Solutions MCQs with Answers – Part 3 (NEET Chemistry)

Explanation: \(\textbf{Recover the initial molality from freezing-point depression:}\) For a dilute molecular nonelectrolyte: \[ \Delta T_f=K_fm \] Therefore: \[ m_i=\frac{\Delta T_f}{K_f} \] \[ m_i=\frac{0.744}{1.86} \] \[ m_i=0.400\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Determine the analytical amount of X:}\) The initial water mass is \(300\,\mathrm g=0.300\,\mathrm{kg}\). \[ n_X=m_iw_{w,i} \] \[ n_X=0.400(0.300) \] \[ n_X=0.120\,\mathrm{mol} \] \(\textbf{Recover the molar mass:}\) \[ M_X=\frac{w_X}{n_X} \] \[ M_X=\frac{18.0}{0.120} \] \[ M_X=150\,\mathrm{g\,mol^{-1}} \] \(\textbf{Reconstruct the diluted solution:}\) Adding \(300\,\mathrm g\) water increases the solvent mass to \(600\,\mathrm g\). The amount of X remains \(0.120\,\mathrm{mol}\). \[ w_{w,f}=0.600\,\mathrm{kg} \] \[ m_f=\frac{0.120}{0.600} \] \[ m_f=0.200\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Apply the ebullioscopic relation:}\) For the diluted solution: \[ \Delta T_{b,f}=K_bm_f \] \[ \Delta T_{b,f}=0.520(0.200) \] \[ \Delta T_{b,f}=0.104\,\mathrm K \] \(\textbf{Calculate the actual boiling temperature:}\) At normal atmospheric pressure, pure water boils at \(100.000^\circ\mathrm C\). \[ T_{b,f}=T_b^\circ+\Delta T_{b,f} \] \[ T_{b,f}=100.000+0.104 \] \[ T_{b,f}=100.104^\circ\mathrm C \] \(\textbf{Check the two-property reasoning:}\) Freezing-point depression determines the original solute molality because the solvent's cryoscopic constant is known. The initial solvent mass then determines the amount and molar mass of X. Dilution halves the molality without altering the solute amount. The final boiling-point elevation must be calculated using \(K_b\), not \(K_f\). The boiling temperature lies above the pure-water value, whereas freezing-point depression would place the freezing temperature below the pure-water value. \(\textbf{Final answer:}\) \[ \boxed{M_X=150\,\mathrm{g\,mol^{-1}},\qquad T_{b,f}=100.104^\circ\mathrm C} \]

Explanation: \(\textbf{Count the conserved molecular solute particles:}\) Urea and glucose remain as separate, nonvolatile molecules. For urea: \[ n_{\text{urea}}=\frac{30.0}{60.0}=0.500\,\mathrm{mol} \] For glucose: \[ n_{\text{glucose}}=\frac{45.0}{180}=0.250\,\mathrm{mol} \] The total dissolved-particle amount is: \[ n_s=0.500+0.250 \] \[ n_s=0.750\,\mathrm{mol} \] \(\textbf{Determine the original molality:}\) \[ m_i=\frac{0.750}{0.300} \] \[ m_i=2.50\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Translate the final boiling temperature:}\) At the specified external pressure, pure water boils at \(100.000^\circ\mathrm C\). The final elevation is: \[ \Delta T_{b,f}=101.950-100.000 \] \[ \Delta T_{b,f}=1.950\,\mathrm K \] \(\textbf{Recover the final molality:}\) \[ \Delta T_b=K_bm \] Therefore: \[ m_f=\frac{1.950}{0.520} \] \[ m_f=3.750\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Infer the remaining solvent mass:}\) No solute is lost, so \(n_s=0.750\,\mathrm{mol}\) remains constant. \[ w_{w,f}=\frac{n_s}{m_f} \] \[ w_{w,f}=\frac{0.750}{3.750} \] \[ w_{w,f}=0.200\,\mathrm{kg} \] \(\textbf{Calculate the evaporated water:}\) \[ w_{\text{lost}}=0.300-0.200 \] \[ w_{\text{lost}}=0.100\,\mathrm{kg} \] \[ w_{\text{lost}}=100\,\mathrm g \] \(\textbf{Connect the changed molality to freezing behaviour:}\) For the same solvent and the same effective solute-particle behaviour: \[ \Delta T_f=K_fm \] The cryoscopic constant cancels when the two solution states are compared. \[ \frac{\Delta T_{f,f}}{\Delta T_{f,i}}=\frac{m_f}{m_i} \] \[ \frac{\Delta T_{f,f}}{\Delta T_{f,i}}=\frac{3.750}{2.50} \] \[ \frac{\Delta T_{f,f}}{\Delta T_{f,i}}=\frac32 \] \(\textbf{Interpret the physical process:}\) As water evaporates, the number of dissolved solute molecules remains constant while the solvent mass decreases. The boiling-point elevation increases, so the temperature of the boiling solution rises under the stated model. The freezing-point depression also increases because it depends on the same total particle molality. No numerical value of \(K_f\) is required for the requested ratio. This result differs from the behaviour of a pure liquid, whose boiling temperature remains fixed at a constant external pressure while both phases coexist. \(\textbf{Final answer:}\) \[ \boxed{w_{\text{lost}}=100\,\mathrm g,\qquad\frac{\Delta T_{f,f}}{\Delta T_{f,i}}=\frac32} \]

Explanation: \(\textbf{Evaluate Statement I:}\) A pure liquid boils when its equilibrium vapour pressure equals the external pressure. At fixed external pressure, the corresponding equilibrium boiling temperature is fixed for a given pure liquid. During boiling, additional heat can supply latent heat for vapour formation rather than continuously increasing the equilibrium temperature. Statement I is correct under the stated conditions. \(\textbf{Evaluate Statement II:}\) The dissolved solute is nonvolatile, so the escaping vapour is essentially solvent. As boiling continues, solvent is removed while the solute remains in the liquid. The remaining solution therefore becomes more concentrated. Under the dilute ideal model: \[ \Delta T_b=K_bm \] An increase in molality produces a greater boiling-point elevation. Consequently, the boiling temperature of the remaining solution can rise during continued solvent loss. Statement II is correct. \(\textbf{Evaluate Statement III:}\) Boiling-point elevation is defined as the boiling temperature of the solution minus that of the pure solvent at the same external pressure. \[ \Delta T_b=T_b-T_b^\circ \] For an ordinary nonvolatile-solute solution, \(T_b\gt T_b^\circ\), so the elevation is positive. Statement III reverses the subtraction order and is incorrect. \(\textbf{Evaluate Statement IV:}\) An increasing boiling temperature can result solely from changing solution composition. The nonvolatile solute need not undergo any chemical reaction. A changing concentration is sufficient to explain the trend through vapour-pressure lowering and the boiling condition. Therefore, Statement IV is correct. \(\textbf{Integrated interpretation:}\) A pure liquid has a fixed equilibrium boiling temperature at a specified external pressure. A solution has a composition-dependent boiling temperature. As solvent evaporates, that composition may change continuously. The comparison distinguishes a physical concentration effect from chemical decomposition and preserves the correct positive sign of boiling-point elevation. Statements I, II and IV are correct.

