Solutions MCQs with Answers – Part 6 (NEET Chemistry)

Explanation: \(\textbf{Identify the experimentally changing particle factor:}\) The analytical amount of NaCl remains unchanged during dilution. However, its experimentally measured effective factor changes from \(1.50\) to \(1.80\). Therefore, the particle correction cannot be treated as constant across the two preparations. \(\textbf{Calculate the original effective molality:}\) \[ m_{\text{eff},1}=i_1m_1 \] \[ m_{\text{eff},1}=1.50(0.200) \] \[ m_{\text{eff},1}=0.300\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Calculate the initial freezing-point depression:}\) \[ \Delta T_{f,1}=i_1K_fm_1 \] \[ \Delta T_{f,1}=1.50(1.86)(0.200) \] \[ \Delta T_{f,1}=0.558\,\mathrm K \] \(\textbf{Reconstruct the diluted analytical concentration:}\) The solute amount remains fixed while the solvent mass doubles. \[ m_2=\frac{m_1}{2} \] \[ m_2=0.100\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Calculate the final effective molality:}\) \[ m_{\text{eff},2}=i_2m_2 \] \[ m_{\text{eff},2}=1.80(0.100) \] \[ m_{\text{eff},2}=0.180\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Determine the final freezing-point depression:}\) \[ \Delta T_{f,2}=1.80(1.86)(0.100) \] \[ \Delta T_{f,2}=0.3348\,\mathrm K \] Therefore: \[ \frac{\Delta T_{f,2}}{\Delta T_{f,1}}=\frac{0.3348}{0.558} \] \[ \frac{\Delta T_{f,2}}{\Delta T_{f,1}}=\frac35 \] \(\textbf{Recover the two apparent molar masses:}\) For a colligative molecular-mass calculation: \[ M_{\text{app}}=\frac{M_{\text{normal}}}{i} \] Initially: \[ M_{\text{app},1}=\frac{58.5}{1.50} \] \[ M_{\text{app},1}=39.0\,\mathrm{g\,mol^{-1}} \] After dilution: \[ M_{\text{app},2}=\frac{58.5}{1.80} \] \[ M_{\text{app},2}=32.5\,\mathrm{g\,mol^{-1}} \] Consequently: \[ \frac{M_{\text{app},2}}{M_{\text{app},1}}=\frac{32.5}{39.0} \] \[ \frac{M_{\text{app},2}}{M_{\text{app},1}}=\frac56 \] \(\textbf{Interpret the opposite experimental trends:}\) The measured van't Hoff factor increases on dilution. Nevertheless, effective particle molality decreases because the analytical molality is halved. The freezing-point depression therefore decreases, although the factor becomes larger. At the same time, the apparent molar mass decreases because it varies inversely with the measured factor. An increasing factor does not by itself imply an increasing colligative effect when the analytical concentration also changes. \(\textbf{Final answer:}\) \[ \boxed{\frac{\Delta T_{f,2}}{\Delta T_{f,1}}=\frac35,\qquad\frac{M_{\text{app},2}}{M_{\text{app},1}}=\frac56} \]

Explanation: \(\textbf{Establish the original association factor:}\) For dimerisation: \[ i=1-\frac{\alpha}{2} \] Initially, the degree of dimerisation is unity. \[ i_1=1-\frac12 \] \[ i_1=0.500 \] \(\textbf{Calculate the initial effective molality:}\) The analytical solute amount is \(0.200\,\mathrm{mol}\). The initial water mass is \(0.500\,\mathrm{kg}\). \[ m_1=\frac{0.200}{0.500} \] \[ m_1=0.400\,\mathrm{mol\,kg^{-1}} \] Therefore: \[ m_{\text{effective},1}=0.500(0.400) \] \[ m_{\text{effective},1}=0.200\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Use the unchanged freezing-point depression:}\) The same solvent and cryoscopic constant apply to both preparations. Equality of depressions therefore requires equality of effective molalities. \[ i_2m_2=i_1m_1 \] \(\textbf{Calculate the final analytical molality:}\) \[ m_2=\frac{0.200}{0.750} \] \[ m_2=\frac4{15}\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Recover the final association factor:}\) \[ i_2\left(\frac4{15}\right)=0.200 \] \[ i_2=0.750 \] For dimerisation: \[ 0.750=1-\frac{\alpha_2}{2} \] \[ \alpha_2=0.500 \] \(\textbf{Count the effective particles at both stages:}\) Initially: \[ n_{\text{effective},1}=0.200(0.500) \] \[ n_{\text{effective},1}=0.100\,\mathrm{mol} \] Finally: \[ n_{\text{effective},2}=0.200(0.750) \] \[ n_{\text{effective},2}=0.150\,\mathrm{mol} \] \(\textbf{Compare osmotic pressures using measured volumes:}\) At unchanged absolute temperature: \[ \frac{\pi_2}{\pi_1}=\frac{n_{\text{effective},2}/V_2}{n_{\text{effective},1}/V_1} \] Substituting: \[ \frac{\pi_2}{\pi_1}=\frac{0.150/0.850}{0.100/0.550} \] \[ \frac{\pi_2}{\pi_1}=\frac{0.150(0.550)}{0.100(0.850)} \] \[ \frac{\pi_2}{\pi_1}=\frac{33}{34} \] \(\textbf{Interpret the different property responses:}\) Dilution decreases analytical molality, but reduced dimerisation increases the number of independently moving particles. Those effects compensate exactly in the freezing-point depression. However, osmotic pressure uses final solution volume rather than solvent mass. The measured volume change does not compensate the increased effective particle amount in precisely the same proportion. Consequently, freezing-point depression remains unchanged while osmotic pressure decreases slightly. \(\textbf{Final answer:}\) \[ \boxed{\alpha_2=50.0\%,\qquad\frac{\pi_2}{\pi_1}=\frac{33}{34}} \]

