatible with differing degrees of dimerisation. \(\textbf{Interpret the compatibility test:}\) The experimental freezing and boiling measurements need not yield identical factors when association changes with temperature. However, temperature dependence cannot override the particle-count limits imposed by the chemical association stoichiometry. Dimerisation can decrease the number of independent solute particles, but it cannot produce more particles than the completely monomeric state. Only record R violates that fundamental limit.

Explanation: \(\textbf{Distinguish ionic species from ionic particles:}\) The mixture contains three different kinds of ions. However, a colligative property depends on the total number of independently moving particles, not merely on the number of different chemical species. A large number of chloride ions does not become one particle because all chloride ions have the same identity. \(\textbf{Count ions supplied by NaCl:}\) \[ \mathrm{NaCl\rightarrow Na^++Cl^-} \] From \(0.0400\,\mathrm{mol}\) NaCl: \[ n_{\mathrm{Na^+}}=0.0400\,\mathrm{mol} \] \[ n_{\mathrm{Cl^-,NaCl}}=0.0400\,\mathrm{mol} \] \(\textbf{Count ions supplied by KCl:}\) \[ \mathrm{KCl\rightarrow K^++Cl^-} \] From \(0.0600\,\mathrm{mol}\) KCl: \[ n_{\mathrm{K^+}}=0.0600\,\mathrm{mol} \] \[ n_{\mathrm{Cl^-,KCl}}=0.0600\,\mathrm{mol} \] \(\textbf{Combine the common-ion contribution:}\) \[ n_{\mathrm{Cl^-,total}}=0.0400+0.0600 \] \[ n_{\mathrm{Cl^-,total}}=0.100\,\mathrm{mol} \] The total dissolved-ion amount is: \[ n_{\text{ions}}=0.0400+0.0600+0.100 \] \[ n_{\text{ions}}=0.200\,\mathrm{mol} \] \(\textbf{Determine effective molarity:}\) \[ C_{\text{effective}}=\frac{0.200}{0.500} \] \[ C_{\text{effective}}=0.400\,\mathrm M \] \(\textbf{Calculate osmotic pressure:}\) \[ \pi=C_{\text{effective}}RT \] \[ \pi=0.400(0.0820)(300) \] \[ \pi=9.840\,\mathrm{atm} \] \(\textbf{Calculate the chloride-ion fraction:}\) \[ f_{\mathrm{Cl^-}}=\frac{n_{\mathrm{Cl^-}}}{n_{\text{ions}}} \] \[ f_{\mathrm{Cl^-}}=\frac{0.100}{0.200} \] \[ f_{\mathrm{Cl^-}}=\frac12 \] \(\textbf{Interpret the common-ion result:}\) The chloride ions from both salts belong to the same chemical species. Their amounts add because each remains an independent dissolved ion under the stated model. The solution contains three ionic species but \(0.200\,\mathrm{mol}\) ionic particles. The distinction between chemical identity and particle count is central to using van't Hoff factors in mixtures. \(\textbf{Final answer:}\) \[ \boxed{\pi=9.840\,\mathrm{atm},\qquad f_{\mathrm{Cl^-}}=\frac12} \]

Explanation: \(\textbf{Recover the effective solute amount from osmotic pressure:}\) \[ \pi=\frac{n_{\text{effective}}RT}{V} \] Therefore: \[ n_{\text{effective}}=\frac{\pi V}{RT} \] \[ n_{\text{effective}}=\frac{2.46(0.200)}{0.0820(300)} \] \[ n_{\text{effective}}=\frac{0.492}{24.6} \] \[ n_{\text{effective}}=0.0200\,\mathrm{mol} \] \(\textbf{Determine the electrolyte particle factor:}\) The original analytical amount is \(0.0125\,\mathrm{mol}\). \[ i=\frac{n_{\text{effective}}}{n_{\text{analytical}}} \] \[ i=\frac{0.0200}{0.0125} \] \[ i=1.60 \] \(\textbf{Recover AB's degree of dissociation:}\) Complete AB dissociation produces two ions. \[ i=1+\alpha \] \[ 1.60=1+\alpha \] \[ \alpha=0.600 \] \(\textbf{Apply the exact relative-lowering measurement:}\) Under the stipulated ideal