Explanation: \(\textbf{Identify the external-pressure requirement:}\) Boiling occurs when the equilibrium vapour pressure equals the applied external pressure. Here, the external pressure is \(496\,\mathrm{mmHg}\), not \(760\,\mathrm{mmHg}\). The boiling temperatures must therefore be determined using this actual pressure. \(\textbf{Compare pure S at }T_1\textbf{:}\) At \(T_1\), pure S has vapour pressure \(500\,\mathrm{mmHg}\). This is already greater than the external pressure. Because the pure-solvent pressure increases continuously with temperature, it reaches \(496\,\mathrm{mmHg}\) at a temperature slightly below \(T_1\). The exact lower temperature cannot be calculated without additional vapour-pressure data. \(\textbf{Determine the solution pressure at }T_1\textbf{:}\) Raoult's law gives: \[ p_{\text{solution}}=x_Sp_S^\circ \] At \(T_1\): \[ p_{\text{solution},1}=0.800(500) \] \[ p_{\text{solution},1}=400\,\mathrm{mmHg} \] This is below the external pressure. The solution consequently does not reach its boiling condition at \(T_1\). \(\textbf{Evaluate the solution at }T_2\textbf{:}\) At \(T_2\), pure S has vapour pressure \(620\,\mathrm{mmHg}\). \[ p_{\text{solution},2}=0.800(620) \] \[ p_{\text{solution},2}=496\,\mathrm{mmHg} \] The solution vapour pressure now equals the external pressure. Thus, the solution reaches its boiling condition at \(T_2\). \(\textbf{Connect the pressure conditions:}\) The nonvolatile solute reduces solvent vapour pressure at a given temperature. Therefore, the solution requires a higher temperature than the pure solvent to reach the same external pressure. In this case, the pure solvent boils below \(T_1\), while the solution boils at \(T_2\). The values do not justify calculating the precise pure-solvent boiling temperature below \(T_1\). \(\textbf{Conclusion:}\) The pure solvent reaches \(496\,\mathrm{mmHg}\) below \(T_1\), whereas the solution reaches that pressure exactly at \(T_2\).

Explanation: \(\textbf{Identify the unpressurised osmotic tendency:}\) The membrane passes water but blocks glucose. At equal hydrostatic pressure and temperature, water tends to move from the lower effective solute concentration toward the higher one. Therefore, before the external pressure is applied, net water movement would be from P to Q. \(\textbf{Calculate the osmotic pressure of P:}\) For dilute glucose solutions: \[ \pi=CRT \] \[ \pi_P=0.100(0.0820)(300) \] \[ \pi_P=2.46\,\mathrm{atm} \] \(\textbf{Calculate the osmotic pressure of Q:}\) \[ \pi_Q=0.200(0.0820)(300) \] \[ \pi_Q=4.92\,\mathrm{atm} \] \(\textbf{Find the balancing mechanical pressure difference:}\) The two sides both contain solutions. Therefore, the additional pressure needed on the more concentrated side to stop net osmosis is the difference between their osmotic pressures. \[ \Delta\pi=\pi_Q-\pi_P \] \[ \Delta\pi=4.92-2.46 \] \[ \Delta\pi=2.46\,\mathrm{atm} \] \(\textbf{Compare the imposed pressure with the balance condition:}\) The actual additional pressure on Q is \(3.00\,\mathrm{atm}\). \[ 3.00\,\mathrm{atm}\gt 2.46\,\mathrm{atm} \] The applied pressure exceeds the value needed merely to stop osmosis from P to Q. Under these conditions, the imposed pressure drives water in the opposite direction. The initial net movement is therefore from Q to P. \(\textbf{Interpret the membrane process:}\) At zero additional pressure, the concentration difference favours water transfer toward Q. At a Q-side pressure excess of \(2.46\,\mathrm{atm}\), the tendencies balance and there is no net water flow. At a larger Q-side pressure excess, reverse osmosis occurs. Glucose remains confined to its original compartment because the membrane is impermeable to it. \(\textbf{Final answer:}\) \[ \boxed{\Delta P_{\text{stop}}=2.46\,\mathrm{atm},\qquad\text{net water movement: Q}\rightarrow\text{P}} \]