Explanation: \(\textbf{Recover the analytical amount of benzoic acid:}\) The normal molar mass refers to the original benzoic acid molecules. \[ n_0=\frac{6.10}{122} \] \[ n_0=0.0500\,\mathrm{mol} \] \(\textbf{Determine analytical molality:}\) The benzene mass is \(250\,\mathrm g=0.250\,\mathrm{kg}\). \[ m_0=\frac{0.0500}{0.250} \] \[ m_0=0.200\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Construct the no-association freezing reference:}\) \[ \Delta T_{f,0}=K_fm_0 \] \[ \Delta T_{f,0}=5.12(0.200) \] \[ \Delta T_{f,0}=1.024\,\mathrm K \] \(\textbf{Recover the particle factor at freezing:}\) \[ i_f=\frac{\Delta T_{f,\text{observed}}}{\Delta T_{f,0}} \] \[ i_f=\frac{0.768}{1.024} \] \[ i_f=0.750 \] \(\textbf{Infer the freezing-temperature dimerisation degree:}\) For dimerisation: \[ i_f=1-\frac{\alpha_f}{2} \] \[ 0.750=1-\frac{\alpha_f}{2} \] \[ \alpha_f=0.500 \] \(\textbf{Construct the no-association boiling reference:}\) \[ \Delta T_{b,0}=K_bm_0 \] \[ \Delta T_{b,0}=2.53(0.200) \] \[ \Delta T_{b,0}=0.5060\,\mathrm K \] \(\textbf{Recover the particle factor at boiling:}\) \[ i_b=\frac{0.4554}{0.5060} \] \[ i_b=0.900 \] \(\textbf{Infer the boiling-temperature dimerisation degree:}\) \[ 0.900=1-\frac{\alpha_b}{2} \] \[ \alpha_b=0.200 \] \(\textbf{Interpret the difference between measurements:}\) The same analytical amount of benzoic acid is used for both properties. However, freezing and boiling measurements occur at different temperatures. The equilibrium population of associated molecules need not remain identical at those temperatures. The supplied observations indicate a greater effective particle count during the boiling measurement. Within the stipulated monomer–dimer model, this corresponds to less extensive dimerisation at the higher measurement temperature. It would be incorrect to force one common van't Hoff factor onto both observations without an assumption that association remains unchanged. \(\textbf{Final answer:}\) \[ \boxed{\alpha_f=50.0\%,\qquad\alpha_b=20.0\%} \]

Explanation: \(\textbf{Recover the effective factor for P:}\) For a molecular solute undergoing association: \[ \pi=iCRT \] Therefore: \[ i_P=\frac{\pi_P}{C_PRT} \] Substituting: \[ i_P=\frac{2.214}{0.100(0.0820)(300)} \] \[ i_P=\frac{2.214}{2.460} \] \[ i_P=0.900 \] \(\textbf{Infer P's degree of dimerisation:}\) For: \[ 2X\rightleftharpoons X_2 \] \[ i=1-\frac{\alpha}{2} \] Thus: \[ 0.900=1-\frac{\alpha_P}{2} \] \[ \alpha_P=0.200 \] \(\textbf{Recover the effective factor for Q:}\) \[ i_Q=\frac{3.690}{0.200(0.0820)(300)} \] \[ i_Q=\frac{3.690}{4.920} \] \[ i_Q=0.750 \] \(\textbf{Infer Q's degree of dimerisation:}\) \[ 0.750=1-\frac{\alpha_Q}{2} \] \[ \alpha_Q=0.500 \] \(\textbf{Test direct proportionality across the preparations:}\) If osmotic pressure were directly proportional to analytical molarity with one constant factor: \[ \frac{\pi_Q}{\pi_P}=\frac{C_Q}{C_P}=2 \] The observed pressure ratio is: \[ \frac{\pi_Q}{\pi_P}=\frac{3.690}{2.214} \] \[ \frac{\pi_Q}{\pi_P}=\frac53 \] Therefore, the proportionality factor relating pressure to analytical molarity is not the same for the two preparations. \(\textbf{Explain the chemical cause:}\) The concentration of Q is twice that of P. However, Q has the greater degree of dimerisation. A larger fraction of its original X molecules therefore belongs to associated species that contribute fewer independent particles. The pressure increases by less than a factor of two because the effective particle factor decreases from \(0.900\) to \(0.750\). \(\textbf{Interpret the graphical implication:}\) The two points cannot both lie on a single straight line through the origin representing a concentration-independent van't Hoff factor. They show that the effective particle behaviour differs between the measured concentrations. The two observations alone do not determine the detailed graph shape at every intermediate concentration. \(\textbf{Final answer:}\) \[ \boxed{\alpha_P=20.0\%,\qquad\alpha_Q=50.0\%,\qquad\frac{\pi_Q}{\pi_P}=\frac53} \]