particle model: \[ L=\frac{n_{\text{effective}}}{n_S+n_{\text{effective}}} \] The measurement gives: \[ \frac1{501}=\frac{0.0200}{n_S+0.0200} \] \(\textbf{Recover the solvent molecular amount:}\) Cross-multiplying: \[ n_S+0.0200=501(0.0200) \] \[ n_S+0.0200=10.020 \] \[ n_S=10.000\,\mathrm{mol} \] \(\textbf{Calculate the solvent molar mass:}\) \[ M_S=\frac{w_S}{n_S} \] \[ M_S=\frac{200}{10.000} \] \[ M_S=20.0\,\mathrm{g\,mol^{-1}} \] \(\textbf{Verify the molecular ratios:}\) The effective solute-to-solvent ratio is: \[ \frac{n_{\text{effective}}}{n_S}=\frac{0.0200}{10.000} \] \[ \frac{n_{\text{effective}}}{n_S}=0.00200 \] Its exact effective solute mole fraction is \(1/501\), consistent with the observation. \(\textbf{Interpret the two independent measurements:}\) Osmotic pressure provides the effective number of dissolved particles per measured solution volume. Because the analytical AB amount is known, this determines its dissociation factor. The exact vapour-pressure lowering then provides the solvent molecular amount. The supplied solvent mass finally gives its molar mass. The measurements reveal two different unknowns because they use the effective particle count in distinct concentration ratios. \(\textbf{Final answer:}\) \[ \boxed{M_S=20.0\,\mathrm{g\,mol^{-1}},\qquad\alpha=60.0\%} \]

Explanation: \(\textbf{Express the general association factor:}\) For association of \(n\) original molecules into one aggregate: \[ i=(1-\alpha)+\frac{\alpha}{n} \] Rearranging: \[ i=1-\alpha\left(1-\frac1n\right) \] \(\textbf{Connect i with apparent molar mass:}\) \[ i=\frac{M_{\text{normal}}}{M_{\text{apparent}}} \] Therefore: \[ \frac1{M_{\text{apparent}}}=\frac{i}{M_{\text{normal}}} \] Substituting: \[ \frac1{M_{\text{apparent}}} =\frac1{M_{\text{normal}}} -\frac{\alpha}{M_{\text{normal}}}\left(1-\frac1n\right) \] \(\textbf{Interpret the graph intercept:}\) At zero association: \[ \frac1{M_{\text{normal}}}=0.01000\,\mathrm{mol\,g^{-1}} \] Therefore: \[ M_{\text{normal}}=\frac1{0.01000} \] \[ M_{\text{normal}}=100\,\mathrm{g\,mol^{-1}} \] \(\textbf{Interpret the graph slope:}\) The magnitude of the slope is: \[ |s|=\frac1{M_{\text{normal}}}\left(1-\frac1n\right) \] Therefore: \[ 0.00750=0.01000\left(1-\frac1n\right) \] \[ 0.750=1-\frac1n \] \[ \frac1n=0.250 \] \[ n=4 \] \(\textbf{Determine the particle factor at the specified association degree:}\) \[ i=1-\alpha\left(1-\frac14\right) \] \[ i=1-0.600(0.750) \] \[ i=0.550 \] \(\textbf{Calculate the apparent molar mass:}\) \[ M_{\text{apparent}}=\frac{100}{0.550} \] \[ M_{\text{apparent}}\approx181.8\,\mathrm{g\,mol^{-1}} \] \(\textbf{Interpret the graphical advantage:}\) The reciprocal-mass representation is linear in association degree for a single fixed aggregate size. Its intercept reveals the normal molar mass independently of association. The slope reveals the fraction of original particles removed per associated molecule. Using both features identifies the aggregate size, after which the apparent mass at another degree can be predicted. \(\textbf{Final answer:}\) \[ \boxed{n=4,\qquad M_{\text{apparent}}\approx181.8\,\mathrm{g\,mol^{-1}}} \]