Explanation: \(\textbf{Compare the initial osmotic tendencies:}\) Both urea and glucose are molecular nonelectrolytes. Each dissolved molecule contributes one effective solute particle. Initially: \[ C_{P,i}=\frac{0.100}{0.500}=0.200\,\mathrm M \] \[ C_{Q,i}=\frac{0.200}{0.500}=0.400\,\mathrm M \] At equal temperature and liquid pressure, the osmotic tendency favours net water movement from P toward Q. \(\textbf{Apply solute conservation:}\) Neither solute can cross the membrane. Therefore: \[ n_{P,f}=0.100\,\mathrm{mol} \] \[ n_{Q,f}=0.200\,\mathrm{mol} \] \(\textbf{Identify the final isotonic condition:}\) At equal temperature and equal mechanical pressure: \[ \pi_P=\pi_Q \] For the two ideal nonelectrolytes: \[ C_PRT=C_QRT \] Thus: \[ C_{P,f}=C_{Q,f}=C_f \] \(\textbf{Use the conserved total solution volume:}\) The final volumes may change separately, but their sum remains \(1.000\,\mathrm L\). Because both final concentrations are \(C_f\): \[ V_{P,f}=\frac{0.100}{C_f} \] \[ V_{Q,f}=\frac{0.200}{C_f} \] Adding: \[ \frac{0.100}{C_f}+\frac{0.200}{C_f}=1.000 \] \[ \frac{0.300}{C_f}=1.000 \] \[ C_f=0.300\,\mathrm{mol\,L^{-1}} \] \(\textbf{Reconstruct the final compartment volumes:}\) \[ V_{P,f}=\frac{0.100}{0.300}=\frac13\,\mathrm L \] \[ V_{Q,f}=\frac{0.200}{0.300}=\frac23\,\mathrm L \] \(\textbf{Determine the transferred solvent volume:}\) P initially contains \(0.500\,\mathrm L\). Its volume decreases to one-third of a litre. \[ V_{\text{transferred}}=0.500-\frac13 \] \[ V_{\text{transferred}}=\frac16\,\mathrm L \] \[ V_{\text{transferred}}\approx167\,\mathrm{mL} \] \(\textbf{Interpret the equilibrium:}\) The solution with the greater initial effective particle concentration gains water. Its concentration decreases, while the concentration in P increases because P loses water. Net movement ceases when the two osmotic pressures become equal under the maintained equal-pressure condition. The final volumes are unequal because the compartments retain different amounts of impermeant solute. The two solutions can be isotonic even though their total solute amounts and chemical identities differ. \(\textbf{Final answer:}\) \[ \boxed{C_f=0.300\,\mathrm M,\qquad V_{\text{P}\rightarrow\text{Q}}\approx167\,\mathrm{mL}} \]

Explanation: \(\textbf{Calculate P's particle concentration:}\) Glucose remains molecularly dissolved. Its amount is: \[ n_P=\frac{18.0}{180} \] \[ n_P=0.100\,\mathrm{mol} \] The final solution volume is \(0.500\,\mathrm L\). \[ C_P=\frac{0.100}{0.500} \] \[ C_P=0.200\,\mathrm M \] \(\textbf{Use the initial isotonic condition:}\) Both solutions are at the same temperature and initially have equal osmotic pressures. For molecular nonelectrolytes: \[ \pi=CRT \] Therefore: \[ C_{Q,i}=C_P=0.200\,\mathrm M \] \(\textbf{Recover the original urea amount:}\) The original volume of Q is \(0.250\,\mathrm L\). \[ n_{Q,i}=C_{Q,i}V_{Q,i} \] \[ n_{Q,i}=0.200(0.250) \] \[ n_{Q,i}=0.0500\,\mathrm{mol} \] \(\textbf{Determine the original urea mass:}\) \[ w_{\text{urea}}=0.0500(60.0) \] \[ w_{\text{urea}}=3.00\,\mathrm g \] \(\textbf{Account for homogeneous withdrawal:}\) Half of Q is removed. Consequently, half its dissolved urea is also removed. \[ n_{Q,\text{remaining}}=\frac{0.0500}{2} \] \[ n_{Q,\text{remaining}}=0.0250\,\mathrm{mol} \] Withdrawal alone does not alter the concentration of the remaining homogeneous portion. The subsequent dilution does. \(\textbf{Calculate Q's final molarity:}\) The remaining urea is diluted to \(500\,\mathrm{mL}=0.500\,\mathrm L\). \[ C_{Q,f}=\frac{0.0250}{0.500} \] \[ C_{Q,f}=0.0500\,\mathrm M \] \(\textbf{Compare the final osmotic pressures:}\) At the common temperature: \[ \frac{\pi_P}{\pi_{Q,f}}=\frac{C_P}{C_{Q,f}} \] \[ \frac{\pi_P}{\pi_{Q,f}}=\frac{0.200}{0.0500}=4 \] P has the higher osmotic pressure. At equal liquid pressures, water consequently moves from the more dilute Q solution toward the more concentrated P solution. \(\textbf{Interpret the reversal:}\) The initial samples were isotonic, so they had no net osmotic tendency. Removing a homogeneous aliquot of Q does not, by itself, change Q's concentration. Diluting the remaining aliquot reduces its particle concentration substantially. The two solutions are no longer isotonic, and the concentration difference now drives water from Q to P. \(\textbf{Final answer:}\) \[ \boxed{w_{\text{urea},i}=3.00\,\mathrm g,\qquad\text{net solvent flow: Q}\rightarrow\text{P}} \]

Explanation: \(\textbf{Recover the protein molarity from osmotic pressure:}\) For a dilute molecular nonelectrolyte: \[ \pi=CRT \] Therefore: \[ C=\frac{\pi}{RT} \] Substituting: \[ C=\frac{0.0246}{0.0820(300)} \] \[ C=\frac{0.0246}{24.6} \] \[ C=0.00100\,\mathrm{mol\,L^{-1}} \] \(\textbf{Determine the dissolved protein amount:}\) The measured solution volume is \(0.250\,\mathrm L\). \[ n_X=CV \] \[ n_X=0.00100(0.250) \] \[ n_X=2.50\times10^{-4}\,\mathrm{mol} \] \(\textbf{Calculate the protein molar mass:}\) \[ M_X=\frac{w_X}{n_X} \] \[ M_X=\frac{2.00}{2.50\times10^{-4}} \] \[ M_X=8000\,\mathrm{g\,mol^{-1}} \] \(\textbf{Reconstruct the separate boiling-point experiment:}\) The same protein mass is now dissolved in \(250\,\mathrm g\) water. The amount of protein remains: \[ n_X=2.50\times10^{-4}\,\mathrm{mol} \] The solvent mass is: \[ w_w=0.250\,\mathrm{kg} \] Therefore: \[ m=\frac{2.50\times10^{-4}}{0.250} \] \[ m=1.00\times10^{-3}\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Calculate the predicted boiling-point elevation:}\) \[ \Delta T_b=K_bm \] \[ \Delta T_b=0.520(1.00\times10^{-3}) \] \[ \Delta T_b=5.20\times10^{-4}\,\mathrm K \] \(\textbf{Interpret the measurement choice:}\) The protein has a large molar mass, so a modest mass contains relatively few molecules. Its dilute solution consequently produces an extremely small boiling-point elevation. Accurately measuring such a small temperature difference can be difficult. Osmotic pressure, by contrast, can be measured at a convenient temperature without boiling or freezing the protein solution. This is one reason osmotic-pressure measurements are useful for determining molar masses of macromolecules. The calculation uses separate solution-volume and solvent-mass bases, which must not be interchanged. \(\textbf{Final answer:}\) \[ \boxed{M_X=8000\,\mathrm{g\,mol^{-1}},\qquad\Delta T_b=5.20\times10^{-4}\,\mathrm K} \]