Explanation: \(\textbf{Determine the total mass of P:}\) The dissolved X mass is: \[ w_X=n_XM_X \] \[ w_X=0.100(60.0) \] \[ w_X=6.00\,\mathrm g \] Therefore: \[ w_{\text{solution},P}=300+6.00 \] \[ w_{\text{solution},P}=306\,\mathrm g \] \(\textbf{Recover the solvent mass in Q:}\) Q contains the same analytical amount of molecular solute but with twice the molar mass. Its dissolved Y mass is: \[ w_Y=0.100(120) \] \[ w_Y=12.0\,\mathrm g \] The total solution mass is also \(306\,\mathrm g\). Thus: \[ w_{w,Q}=306-12.0 \] \[ w_{w,Q}=294\,\mathrm g \] \(\textbf{Compare the two molalities:}\) For P: \[ m_P=\frac{0.100}{0.300} \] \[ m_P=\frac13\,\mathrm{mol\,kg^{-1}} \] For Q: \[ m_Q=\frac{0.100}{0.294} \] \[ m_Q=\frac{50}{147}\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Determine the depression ratio:}\) Both solutions contain water and molecular nonelectrolytes. Therefore: \[ \frac{\Delta T_{f,Q}}{\Delta T_{f,P}}=\frac{m_Q}{m_P} \] \[ \frac{\Delta T_{f,Q}}{\Delta T_{f,P}}=\frac{0.100/0.294}{0.100/0.300} \] \[ \frac{\Delta T_{f,Q}}{\Delta T_{f,P}}=\frac{300}{294} \] \[ \frac{\Delta T_{f,Q}}{\Delta T_{f,P}}=\frac{50}{49} \] \(\textbf{Compare the osmotic concentrations:}\) Both solutions contain \(0.100\,\mathrm{mol}\) independent solute particles. Both also have final volume \(0.320\,\mathrm L\). Therefore: \[ C_P=C_Q=\frac{0.100}{0.320} \] \[ C_P=C_Q=0.3125\,\mathrm M \] \(\textbf{Determine the osmotic-pressure ratio:}\) At equal absolute temperature: \[ \pi=CRT \] Consequently: \[ \frac{\pi_Q}{\pi_P}=1 \] \(\textbf{Interpret the surprising combination:}\) The two solutions have equal total masses, equal dissolved molecular amounts and equal final volumes. Nevertheless, the solvent masses are not equal because Y has the greater molar mass. Therefore, the solution with Y has the greater molality and freezing-point depression. Osmotic pressure remains equal because the solute particle amounts per litre are equal. Equal total solution mass does not guarantee equal solvent mass, even when the analytical numbers of solute molecules match. \(\textbf{Final answer:}\) \[ \boxed{\frac{\Delta T_{f,Q}}{\Delta T_{f,P}}=\frac{50}{49},\qquad\frac{\pi_Q}{\pi_P}=1} \]

Explanation: \(\textbf{Determine X's analytical molecular amount:}\) \[ n_X=\frac{12.0}{120} \] \[ n_X=0.100\,\mathrm{mol} \] \(\textbf{Apply the dimerisation factor:}\) For \(50.0\%\) dimerisation: \[ i_X=1-\frac{0.500}{2} \] \[ i_X=0.750 \] Therefore: \[ n_{X,\text{effective}}=0.100(0.750) \] \[ n_{X,\text{effective}}=0.0750\,\mathrm{mol} \] \(\textbf{Calculate X's apparent molar mass:}\) \[ M_{\text{app},X}=\frac{M_X}{i_X} \] \[ M_{\text{app},X}=\frac{120}{0.750} \] \[ M_{\text{app},X}=160\,\mathrm{g\,mol^{-1}} \] \(\textbf{Determine AB's analytical amount:}\) \[ n_{AB}=\frac{12.0}{240} \] \[ n_{AB}=0.0500\,\mathrm{mol} \] \(\textbf{Apply the dissociation factor:}\) AB gives two ions per dissociated formula unit. \[ i_{AB}=1+\alpha \] \[ i_{AB}=1+0.500 \] \[ i_{AB}=1.500 \] Thus: \[ n_{AB,\text{effective}}=0.0500(1.500) \] \[ n_{AB,\text{effective}}=0.0750\,\mathrm{mol} \] \(\textbf{Calculate AB's apparent molar mass:}\) \[ M_{\text{app},AB}=\frac{240}{1.500} \] \[ M_{\text{app},AB}=160\,\mathrm{g\,mol^{-1}} \] \(\textbf{Compare the cryoscopic responses:}\) The solvent mass is \(0.200\,\mathrm{kg}\) in both preparations. Both contain \(0.0750\,\mathrm{mol}\) effective independent solute particles. Therefore: \[ \Delta T_f=K_f\frac{n_{\text{effective}}}{w_S} \] \[ \Delta T_f=5.00\left(\frac{0.0750}{0.200}\right) \] \[ \Delta T_f=1.875\,\mathrm K \] \(\textbf{Interpret the equal apparent masses:}\) Association makes X appear heavier than its normal molecular mass. Dissociation makes AB appear lighter than its normal formula-unit molar mass. Despite those opposite mechanisms, the different starting molar masses and particle factors produce the same apparent molar mass. Equal effective particle amounts then give equal freezing-point depressions at the same solvent mass. Consequently, equality of a colligatively measured apparent molar mass does not establish equality of chemical identity or molecular behaviour. \(\textbf{Final answer:}\) \[ \boxed{M_{\text{app},X}=M_{\text{app},AB}=160\,\mathrm{g\,mol^{-1}},\qquad\Delta T_{f,P}=\Delta T_{f,Q}=1.875\,\mathrm K} \]