Explanation: \(\textbf{Use the purified solution to isolate protein particles:}\) After dialysis, the only dissolved solute is the retained molecular protein. Its osmotic pressure therefore satisfies: \[ \pi_P=C_PRT \] \(\textbf{Recover protein molarity:}\) \[ C_P=\frac{\pi_P}{RT} \] \[ C_P=\frac{0.0492}{0.0820(300)} \] \[ C_P=\frac{0.0492}{24.6} \] \[ C_P=0.00200\,\mathrm M \] \(\textbf{Calculate the protein molar mass:}\) The protein mass concentration is \(4.00\,\mathrm{g\,L^{-1}}\). \[ M_P=\frac{4.00\,\mathrm{g\,L^{-1}}}{0.00200\,\mathrm{mol\,L^{-1}}} \] \[ M_P=2000\,\mathrm{g\,mol^{-1}} \] \(\textbf{Determine the original total effective concentration:}\) Before dialysis: \[ \pi_{\text{before}}=(C_P+C_{\text{salt,effective}})RT \] Therefore: \[ C_{\text{effective,before}}=\frac{0.1968}{24.6} \] \[ C_{\text{effective,before}}=0.00800\,\mathrm M \] \(\textbf{Isolate the salt contribution:}\) \[ C_{\text{salt,effective}}=0.00800-0.00200 \] \[ C_{\text{salt,effective}}=0.00600\,\mathrm M \] \(\textbf{Recover analytical NaCl molarity:}\) Complete NaCl dissociation produces two ions. \[ C_{\text{salt,effective}}=2C_{\mathrm{NaCl}} \] \[ C_{\mathrm{NaCl}}=\frac{0.00600}{2} \] \[ C_{\mathrm{NaCl}}=0.00300\,\mathrm M \] \(\textbf{Convert NaCl molarity into mass concentration:}\) \[ c_{\mathrm{NaCl}}=C_{\mathrm{NaCl}}M_{\mathrm{NaCl}} \] \[ c_{\mathrm{NaCl}}=0.00300(58.5) \] \[ c_{\mathrm{NaCl}}=0.1755\,\mathrm{g\,L^{-1}} \] \(\textbf{Interpret the purification comparison:}\) The purified measurement isolates the osmotic contribution of the protein because the ions have been removed. The difference between original and purified pressures then determines the effective concentration contributed by NaCl. The factor of two converts this ionic concentration into analytical salt concentration. This two-stage experiment separates an unknown protein molecular mass from an initially unknown electrolyte impurity. \(\textbf{Final answer:}\) \[ \boxed{M_P=2000\,\mathrm{g\,mol^{-1}},\qquad c_{\mathrm{NaCl}}=0.1755\,\mathrm{g\,L^{-1}}} \]

Explanation: \(\textbf{Interpret the observed solvent direction:}\) The membrane allows water to move but retains the dissolved ions and formula units. Ordinary osmosis favours solvent movement from pure water into the solution. The observed movement from solution toward pure water is reverse osmosis. Therefore, the applied solution-side pressure exceeds the solution's osmotic stopping pressure. \(\textbf{Express the dissociation factor:}\) For: \[ \mathrm{CaCl_2\rightleftharpoons Ca^{2+}+2Cl^-} \] each completely dissociated formula unit produces three ions. Under partial dissociation: \[ i=1+2\alpha \] \(\textbf{Construct the osmotic-pressure relation:}\) \[ \pi=iCRT \] Substituting: \[ \pi=(1+2\alpha)(0.100)(0.0820)(300) \] \[ \pi=2.46(1+2\alpha)\,\mathrm{atm} \] \(\textbf{Use the reverse-osmosis observation:}\) Because the applied pressure exceeds the osmotic pressure: \[ 4.92\gt 2.46(1+2\alpha) \] Dividing by \(2.46\): \[ 2.00\gt 1+2\alpha \] \[ 1.00\gt 2\alpha \] \[ \alpha\lt 0.500 \] \(\textbf{Distinguish the three pressure regimes:}\) At \(\alpha=0.500\), the solution osmotic pressure would equal \(4.92\,\mathrm{atm}\), producing the stopping condition. At greater dissociation, the osmotic pressure would exceed the applied pressure, allowing ordinary osmotic movement. At smaller dissociation, the applied pressure exceeds the osmotic pressure and drives reverse osmosis. \(\textbf{Interpret the inequality:}\) The flow-direction observation does not determine one exact degree of dissociation. Instead, it places a strict upper bound on the degree under the stipulated chemical model. This is a reverse inference from a physical observation to the effective solute-particle count. \(\textbf{Final answer:}\) \[ \boxed{0\leq\alpha\lt 0.500} \]