Explanation: \(\textbf{Identify the effective particle concentrations:}\) Osmotic pressure depends on the concentration of independent dissolved particles. Urea remains molecular and contributes one particle per dissolved molecule. Calcium chloride is stipulated to dissociate completely: \[ \mathrm{CaCl_2\rightarrow Ca^{2+}+2Cl^-} \] Thus, each formula unit of calcium chloride contributes three effective particles. \(\textbf{Calculate P's effective concentration:}\) For the urea solution: \[ C_{\text{effective,P}}=0.120\,\mathrm M \] At the common temperature: \[ \pi_P=0.120RT \] \(\textbf{Calculate Q's initial effective concentration:}\) The analytical molarity of calcium chloride is \(0.0500\,\mathrm M\). After complete dissociation: \[ C_{\text{effective,Q},i}=3(0.0500) \] \[ C_{\text{effective,Q},i}=0.150\,\mathrm M \] Therefore: \[ \pi_{Q,i}=0.150RT \] \(\textbf{Predict the first net solvent movement:}\) Initially: \[ \pi_{Q,i}\gt \pi_P \] At equal liquid pressure, water tends to move toward the compartment with the greater effective particle concentration. Hence, the first net movement is from P to Q. \(\textbf{Apply the separate dilution of Q:}\) In the second experiment, Q's volume doubles while the amount of calcium chloride remains fixed. Its analytical molarity is therefore halved: \[ C_{\mathrm{CaCl_2},f}=\frac{0.0500}{2} \] \[ C_{\mathrm{CaCl_2},f}=0.0250\,\mathrm M \] With complete dissociation: \[ C_{\text{effective,Q},f}=3(0.0250) \] \[ C_{\text{effective,Q},f}=0.0750\,\mathrm M \] \(\textbf{Predict the second net solvent movement:}\) After dilution: \[ C_{\text{effective,Q},f}\lt C_{\text{effective,P}} \] Consequently: \[ \pi_{Q,f}\lt \pi_P \] The net water movement is now from Q to P. \(\textbf{Explain the reversal:}\) The analytical molarity of calcium chloride is smaller than that of urea in both experiments. However, its complete dissociation initially gives Q the greater effective particle concentration. Diluting Q sufficiently reduces this particle concentration below that of P. The osmotic direction consequently reverses even though the chemical identities remain unchanged. The comparison must use effective dissolved-particle concentrations rather than analytical formula-unit molarities alone. \(\textbf{Conclusion:}\) Water initially moves from P to Q, but after Q is diluted to twice its original volume, the net movement is from Q to P.

Explanation: \(\textbf{Connect the graph slope with absolute temperature:}\) For a dilute molecular nonelectrolyte: \[ \pi=CRT \] At fixed temperature, osmotic pressure is directly proportional to molarity. A graph of \(\pi\) against \(C\) therefore passes through the origin. Its slope is: \[ s=RT \] \(\textbf{Verify the original slope:}\) At \(300\,\mathrm K\): \[ s_1=R(300) \] The measured slope is \(24.6\,\mathrm{L\,atm\,mol^{-1}}\), which corresponds to: \[ R=\frac{24.6}{300} \] \[ R=0.0820\,\mathrm{L\,atm\,mol^{-1}\,K^{-1}} \] \(\textbf{Calculate the new graph slope:}\) At \(330\,\mathrm K\): \[ s_2=R(330) \] \[ s_2=0.0820(330) \] \[ s_2=27.06\,\mathrm{L\,atm\,mol^{-1}} \] \(\textbf{Account for the sample's simultaneous expansion:}\) The amount of solute remains constant. The new solution volume is: \[ V_2=1.20V_1 \] Because: \[ C=\frac nV \] the sample's final molarity is: \[ C_2=\frac{C_1}{1.20} \] \(\textbf{Combine concentration and temperature changes:}\) For each state: \[ \pi_1=C_1RT_1 \] \[ \pi_2=C_2RT_2 \] Therefore: \[ \frac{\pi_2}{\pi_1}=\frac{C_2}{C_1}\frac{T_2}{T_1} \] Substituting: \[ \frac{\pi_2}{\pi_1}=\frac1{1.20}\times\frac{330}{300} \] \[ \frac{\pi_2}{\pi_1}=\frac56\times\frac{11}{10} \] \[ \frac{\pi_2}{\pi_1}=\frac{11}{12} \] \(\textbf{Interpret the competing effects:}\) Increasing temperature makes the \(\pi\)-versus-\(C\) graph steeper. However, the particular heated sample also expands, reducing its molarity. Its concentration decrease more than offsets its temperature increase. Consequently, its osmotic pressure decreases slightly even though the graph slope at the higher temperature is greater. A property of the entire graph and the behaviour of one changing-composition sample must therefore be distinguished. \(\textbf{Final answer:}\) \[ \boxed{s_2=27.06\,\mathrm{L\,atm\,mol^{-1}},\qquad\frac{\pi_2}{\pi_1}=\frac{11}{12}} \]