Explanation: \(\textbf{Separate property deviations from mass deviations:}\) A van't Hoff factor is directly proportional to an observed colligative property under the corrected dilute relation. However, the apparent molar mass obtained from an uncorrected colligative expression varies inversely with that factor. Therefore, equal numerical changes in the factor need not produce equal percentage changes in apparent molar mass. \(\textbf{Determine X's apparent mass:}\) The normal molar mass is: \[ M_X=120\,\mathrm{g\,mol^{-1}} \] For X: \[ i_X=0.750 \] Therefore: \[ M_{\text{app},X}=\frac{M_X}{i_X} \] \[ M_{\text{app},X}=\frac{120}{0.750} \] \[ M_{\text{app},X}=160\,\mathrm{g\,mol^{-1}} \] \(\textbf{Calculate X's percentage mass error:}\) \[ \%\text{ error}_X=\frac{160-120}{120}\times100 \] \[ \%\text{ error}_X=33.3\% \] The apparent molar mass is overestimated. \(\textbf{Determine Y's apparent mass:}\) For Y: \[ i_Y=1.250 \] Therefore: \[ M_{\text{app},Y}=\frac{120}{1.250} \] \[ M_{\text{app},Y}=96.0\,\mathrm{g\,mol^{-1}} \] \(\textbf{Calculate Y's percentage mass error:}\) \[ \%\text{ error}_Y=\frac{96.0-120}{120}\times100 \] \[ \%\text{ error}_Y=-20.0\% \] Thus, Y's apparent molar mass is underestimated by \(20.0\%\). \(\textbf{Compare the colligative-property changes:}\) X has \(i_X=0.750\), so its observed property is \(25.0\%\) below the unassociated reference prediction. Y has \(i_Y=1.250\), so its observed property is \(25.0\%\) above its undissociated reference prediction. Those property deviations have equal magnitudes. The molecular-mass deviations do not because the reciprocal transformation is nonlinear. \(\textbf{Interpret the misconception:}\) The normal-to-apparent mass relation is a division by \(i\), not multiplication by \(i\). A reduction in particle count therefore produces a proportionally larger increase in apparent molecular mass than the corresponding decrease in the measured property. Similarly, an increase in particle count produces a smaller percentage decrease in apparent molar mass than the percentage increase in the measured property. The student's proposed symmetry is valid for property deviations about the undissociated reference, but not for the associated apparent molar-mass errors. \(\textbf{Final answer:}\) \[ \boxed{M_{\text{app},X}=160\,\mathrm{g\,mol^{-1}}\ (+33.3\%),\qquad M_{\text{app},Y}=96.0\,\mathrm{g\,mol^{-1}}\ (-20.0\%)} \]

Explanation: \(\textbf{Express the exact relative-lowering relation:}\) The dissolved electrolyte produces an effective number of independently moving solute particles. Under the stipulated ideal model: \[ L=\frac{n_{\text{effective}}}{n_w+n_{\text{effective}}} \] The measurements give: \[ L=\frac1{46} \] \[ n_w=9.00\,\mathrm{mol} \] \(\textbf{Recover the effective dissolved-particle amount:}\) Rearrange the mole-fraction relation: \[ \frac{n_{\text{effective}}}{n_w}=\frac{L}{1-L} \] Substituting: \[ \frac{n_{\text{effective}}}{9.00}=\frac{1/46}{45/46} \] \[ \frac{n_{\text{effective}}}{9.00}=\frac1{45} \] \[ n_{\text{effective}}=0.200\,\mathrm{mol} \] \(\textbf{Determine the electrolyte factor:}\) The original analytical electrolyte amount is \(0.100\,\mathrm{mol}\). \[ i=\frac{n_{\text{effective}}}{n_{\text{analytical}}} \] \[ i=\frac{0.200}{0.100} \] \[ i=2.00 \] \(\textbf{Infer the dissociation degree:}\) Complete dissociation would produce three ions per formula unit. For partial dissociation: \[ i=1+2\alpha \] Therefore: \[ 2.00=1+2\alpha \] \[ \alpha=0.500 \] \(\textbf{Convert the solvent amount into mass:}\) The solution contains \(9.00\,\mathrm{mol}\) water. \[ w_w=9.00(18.0) \] \[ w_w=162\,\mathrm g \] \[ w_w=0.162\,\mathrm{kg} \] \(\textbf{Calculate the analytical molality:}\) \[ m=\frac{0.100}{0.162} \] \[ m\approx0.61728\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Apply the corrected boiling-point relation:}\) \[ \Delta T_b=iK_bm \] \[ \Delta T_b=2.00(0.520)\left(\frac{0.100}{0.162}\right) \] \[ \Delta T_b\approx0.64198\,\mathrm K \] \(\textbf{Interpret the cross-property inference:}\) The exact relative vapour-pressure lowering identifies the effective solute mole fraction. Because the solvent molecular amount is independently known, that fraction determines the effective particle amount. The analytical electrolyte amount then gives its van't Hoff factor and the degree of dissociation under the stipulated three-ion model. For boiling-point elevation, the same effective factor must be combined with analytical molality, which uses kilograms of water rather than total solution moles. The two properties consequently reveal and use the same particle inventory through different concentration expressions. \(\textbf{Final answer:}\) \[ \boxed{\alpha=50.0\%,\qquad\Delta T_b\approx0.642\,\mathrm K} \]