Explanation: \(\textbf{Identify the persistent osmotic difference for membrane I:}\) Membrane I allows glucose to cross. After glucose equilibration, its concentrations are equal on the two sides. Therefore, glucose does not contribute to the persistent concentration difference across membrane I. Sucrose remains impermeant. \(\textbf{Determine membrane I's impermeant concentration difference:}\) \[ \Delta C_I=0.250-0.120 \] \[ \Delta C_I=0.130\,\mathrm M \] Q has more impermeant sucrose. At equal mechanical pressures, water therefore moves from P toward Q. \(\textbf{Determine membrane I's stopping pressure:}\) \[ \Delta\pi_I=\Delta C_IRT \] \[ \Delta\pi_I=0.130(0.0820)(300) \] \[ \Delta\pi_I=3.198\,\mathrm{atm} \] \(\textbf{Identify the persistent osmotic difference for membrane II:}\) Membrane II permits sucrose to cross. After sucrose equilibration, its concentrations become equal on both sides. Glucose is now the impermeant solute responsible for the persistent difference. \(\textbf{Determine membrane II's concentration difference:}\) \[ \Delta C_{II}=0.200-0.0500 \] \[ \Delta C_{II}=0.150\,\mathrm M \] P has the greater concentration of impermeant glucose. At equal mechanical pressures, water moves from Q toward P. \(\textbf{Determine membrane II's stopping pressure:}\) \[ \Delta\pi_{II}=0.150(0.0820)(300) \] \[ \Delta\pi_{II}=3.690\,\mathrm{atm} \] \(\textbf{Compare the two pressure differences:}\) \[ \frac{\Delta\pi_I}{\Delta\pi_{II}}=\frac{0.130}{0.150} \] \[ \frac{\Delta\pi_I}{\Delta\pi_{II}}=\frac{13}{15} \] \(\textbf{Interpret the membrane-dependent reversal:}\) The original chemical compositions of P and Q are identical in both experiments. Only the identity of the membrane-permeant solute changes. Consequently, a different concentration difference persists after equilibration. The same pair of solutions can therefore exhibit opposite net water-flow directions with different selectively permeable membranes. Tonicity and sustained osmotic movement must be interpreted using the actual permeability properties of the separating membrane. \(\textbf{Final answer:}\) \[ \boxed{\text{I: P}\rightarrow\text{Q},\qquad\text{II: Q}\rightarrow\text{P},\qquad\frac{\Delta\pi_I}{\Delta\pi_{II}}=\frac{13}{15}} \]

Explanation: \(\textbf{Evaluate Statement I:}\) Dimerisation is explicitly described as a reversible equilibrium process. The final analytical concentration, solvent, temperature and chemical system are identical for both preparations. Both are also allowed sufficient time to reach equilibrium. The final association equilibrium is therefore determined by the same final state conditions. The preparation route does not independently determine a different equilibrium degree once those conditions are identical. Statement I is correct. \(\textbf{Evaluate Statement II:}\) The osmotic pressure depends on the concentration of independently moving dissolved species. Because both solutions reach the same equilibrium association degree at the same analytical concentration, their effective particle concentrations are equal. Their temperatures are also equal. Consequently, their ideal