Explanation: \(\textbf{Distinguish the concentration scales:}\) Freezing-point depression is proportional to molality for the stated molecular-nonelectrolyte model. Osmotic pressure, however, is proportional to molarity at fixed absolute temperature. The two samples have equal molality but different densities. Their solution volumes per kilogram of solvent are therefore different. \(\textbf{Choose a common solvent-mass basis:}\) Consider \(1.000\,\mathrm{kg}\) water in each preparation. At molality \(1.00\,\mathrm{mol\,kg^{-1}}\): \[ n_X=1.00\,\mathrm{mol} \] The corresponding solute mass is: \[ w_X=1.00(60.0)=60.0\,\mathrm g \] Hence, the total solution mass on this basis is: \[ w_{\text{solution}}=1000+60.0 \] \[ w_{\text{solution}}=1060\,\mathrm g \] \(\textbf{Reconstruct the volume of P:}\) \[ V_P=\frac{1060}{1.05} \] \[ V_P=1009.52\,\mathrm{mL} \] \[ V_P=1.00952\,\mathrm L \] Therefore: \[ C_P=\frac{1.00}{1.00952} \] \[ C_P\approx0.99057\,\mathrm M \] \(\textbf{Reconstruct the volume of Q:}\) \[ V_Q=\frac{1060}{1.12} \] \[ V_Q=946.43\,\mathrm{mL} \] \[ V_Q=0.94643\,\mathrm L \] Thus: \[ C_Q=\frac{1.00}{0.94643} \] \[ C_Q\approx1.05660\,\mathrm M \] \(\textbf{Compare osmotic pressures:}\) At equal absolute temperature: \[ \frac{\pi_Q}{\pi_P}=\frac{C_QRT}{C_PRT} \] \[ \frac{\pi_Q}{\pi_P}=\frac{C_Q}{C_P} \] Using the common solution-mass and solute-amount basis, the ratio also equals the density ratio. \[ \frac{\pi_Q}{\pi_P}=\frac{1.12}{1.05} \] \[ \frac{\pi_Q}{\pi_P}=\frac{16}{15} \] \(\textbf{Compare freezing-point depressions:}\) For the same solvent: \[ \Delta T_f=K_fm \] Both samples have molality \(1.00\,\mathrm{mol\,kg^{-1}}\). Therefore: \[ \Delta T_{f,P}=\Delta T_{f,Q} \] \(\textbf{Interpret the experimental distinction:}\) Q contains the same number of solute molecules per kilogram of solvent as P. However, its greater solution density corresponds to a smaller total solution volume for the chosen mass basis. It consequently has greater molarity and greater osmotic pressure. Its freezing-point depression remains equal to P's because the solvent identity and molality are unchanged. Thus, solutions with equal freezing-point depression need not have equal osmotic pressure. \(\textbf{Final answer:}\) \[ \boxed{\frac{\pi_Q}{\pi_P}=\frac{16}{15},\qquad\Delta T_{f,Q}=\Delta T_{f,P}} \]

Explanation: \(\textbf{Identify the membrane's selective behaviour:}\) The membrane allows two species to move through it: water and urea. It blocks sucrose completely. Therefore, the initially unequal urea concentrations can change through urea diffusion. The sucrose concentrations, by contrast, remain associated with their respective reservoirs. \(\textbf{Interpret the stipulated urea equilibration:}\) The problem explicitly states that urea has become equally concentrated on the two sides. Its contribution to the ideal solution particle concentration is therefore the same in both reservoirs at that stage. The equal urea contributions do not create a residual difference favouring one direction of water movement. \(\textbf{Compare the impermeant solute concentrations:}\) P contains \(0.100\,\mathrm M\) sucrose. Q contains \(0.200\,\mathrm M\) sucrose. Sucrose remains molecularly dissolved and cannot traverse the membrane. Therefore, Q retains the greater effective concentration of membrane-impermeant solute. \(\textbf{Determine the direction of water movement:}\) At equal mechanical pressure, the persistent concentration difference associated with impermeant sucrose favours net water movement from P toward Q. This is the direction expected when water tends to dilute the compartment containing more impermeant solute. The equalised urea contribution does not reverse that direction. \(\textbf{Evaluate the alternative conclusions:}\) The initial urea concentration difference is not a valid basis for determining the stated final-stage water movement because urea has already equilibrated. Equality of the concentration of one permeant solute does not establish equality of all solute contributions. Likewise, the molecular size of sucrose is not the determining colligative variable. The important condition is that sucrose is impermeant and occurs at different concentrations on the two sides. \(\textbf{Distinguish this arrangement from a standard semipermeable membrane:}\) In the usual introductory osmotic-pressure arrangement, the membrane passes solvent while restricting the solute. Here, urea is explicitly allowed to pass, so its concentration difference is not permanently maintained. The remaining impermeant sucrose gradient supplies the persistent osmotic driving tendency under the specified idealisation. \(\textbf{Conclusion:}\) After urea concentrations become equal, net water movement is from P to Q.

Explanation: \(\textbf{Determine the initial osmotic direction:}\) The membrane passes water but blocks glucose. At equal hydrostatic pressures, net solvent movement is from the pure-water side toward the glucose-solution side. Consequently, the glucose-solution column rises relative to the pure-water column. A hydrostatic pressure difference develops as the two liquid levels separate. \(\textbf{Calculate the osmotic pressure:}\) For a dilute molecular nonelectrolyte: \[ \pi=CRT \] Substitution gives: \[ \pi=0.00400(0.0820)(300) \] \[ \pi=0.0984\,\mathrm{atm} \] \(\textbf{Convert the pressure into SI units:}\) \[ \pi=0.0984(1.00\times10^5) \] \[ \pi=9840\,\mathrm{Pa} \] \(\textbf{Identify the hydrostatic balancing condition:}\) At equilibrium, the pressure difference between the columns balances the osmotic pressure. \[ \Delta P_{\text{hydrostatic}}=\rho gh \] Therefore: \[ \rho gh=\pi \] \(\textbf{Solve for the liquid-level difference:}\) \[ h=\frac{\pi}{\rho g} \] Substituting: \[ h=\frac{9840}{1000(9.84)} \] \[ h=\frac{9840}{9840} \] \[ h=1.00\,\mathrm m \] \(\textbf{Identify the higher limb:}\) Water enters the glucose-containing compartment before hydrostatic balance is reached. The glucose-solution limb therefore has the higher liquid level. The pressure produced by this difference opposes further entry of water. At the calculated height, the two opposing tendencies balance and net solvent movement ceases. \(\textbf{Check the physical units:}\) The hydrostatic expression uses density in \(\mathrm{kg\,m^{-3}}\), gravitational acceleration in \(\mathrm{m\,s^{-2}}\), and height in metres. Its result is pressure in pascals. Therefore, the osmotic pressure must be converted from atmospheres before substitution. Using \(0.0984\) directly with the SI density and acceleration would introduce a large unit error. \(\textbf{Final answer:}\) \[ h=1.00\,\mathrm m,\qquad\text{higher level: glucose-solution limb} \]