Explanation: \(\textbf{Identify the initial osmotic imbalance:}\) With the urea channels initially closed, dissolved sucrose and urea both contribute to the osmotic-pressure difference. Because both compartments have equal volumes and temperatures, their effective molarities can be compared directly. For P: \[ C_{P,0}=\frac{0.100+0.300}{1.000} \] \[ C_{P,0}=0.400\,\mathrm M \] For Q: \[ C_{Q,0}=\frac{0.180+0.0600}{1.000} \] \[ C_{Q,0}=0.240\,\mathrm M \] At equal mechanical pressures, water initially tends to move from Q toward P. \(\textbf{Represent the transfer of permeant urea:}\) Let \(x\) moles of urea move from P to Q while the compartment volumes are constrained. Sucrose cannot cross the membrane. The effective molarities then become: \[ C_P=\frac{0.100+0.300-x}{1.000}=0.400-x \] \[ C_Q=\frac{0.180+0.0600+x}{1.000}=0.240+x \] Here, \(x\) is the numerical amount in moles and each concentration is expressed in \(\mathrm{mol\,L^{-1}}\). \(\textbf{Locate the osmotic-pressure equality:}\) At the same temperature, equal osmotic pressures require equal effective particle molarities. \[ 0.400-x=0.240+x \] \[ 2x=0.160 \] \[ x=0.0800\,\mathrm{mol} \] At this intermediate state, the instantaneous osmotic-pressure difference is zero. \(\textbf{Determine the final urea distribution:}\) The total urea amount is conserved: \[ n_{\mathrm{urea,total}}=0.300+0.0600 \] \[ n_{\mathrm{urea,total}}=0.360\,\mathrm{mol} \] Because the compartment volumes are equal, complete urea equilibration gives: \[ n_{\mathrm{urea},P}=n_{\mathrm{urea},Q}=0.180\,\mathrm{mol} \] Thus, \(0.120\,\mathrm{mol}\) urea has ultimately transferred from P to Q. \(\textbf{Compare the final osmotic concentrations:}\) \[ C_{P,f}=0.100+0.180=0.280\,\mathrm M \] \[ C_{Q,f}=0.180+0.180=0.360\,\mathrm M \] Q now has the greater effective concentration. At equal mechanical pressures, water consequently tends to move from P toward Q. \(\textbf{Interpret the reversal:}\) The initially larger total solute concentration in P is caused mainly by its excess urea. Urea can cross the membrane, whereas sucrose remains confined. As urea redistributes, its concentration difference decreases and eventually disappears. The persistent sucrose difference then determines the preferred long-term direction of water movement. This experiment demonstrates why initial total osmotic pressure and the osmotic effect remaining after permeant-solute equilibration must be distinguished. \(\textbf{Final answer:}\) \[ \boxed{x=0.0800\,\mathrm{mol};\quad\text{initial flow Q}\rightarrow\text{P},\quad\text{final flow P}\rightarrow\text{Q}} \]

Explanation: \(\textbf{Count the ions before sulfate addition:}\) Complete barium chloride dissociation gives: \[ \mathrm{BaCl_2\rightarrow Ba^{2+}+2Cl^-} \] Initially: \[ n_{\mathrm{Ba^{2+}}}=0.100\,\mathrm{mol} \] \[ n_{\mathrm{Cl^-}}=0.200\,\mathrm{mol} \] Therefore: \[ n_{\mathrm{effective},0}=0.300\,\mathrm{mol} \] \(\textbf{Account for precipitation before the endpoint:}\) Let \(x\) moles of sodium sulfate be added, with \(0\leq x\leq0.100\). The added salt introduces \(3x\) moles of ions before reaction. Precipitation removes \(x\) moles of barium ions and \(x\) moles of sulfate ions. The net increase in dissolved particles is therefore \(x\), not \(3x\). \[ n_{\mathrm{effective}}=0.300+3x-2x \] \[ n_{\mathrm{effective}}=0.300+x \] \(\textbf{Determine the initial graph slope:}\) Because the water mass is \(1.000\,\mathrm{kg}\): \[ \Delta T_f=1.86(0.300+x) \] The slope with respect to the amount of sodium sulfate is: \[ s_1=1.86\,\mathrm{K\,mol^{-1}} \] \(\textbf{Identify the precipitation endpoint:}\) All \(0.100\,\mathrm{mol}\) barium ions have precipitated after adding \(0.100\,\mathrm{mol}\) sodium sulfate. Beyond this point, additional sulfate cannot remove more barium ions. Each extra sodium sulfate formula unit therefore contributes three independent dissolved ions. \(\textbf{Construct the particle balance after the endpoint:}\) For \(x\geq0.100\): \[ n_{\mathrm{effective}}=0.300+3x-2(0.100) \] \[ n_{\mathrm{effective}}=0.100+3x \] Consequently: \[ \Delta T_f=1.86(0.100+3x) \] The new slope is: \[ s_2=3(1.86)=5.58\,\mathrm{K\,mol^{-1}} \] \(\textbf{Calculate the specified final depression:}\) For \(x=0.150\,\mathrm{mol}\): \[ n_{\mathrm{effective}}=0.100+3(0.150) \] \[ n_{\mathrm{effective}}=0.550\,\mathrm{mol} \] \[ \Delta T_f=1.86(0.550) \] \[ \Delta T_f=1.023\,\mathrm K \] \(\textbf{Interpret the slope change:}\) Before the endpoint, sulfate addition simultaneously supplies sodium ions and removes barium ions through precipitation. After the endpoint, no barium ions remain available for further precipitation. The same amount of added sodium sulfate therefore produces a greater increase in dissolved-particle number. The graph becomes steeper because the chemical particle balance changes at the limiting-reagent endpoint. \(\textbf{Final answer:}\) \[ \boxed{s_1=1.86\,\mathrm{K\,mol^{-1}},\quad s_2=5.58\,\mathrm{K\,mol^{-1}},\quad\Delta T_f=1.023\,\mathrm K} \]