osmotic pressures are equal. Statement II is correct. \(\textbf{Evaluate Statement III:}\) The degree of association at any stage is determined by the equilibrium state at that stage. An arithmetic average of intermediate degrees has no general chemical-equilibrium basis. Intermediate concentrations and association states need not contribute equally to the final equilibrium. Once the final composition and temperature are established, the association distribution must satisfy the final equilibrium condition. Statement III is incorrect. \(\textbf{Evaluate Statement IV:}\) The apparent molar mass from a colligative measurement is related to the effective particle factor. Equal final association degrees give equal final particle factors. The two preparations therefore have the same apparent molar mass under the stated ideal model. Their different preparation histories do not require different final apparent masses. Statement IV is incorrect. \(\textbf{Integrated interpretation:}\) Analytical composition and chemical equilibrium must be distinguished from the steps used to reach that composition. A reversible molecular association can respond to intermediate dilution or concentration changes. However, the final equilibrium distribution is fixed by the final thermodynamic state under the assumed model. This distinction prevents a mistaken averaging of intermediate particle factors. Statements I and II are correct.

Explanation: \(\textbf{Identify the volume-percentage basis:}\) The preparation contains \(10.0\,\mathrm{mL}\) pure liquid X in \(500\,\mathrm{mL}\) final solution. Therefore: \[ \%\,(v/v)=\frac{10.0}{500}\times100 \] \[ \%\,(v/v)=2.00\% \] This does not establish the mass percentage because the component and solution densities differ. \(\textbf{Convert the component volume into mass:}\) \[ w_X=\rho_XV_X \] \[ w_X=1.15(10.0) \] \[ w_X=11.5\,\mathrm g \] \(\textbf{Determine the final solution mass:}\) Use the measured final volume and solution density: \[ w_{\text{solution}}=\rho_{\text{solution}}V_{\text{solution}} \] \[ w_{\text{solution}}=1.010(500) \] \[ w_{\text{solution}}=505\,\mathrm g \] \(\textbf{Calculate the solute mass percentage:}\) \[ \%\,(w/w)=\frac{w_X}{w_{\text{solution}}}\times100 \] \[ \%\,(w/w)=\frac{11.5}{505}\times100 \] \[ \%\,(w/w)\approx2.28\% \] \(\textbf{Recover the solvent mass:}\) The solution contains only X and water. \[ w_w=505-11.5 \] \[ w_w=493.5\,\mathrm g \] \[ w_w=0.4935\,\mathrm{kg} \] \(\textbf{Determine the solute molecular amount:}\) \[ n_X=\frac{11.5}{92.0} \] \[ n_X=0.125\,\mathrm{mol} \] \(\textbf{Calculate molality:}\) \[ m=\frac{0.125}{0.4935} \] \[ m\approx0.25329\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Calculate freezing-point depression:}\) \[ \Delta T_f=K_fm \] \[ \Delta T_f=1.86(0.25329) \] \[ \Delta T_f\approx0.471\,\mathrm K \] \(\textbf{Interpret the density correction:}\) Volume percentage uses the volume of pure solute introduced and the final solution volume. Mass percentage uses the solute mass and total final solution mass. Molality requires a further change of denominator, from total solution mass to solvent mass. The three concentration measures are consequently not interchangeable, even though they describe the same preparation. \(\textbf{Final answer:}\) \[ \boxed{\%\,(w/w)\approx2.28\%,\qquad\Delta T_f\approx0.471\,\mathrm K} \]