Explanation: \(\textbf{Use osmotic pressure to determine solute molarity:}\) For the molecular nonelectrolyte: \[ \pi=CRT \] Therefore: \[ C=\frac{\pi}{RT} \] Substituting: \[ C=\frac{2.46}{0.0820(300)} \] \[ C=\frac{2.46}{24.6} \] \[ C=0.100\,\mathrm{mol\,L^{-1}} \] \(\textbf{Recover the amount of dissolved X:}\) The final solution volume is \(1.000\,\mathrm L\). \[ n_X=CV \] \[ n_X=0.100(1.000) \] \[ n_X=0.100\,\mathrm{mol} \] \(\textbf{Use freezing-point depression to determine molality:}\) For the same molecular nonelectrolyte: \[ \Delta T_f=K_fm \] Thus: \[ m=\frac{\Delta T_f}{K_f} \] \[ m=\frac{0.186}{1.86} \] \[ m=0.100\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Reconstruct the solvent mass:}\) By the definition of molality: \[ m=\frac{n_X}{w_w(\mathrm{kg})} \] Therefore: \[ w_w(\mathrm{kg})=\frac{n_X}{m} \] \[ w_w=\frac{0.100}{0.100} \] \[ w_w=1.000\,\mathrm{kg} \] \(\textbf{Calculate the dissolved solute mass:}\) \[ w_X=n_XM_X \] \[ w_X=0.100(100) \] \[ w_X=10.0\,\mathrm g \] \(\textbf{Determine the total solution mass:}\) The water mass is \(1000\,\mathrm g\). \[ w_{\text{solution}}=1000+10.0 \] \[ w_{\text{solution}}=1010\,\mathrm g \] \(\textbf{Calculate density from total mass and measured volume:}\) \[ \rho=\frac{w_{\text{solution}}}{V_{\text{solution}}} \] \[ \rho=\frac{1010}{1000} \] \[ \rho=1.010\,\mathrm{g\,mL^{-1}} \] \(\textbf{Interpret the combined measurement:}\) Osmotic pressure measures the number of dissolved particles per litre of solution. Freezing-point depression measures the number of dissolved particles per kilogram of solvent. Combining the two experiments therefore reveals both the solute amount and the solvent mass. The known solute molar mass then permits reconstruction of the total solution mass. Only after this mass balance can the density be calculated from the measured solution volume. \(\textbf{Final answer:}\) \[ \boxed{w_w=1.000\,\mathrm{kg},\qquad\rho=1.010\,\mathrm{g\,mL^{-1}}} \]

Explanation: \(\textbf{Recover the molality of X from the freezing experiment:}\) For a molecular nonelectrolyte: \[ \Delta T_f=K_fm \] Therefore: \[ m_P=\frac{\Delta T_f}{K_f} \] Substituting: \[ m_P=\frac{0.186}{1.86} \] \[ m_P=0.100\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Determine the amount of X:}\) P contains \(200\,\mathrm g=0.200\,\mathrm{kg}\) water. \[ n_X=m_Pw_w \] \[ n_X=0.100(0.200) \] \[ n_X=0.0200\,\mathrm{mol} \] \(\textbf{Calculate the molarity of P:}\) The stated final solution volume is \(250\,\mathrm{mL}=0.250\,\mathrm L\). \[ C_P=\frac{n_X}{V_P} \] \[ C_P=\frac{0.0200}{0.250} \] \[ C_P=0.0800\,\mathrm M \] \(\textbf{Compare P with the glucose solution:}\) Solution Q contains glucose at: \[ C_Q=0.0800\,\mathrm M \] Both solutes are stipulated to remain as independent molecules. Neither contributes additional ions through dissociation. The effective dissolved-particle molarities are therefore equal. \(\textbf{Apply the osmotic-pressure relation:}\) For each solution: \[ \pi=CRT \] At the common temperature: \[ \pi_P=0.0800RT \] \[ \pi_Q=0.0800RT \] Consequently: \[ \pi_P=\pi_Q \] \(\textbf{Interpret the membrane observation:}\) At equal mechanical pressure, equal osmotic pressures imply no net osmotic tendency between the compartments. Water molecules can still cross the membrane in both directions, but the opposing transfers balance on average. Therefore, the stipulated ideal model does not predict sustained directional movement from P to Q. \(\textbf{Evaluate the proposed explanations:}\) The relative molar masses of the two solutes do not independently determine osmotic direction. Likewise, a positive freezing-point depression demonstrates the presence of dissolved particles but does not establish that one solution has greater osmotic pressure than another. The appropriate comparison requires effective particle molarity at the same temperature. The freezing measurement, water mass and final solution volume together establish that P has exactly the same effective molarity as Q. \(\textbf{Conclusion:}\) The two solutions should be isotonic under the stated conditions. The reported net movement is inconsistent with the supplied ideal measurements and membrane assumptions.

Explanation: \(\textbf{Use P to determine the solvent constant:}\) The two solutions use the same solvent, so its ebullioscopic constant is common to both. For a dilute molecular nonelectrolyte: \[ \Delta T_b=K_bm \] First calculate the solute amount in P. \[ n_P=\frac{2.00}{50.0} \] \[ n_P=0.0400\,\mathrm{mol} \] The solvent mass is: \[ w_{S,P}=0.200\,\mathrm{kg} \] Therefore: \[ m_P=\frac{0.0400}{0.200} \] \[ m_P=0.200\,\mathrm{mol\,kg^{-1}} \] The experimental boiling-point elevation gives: \[ K_b=\frac{\Delta T_{b,P}}{m_P} \] \[ K_b=\frac{0.400}{0.200} \] \[ K_b=2.00\,\mathrm{K\,kg\,mol^{-1}} \] \(\textbf{Recover the molecular amount of X:}\) For Q: \[ \Delta T_{b,Q}=K_bm_Q \] Thus: \[ m_Q=\frac{0.600}{2.00} \] \[ m_Q=0.300\,\mathrm{mol\,kg^{-1}} \] The initial solvent mass in Q is \(0.150\,\mathrm{kg}\). \[ n_X=m_Qw_{S,Q} \] \[ n_X=0.300(0.150) \] \[ n_X=0.0450\,\mathrm{mol} \] \(\textbf{Determine the molar mass of X:}\) \[ M_X=\frac{w_X}{n_X} \] \[ M_X=\frac{9.00}{0.0450} \] \[ M_X=200\,\mathrm{g\,mol^{-1}} \] \(\textbf{Reconstruct the diluted Q preparation:}\) The added solvent increases Q's total solvent mass to: \[ w_{S,Q,f}=150+150 \] \[ w_{S,Q,f}=300\,\mathrm g=0.300\,\mathrm{kg} \] The amount of X remains \(0.0450\,\mathrm{mol}\). \[ m_{Q,f}=\frac{0.0450}{0.300} \] \[ m_{Q,f}=0.150\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Calculate the revised boiling-point elevation:}\) \[ \Delta T_{b,Q,f}=2.00(0.150) \] \[ \Delta T_{b,Q,f}=0.300\,\mathrm K \] \(\textbf{Interpret the experimental sequence:}\) Solution P establishes the solvent-specific constant without requiring the boiling temperature of pure S. That constant converts Q's initial elevation into molality. The known mass of X then determines its molar mass. Finally, doubling Q's solvent mass at fixed solute amount halves its boiling-point elevation. The three measurements are connected through one solvent property but require separate concentration calculations. \(\textbf{Final answer:}\) \[ \boxed{K_b=2.00\,\mathrm{K\,kg\,mol^{-1}},\quad M_X=200\,\mathrm{g\,mol^{-1}},\quad\Delta T_{b,Q,f}=0.300\,\mathrm K} \]