Explanation: \(\textbf{Establish the original solute molality:}\) Initially: \[ n_{X,1}=0.200\,\mathrm{mol} \] \[ w_{S,1}=0.500\,\mathrm{kg} \] Therefore: \[ m_1=\frac{0.200}{0.500} \] \[ m_1=0.400\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Calculate the initial boiling-point elevation:}\) \[ \Delta T_{b,1}=K_bm_1 \] \[ \Delta T_{b,1}=2.00(0.400) \] \[ \Delta T_{b,1}=0.800\,\mathrm K \] \(\textbf{Account for solvent evaporation:}\) Only pure solvent escapes during evaporation. \[ w_{S,2}=500-100 \] \[ w_{S,2}=400\,\mathrm g \] \[ w_{S,2}=0.400\,\mathrm{kg} \] \(\textbf{Account separately for solute crystallisation:}\) The crystallised \(8.00\,\mathrm g\) X is no longer part of the dissolved-particle inventory. The original dissolved mass was: \[ w_{X,1}=0.200(100)=20.0\,\mathrm g \] The remaining dissolved mass is: \[ w_{X,2}=20.0-8.00 \] \[ w_{X,2}=12.0\,\mathrm g \] Therefore: \[ n_{X,2}=\frac{12.0}{100} \] \[ n_{X,2}=0.120\,\mathrm{mol} \] \(\textbf{Reconstruct the final liquid molality:}\) \[ m_2=\frac{0.120}{0.400} \] \[ m_2=0.300\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Calculate the final boiling-point elevation:}\) \[ \Delta T_{b,2}=2.00(0.300) \] \[ \Delta T_{b,2}=0.600\,\mathrm K \] \(\textbf{Explain the competing concentration effects:}\) Evaporation alone would increase molality by decreasing the solvent mass. Crystallisation acts in the opposite direction by removing dissolved solute molecules. Here, the dissolved solute amount falls by \(40.0\%\), while the solvent mass falls by only \(20.0\%\). The overall molality consequently decreases. Therefore, the usual prediction that continued evaporation necessarily raises a solution's boiling point is not valid when substantial solute simultaneously leaves the liquid phase. \(\textbf{Final answer:}\) \[ \boxed{\Delta T_b:0.800\,\mathrm K\longrightarrow0.600\,\mathrm K} \]

Explanation: \(\textbf{Determine the first equilibrium state:}\) The liquid mole fractions are: \[ x_A=0.400,\qquad x_B=0.600 \] At \(T_1\), Raoult's law gives: \[ p_{A,1}=0.400(150)=60.0\,\mathrm{mmHg} \] \[ p_{B,1}=0.600(100)=60.0\,\mathrm{mmHg} \] Therefore: \[ P_1=60.0+60.0=120\,\mathrm{mmHg} \] \(\textbf{Calculate the first vapour composition:}\) \[ y_{A,1}=\frac{60.0}{120} \] \[ y_{A,1}=0.500 \] Because: \[ y_{A,1}=0.500\gt x_A=0.400 \] A is enriched in the vapour at \(T_1\). \(\textbf{Recalculate partial pressures at the higher temperature:}\) The liquid composition remains unchanged, but the pure-component vapour pressures change. \[ p_{A,2}=0.400(200) \] \[ p_{A,2}=80.0\,\mathrm{mmHg} \] \[ p_{B,2}=0.600(300) \] \[ p_{B,2}=180\,\mathrm{mmHg} \] Therefore: \[ P_2=80.0+180 \] \[ P_2=260\,\mathrm{mmHg} \] \(\textbf{Determine the second vapour composition:}\) \[ y_{A,2}=\frac{80.0}{260} \] \[ y_{A,2}=\frac4{13}\approx0.3077 \] Now: \[ y_{A,2}\lt x_A \] Thus, B is enriched in the vapour at \(T_2\). \(\textbf{Interpret the volatility reversal:}\) At \(T_1\), pure A has the greater vapour pressure. At \(T_2\), pure B has the greater vapour pressure. Both pure-component pressures increase with temperature, but B's increase is sufficiently large to reverse their relative volatility. Consequently, the component enriched in the equilibrium vapour changes even though the liquid composition is fixed. The total pressure increases at the same time. \(\textbf{Final answer:}\) \[ \boxed{P_1=120,\quad P_2=260\,\mathrm{mmHg};\qquad y_{A,1}=\frac12,\quad y_{A,2}=\frac4{13}} \]

Explanation: \(\textbf{Identify the required pressure reduction:}\) At the fixed bath temperature, pure S has vapour pressure: \[ p_S^\circ=101\,\mathrm{mmHg} \] The external pressure is: \[ P_{\text{ext}}=100\,\mathrm{mmHg} \] To reach the stated threshold, the solution must satisfy: \[ p_S=100\,\mathrm{mmHg} \] \(\textbf{Apply the solvent mole-fraction relation:}\) For the stipulated ideal effective-particle model: \[ p_S=x_Sp_S^\circ \] Therefore: \[ x_S=\frac{100}{101} \] \(\textbf{Express the mole fraction through particle amounts:}\) Let \(n_X\) represent the analytical moles of X introduced. For \(50.0\%\) dimerisation: \[ i=1-\frac{\alpha}{2} \] \[ i=1-\frac{0.500}{2} \] \[ i=0.750 \] The effective solute-particle amount is: \[ n_{\text{effective}}=0.750n_X \] \(\textbf{Impose the boiling-threshold composition:}\) The solvent amount is \(100\,\mathrm{mol}\). \[ \frac{100}{100+0.750n_X}=\frac{100}{101} \] Cross-multiplying: \[ 100+0.750n_X=101 \] \[ 0.750n_X=1.00 \] \[ n_X=\frac43\,\mathrm{mol} \] \(\textbf{Convert analytical amount into mass:}\) \[ w_X=n_XM_X \] \[ w_X=\frac43(120) \] \[ w_X=160\,\mathrm g \] \(\textbf{Verify the vapour-pressure condition:}\) The effective solute amount is: \[ n_{\text{effective}}=0.750\left(\frac43\right)=1.00\,\mathrm{mol} \] Thus: \[ p_S=101\left(\frac{100}{100+1.00}\right) \] \[ p_S=100\,\mathrm{mmHg} \] \(\textbf{Interpret the combined effect:}\) The external pressure is fixed, so boiling is controlled by whether the equilibrium vapour pressure reaches that pressure at the bath temperature. Adding nonvolatile X lowers the solvent mole fraction. Dimerisation reduces the effective number of X particles relative to the analytical amount introduced. Consequently, more X is required than would be needed if every original molecule remained independent. At \(160\,\mathrm g\), the solution reaches the specified pressure threshold; additional X would suppress boiling at this fixed bath temperature. \(\textbf{Final answer:}\) \[ \boxed{w_X=160\,\mathrm g} \]