Explanation: \(\textbf{Determine the pure-solvent pressure slope:}\) The pressure changes from \(700\) to \(900\,\mathrm{mmHg}\) over a temperature interval of \(10.0\,\mathrm K\). \[ s=\frac{900-700}{100.0-90.0} \] \[ s=20.0\,\mathrm{mmHg\,K^{-1}} \] The stipulated linear relationship within this interval is: \[ p_S^\circ=700+20.0(T-90.0) \] Here, \(T\) is inserted numerically in degrees Celsius. \(\textbf{Find the pure-solvent boiling temperature:}\) Boiling occurs when equilibrium vapour pressure equals external pressure. \[ p_S^\circ=720\,\mathrm{mmHg} \] Therefore: \[ 720=700+20.0(T_b^\circ-90.0) \] \[ 20.0=20.0(T_b^\circ-90.0) \] \[ T_b^\circ=91.0^\circ\mathrm C \] \(\textbf{Determine the pure-solvent pressure required for solution boiling:}\) For the ideal solution: \[ p_{\text{solution}}=x_Sp_S^\circ \] The solvent mole fraction is: \[ x_S=0.900 \] At the solution's boiling temperature: \[ 720=0.900p_S^\circ \] Therefore: \[ p_S^\circ=\frac{720}{0.900} \] \[ p_S^\circ=800\,\mathrm{mmHg} \] \(\textbf{Find the solution boiling temperature:}\) Use the experimental pure-solvent pressure line. \[ 800=700+20.0(T_b-90.0) \] \[ 100=20.0(T_b-90.0) \] \[ T_b=95.0^\circ\mathrm C \] \(\textbf{Calculate the boiling-point elevation:}\) \[ \Delta T_b=T_b-T_b^\circ \] \[ \Delta T_b=95.0-91.0 \] \[ \Delta T_b=4.00\,\mathrm K \] \(\textbf{Interpret the two pressure curves:}\) At the pure-solvent boiling temperature, the solution pressure is only \(90.0\%\) of the pure-solvent pressure. The solution therefore requires additional heating to reach the same external pressure. The observed elevation follows from the pressure-temperature graph and the solvent mole fraction. A separate numerical value of \(K_b\) is unnecessary because sufficient vapour-pressure data are provided. The linear pressure-temperature relation is used only within the explicitly stated experimental interval. \(\textbf{Final answer:}\) \[ \boxed{T_b^\circ=91.0^\circ\mathrm C,\quad T_b=95.0^\circ\mathrm C,\quad\Delta T_b=4.00\,\mathrm K} \]

Explanation: \(\textbf{Determine the initial boiling-point elevation:}\) The initial amount of urea is \(0.100\,\mathrm{mol}\). The water mass is \(0.500\,\mathrm{kg}\). \[ m_i=\frac{0.100}{0.500} \] \[ m_i=0.200\,\mathrm{mol\,kg^{-1}} \] For molecular urea: \[ \Delta T_{b,i}=K_bm_i \] \[ \Delta T_{b,i}=0.520(0.200) \] \[ \Delta T_{b,i}=0.104\,\mathrm K \] \(\textbf{Apply the required elevation condition:}\) The desired final elevation is twice the initial value. \[ \Delta T_{b,f}=2(0.104) \] \[ \Delta T_{b,f}=0.208\,\mathrm K \] \(\textbf{Reconstruct the remaining solvent mass:}\) Only water evaporates during boiling. \[ w_{w,f}=500-100 \] \[ w_{w,f}=400\,\mathrm g \] \[ w_{w,f}=0.400\,\mathrm{kg} \] The original \(0.100\,\mathrm{mol}\) urea remains dissolved. \(\textbf{Determine the total particle molality required:}\) Both urea and sucrose are molecular nonelectrolytes. Their individual dissolved molecular amounts contribute additively. \[ m_{\text{total},f}=\frac{\Delta T_{b,f}}{K_b} \] \[ m_{\text{total},f}=\frac{0.208}{0.520} \] \[ m_{\text{total},f}=0.400\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Recover the required total dissolved amount:}\) \[ n_{\text{total solute},f}=m_{\text{total},f}w_{w,f} \] \[ n_{\text{total solute},f}=0.400(0.400) \] \[ n_{\text{total solute},f}=0.160\,\mathrm{mol} \] \(\textbf{Determine the sucrose amount:}\) The urea contributes \(0.100\,\mathrm{mol}\) of this total. Therefore: \[ n_{\text{sucrose}}=0.160-0.100 \] \[ n_{\text{sucrose}}=0.0600\,\mathrm{mol} \] \(\textbf{Convert the amount into mass:}\) \[ w_{\text{sucrose}}=n_{\text{sucrose}}M_{\text{sucrose}} \] \[ w_{\text{sucrose}}=0.0600(342) \] \[ w_{\text{sucrose}}=20.52\,\mathrm g \] \(\textbf{Verify the final condition:}\) The final solution contains \(0.160\,\mathrm{mol}\) of molecular solute particles in \(0.400\,\mathrm{kg}\) water. \[ m_{\text{total},f}=\frac{0.160}{0.400}=0.400\,\mathrm{mol\,kg^{-1}} \] \[ \Delta T_{b,f}=0.520(0.400)=0.208\,\mathrm K \] The target elevation is reproduced. \(\textbf{Interpret the two concentration changes:}\) Evaporation increases molality because solvent mass decreases while solute amount remains fixed. Adding sucrose increases molality again because additional solute molecules enter the liquid. The two effects must be combined through the final solvent mass and total dissolved-particle amount. Simply doubling the original mass of urea or doubling its analytical molality would not reproduce the stated preparation. \(\textbf{Final answer:}\) \[ \boxed{w_{\text{sucrose}}=20.52\,\mathrm g} \]