Explanation: \(\textbf{Determine the initial molecular inventory:}\) Before association, X and Y would contribute: \[ n_{\text{initial}}=0.300+0.150 \] \[ n_{\text{initial}}=0.450\,\mathrm{mol} \] Association decreases the number of independently moving dissolved species. However, the particle reduction differs for the two reactions because their stoichiometries differ. \(\textbf{Recover the final effective particle amount:}\) \[ \Delta T_f=K_f\frac{n_{\text{effective}}}{w_S} \] Therefore: \[ n_{\text{effective}}=\frac{\Delta T_fw_S}{K_f} \] \[ n_{\text{effective}}=\frac{1.20(0.500)}{2.00} \] \[ n_{\text{effective}}=0.300\,\mathrm{mol} \] \(\textbf{Assign independent reaction extents:}\) Let \(a\) moles of XY and \(b\) moles of \(\mathrm{X_2Y}\) form. The remaining free species amounts are: \[ n_{X,\text{free}}=0.300-a-2b \] \[ n_{Y,\text{free}}=0.150-a-b \] \(\textbf{Apply the free-Y measurement:}\) Exactly \(0.0500\,\mathrm{mol}\) Y remains free. \[ 0.150-a-b=0.0500 \] \[ a+b=0.100 \] \(\textbf{Construct the effective particle balance:}\) The final particles comprise free X, free Y, XY and \(\mathrm{X_2Y}\). \[ n_{\text{effective}}=(0.300-a-2b)+(0.150-a-b)+a+b \] Simplifying: \[ n_{\text{effective}}=0.450-a-2b \] The freezing measurement requires: \[ 0.300=0.450-a-2b \] \[ a+2b=0.150 \] \(\textbf{Solve the simultaneous chemical constraints:}\) Subtracting \(a+b=0.100\) from \(a+2b=0.150\): \[ b=0.0500\,\mathrm{mol} \] Then: \[ a=0.100-0.0500 \] \[ a=0.0500\,\mathrm{mol} \] \(\textbf{Verify the final species inventory:}\) \[ n_{X,\text{free}}=0.300-0.0500-2(0.0500) \] \[ n_{X,\text{free}}=0.150\,\mathrm{mol} \] The final species amounts are \(0.150\) free X, \(0.0500\) free Y, \(0.0500\) XY and \(0.0500\,\mathrm{mol}\) \(\mathrm{X_2Y}\). Their sum is \(0.300\,\mathrm{mol}\), matching the freezing measurement. \(\textbf{Interpret the dual constraint:}\) A single colligative measurement gives the combined reduction in dissolved-particle count. It does not independently identify how much of each associated species forms. The free-Y measurement supplies a second independent relation. Together, the two measurements distinguish one-to-one association from the association consuming two X molecules per Y molecule. \(\textbf{Final answer:}\) \[ \boxed{n_{XY}=0.0500\,\mathrm{mol},\qquad n_{X_2Y}=0.0500\,\mathrm{mol}} \]

Explanation: \(\textbf{Construct the exact Raoult expression:}\) The amount of solvent remains fixed at \(10.0\,\mathrm{mol}\). For a nonvolatile molecular solute: \[ p=p^\circ\frac{n_S}{n_S+n_X} \] The solute amount is related to its added mass by: \[ n_X=\frac{w}{M_X} \] Therefore: \[ p=100\frac{10.0}{10.0+w/M_X} \] \(\textbf{Transform the pressure expression:}\) Taking reciprocals: \[ \frac1p=\frac{10.0+w/M_X}{100(10.0)} \] Separating the terms: \[ \frac1p=\frac1{100}+\frac{w}{100(10.0)M_X} \] Thus, a graph of \(1/p\) against added solute mass is a straight line under the stipulated model. \(\textbf{Interpret the graph intercept:}\) At zero added solute mass: \[ \frac1p=\frac1{p^\circ} \] The observed intercept is: \[ \frac1{100}=0.01000\,\mathrm{mmHg^{-1}} \] This agrees with the supplied pure-solvent pressure. \(\textbf{Use the graph slope to determine molecular mass:}\) The theoretical slope is: \[ s=\frac1{p^\circ n_SM_X} \] Therefore: \[ 0.0000100=\frac1{100(10.0)M_X} \] \[ M_X=\frac1{0.0000100(100)(10.0)} \] \[ M_X=100\,\mathrm{g\,mol^{-1}} \] \(\textbf{Predict the pressure after adding }20.0\,\mathrm g\textbf{:}\) The added amount is: \[ n_X=\frac{20.0}{100} \] \[ n_X=0.200\,\mathrm{mol} \] Using Raoult's law: \[ p=100\frac{10.0}{10.0+0.200} \] \[ p=\frac{5000}{51} \] \[ p\approx98.04\,\mathrm{mmHg} \] \(\textbf{Interpret the graphical transformation:}\) The pressure itself is not a linear function of the added solute mass at finite concentration. Its reciprocal becomes linear because the amount of solvent is fixed. The intercept identifies the pure-solvent pressure, while the slope contains the unknown solute molar mass. This makes the transformed plot useful for extracting molecular information without applying the dilute approximation \(n_X\ll n_S\). \(\textbf{Final answer:}\) \[ \boxed{M_X=100\,\mathrm{g\,mol^{-1}},\qquad p\approx98.04\,\mathrm{mmHg}} \]