Explanation: \(\textbf{Identify the two quantities being compared:}\) Boiling-point elevation measures the increase above the boiling temperature of the pure solvent. Actual boiling temperature includes both the pure-solvent boiling temperature and the elevation. For a dilute molecular nonelectrolyte: \[ \Delta T_b=K_bm \] \[ T_b=T_b^\circ+\Delta T_b \] \(\textbf{Calculate the molality and elevation of P:}\) \[ m_P=\frac{0.300}{0.200} \] \[ m_P=1.50\,\mathrm{mol\,kg^{-1}} \] \[ \Delta T_{b,P}=0.800(1.50) \] \[ \Delta T_{b,P}=1.20\,\mathrm K \] Its actual boiling temperature is: \[ T_{b,P}=78.0+1.20 \] \[ T_{b,P}=79.20^\circ\mathrm C \] \(\textbf{Calculate the molality and elevation of Q:}\) \[ m_Q=\frac{0.300}{0.300} \] \[ m_Q=1.00\,\mathrm{mol\,kg^{-1}} \] \[ \Delta T_{b,Q}=1.20(1.00) \] \[ \Delta T_{b,Q}=1.20\,\mathrm K \] Therefore: \[ T_{b,Q}=80.0+1.20 \] \[ T_{b,Q}=81.20^\circ\mathrm C \] \(\textbf{Calculate the molality and elevation of R:}\) \[ m_R=\frac{0.200}{0.200} \] \[ m_R=1.00\,\mathrm{mol\,kg^{-1}} \] \[ \Delta T_{b,R}=1.50(1.00) \] \[ \Delta T_{b,R}=1.50\,\mathrm K \] Therefore: \[ T_{b,R}=85.0+1.50 \] \[ T_{b,R}=86.50^\circ\mathrm C \] \(\textbf{Compare the calculated elevations:}\) \[ \Delta T_{b,R}=1.50\,\mathrm K \] \[ \Delta T_{b,P}=\Delta T_{b,Q}=1.20\,\mathrm K \] Thus, the elevation ordering is: \[ R\gt P=Q \] \(\textbf{Compare actual boiling temperatures:}\) The calculated boiling temperatures are \(79.20^\circ\mathrm C\), \(81.20^\circ\mathrm C\) and \(86.50^\circ\mathrm C\). Their ordering is: \[ R\gt Q\gt P \] \(\textbf{Interpret the solvent dependence:}\) P and Q have different molalities and different ebullioscopic constants, yet their boiling-point elevations are equal. Their actual boiling temperatures differ because their pure-solvent boiling temperatures differ. R has both the greatest elevation and the greatest actual boiling temperature. The comparisons therefore require two distinct calculations rather than ranking the solutions from concentration alone. \(\textbf{Conclusion:}\) The elevations follow \(R\gt P=Q\), whereas the actual boiling temperatures follow \(R\gt Q\gt P\).

Explanation: \(\textbf{Separate solution mass from solvent mass:}\) The given \(10.0\%\) concentration is by mass. Therefore, \(10.0\%\) of the total solution mass is urea. \[ w_{\text{urea}}=\frac{10.0}{100}(500) \] \[ w_{\text{urea}}=50.0\,\mathrm g \] The remaining mass is water. \[ w_w=500-50.0 \] \[ w_w=450\,\mathrm g \] \[ w_w=0.450\,\mathrm{kg} \] \(\textbf{Calculate the solute amount:}\) \[ n_{\text{urea}}=\frac{50.0}{60.0} \] \[ n_{\text{urea}}=\frac56\,\mathrm{mol} \] \(\textbf{Determine the correct molality:}\) Molality uses solvent mass alone. \[ m_{\text{correct}}=\frac{5/6}{0.450} \] \[ m_{\text{correct}}=\frac{50}{27} \] \[ m_{\text{correct}}=1.85185\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Calculate the correct boiling-point elevation:}\) \[ \Delta T_{b,\text{correct}}=K_bm_{\text{correct}} \] \[ \Delta T_{b,\text{correct}}=0.520(1.85185) \] \[ \Delta T_{b,\text{correct}}=0.962963\,\mathrm K \] \(\textbf{Reconstruct the student's incorrect calculation:}\) The student uses \(0.500\,\mathrm{kg}\) instead of \(0.450\,\mathrm{kg}\). \[ m_{\text{student}}=\frac{5/6}{0.500} \] \[ m_{\text{student}}=1.66667\,\mathrm{mol\,kg^{-1}} \] The corresponding reported elevation would be: \[ \Delta T_{b,\text{student}}=0.520(1.66667) \] \[ \Delta T_{b,\text{student}}=0.866667\,\mathrm K \] \(\textbf{Quantify the percentage underestimation:}\) The correct elevation is the reference. \[ \%\text{ error}=\frac{0.962963-0.866667}{0.962963}\times100 \] \[ \%\text{ error}=10.0\% \] \(\textbf{Explain the direction of error:}\) The student's denominator is greater than the actual solvent mass. Therefore, the calculated molality is smaller than the correct molality. The predicted boiling-point elevation is correspondingly underestimated. The error is numerically equal to the solute mass percentage in this particular calculation because the student uses total solution mass in place of solvent mass. That correspondence is a consequence of the defined masses, not a replacement for the molality calculation. \(\textbf{Final answer:}\) \[ \boxed{\Delta T_{b,\text{correct}}\approx0.963\,\mathrm K,\qquad\text{underestimation}=10.0\%} \]