Explanation: \(\textbf{Evaluate the Assertion:}\) Negative deviation means that the measured equilibrium pressure is lower than the ideal Raoult-law prediction for the same temperature and composition. The comparison described by the chemist does not use the same temperature for the actual and ideal pressures. The real mixture has warmed because heat was released during mixing. At a sufficiently higher temperature, its vapour pressure could exceed the ideal pressure predicted using pure-liquid data at the original lower temperature. Such an observation would not contradict negative deviation under a properly matched isothermal comparison. The Assertion is true. \(\textbf{Evaluate the Reason:}\) Equilibrium vapour pressure generally increases with temperature. Raoult's law uses pure-component vapour pressures evaluated at the temperature of the mixture under investigation. Consequently, a comparison involving pure-component pressures at one temperature and mixture pressure at another does not isolate the effect of non-ideality. The appropriate ideal reference must be reconstructed at the actual mixture temperature and composition. The Reason is true. \(\textbf{Establish the explanatory connection:}\) The initial exothermic mixing process changes the temperature before the pressure comparison is made. Temperature and intermolecular-interaction effects are therefore combined in the immediate experimental observation. A higher absolute measured pressure cannot, by itself, be interpreted as positive deviation when the ideal reference is taken from a different temperature. The Reason directly explains the possibility described in the Assertion. Both A and R are true, and R correctly explains A.

Explanation: \(\textbf{Identify the quantity common to both measurements:}\) Osmotic pressure and freezing-point depression depend on the number of independently moving dissolved particles. The problem explicitly states that the effective particle amount is the same during both measurements. Therefore, it can be eliminated between their governing relations. \(\textbf{Use osmotic pressure to reconstruct particle amount:}\) \[ \pi=\frac{n_{\text{effective}}RT}{V} \] Rearranging: \[ n_{\text{effective}}=\frac{\pi V}{RT} \] Substituting: \[ n_{\text{effective}}=\frac{4.92(0.500)}{0.0820(300)} \] \[ n_{\text{effective}}=\frac{2.46}{24.6} \] \[ n_{\text{effective}}=0.100\,\mathrm{mol} \] \(\textbf{Translate the freezing measurement:}\) The solvent mass is: \[ w_S=0.400\,\mathrm{kg} \] The corrected cryoscopic relation is: \[ \Delta T_f=K_f\frac{n_{\text{effective}}}{w_S} \] Therefore: \[ K_f=\frac{\Delta T_fw_S}{n_{\text{effective}}} \] \(\textbf{Determine the solvent constant:}\) \[ K_f=\frac{0.600(0.400)}{0.100} \] \[ K_f=2.40\,\mathrm{K\,kg\,mol^{-1}} \] \(\textbf{Show the cancellation explicitly:}\) Combining the equations gives: \[ K_f=\frac{\Delta T_fw_SRT}{\pi V} \] The analytical solute amount and van't Hoff factor do not appear in this expression. Neither does the solute's normal molar mass. \(\textbf{Interpret the experimental requirement:}\) This cancellation is valid only when both measurements correspond to the same effective particle amount. If association or dissociation changes substantially between the measurement conditions, the cancellation would no longer be justified. Under the stipulated consistency condition, osmotic pressure supplies the effective particle inventory directly. Freezing-point depression then identifies the solvent-specific cryoscopic constant. \(\textbf{Final answer:}\) \[ \boxed{K_f=2.40\,\mathrm{K\,kg\,mol^{-1}}} \]

Explanation: \(\textbf{Evaluate Statement I:}\) Equilibrium vapour pressure of a pure liquid is determined by the identity of the liquid and the temperature. The final pressure does not depend on the exposed liquid surface area, provided the liquid remains and equilibrium is established. P and Q contain the same liquid at the same temperature. Therefore, their final saturated vapour pressures are equal. Statement I is correct. \(\textbf{Evaluate Statement II:}\) Evaporation takes place at the liquid surface. A greater exposed surface can provide more sites for molecules to escape into the vapour. Under otherwise comparable conditions, Q may therefore approach its equilibrium pressure more quickly than P. This affects the rate of approach to equilibrium rather than the pressure defining the final equilibrium state. Statement II is correct. \(\textbf{Evaluate Statement III:}\) Increasing vapour-space volume does not automatically reduce the equilibrium vapour pressure while sufficient liquid remains. Additional liquid evaporates until the saturated pressure appropriate to the fixed temperature is attained. R therefore has the same equilibrium vapour pressure as P. Statement III is incorrect. \(\textbf{Evaluate Statement IV:}\) For saturated vapour behaving ideally: \[ n=\frac{p^\circ V}{RT} \] At the same temperature and saturated pressure, the amount of vapour is proportional to vapour-space volume. Because R has twice P's vapour-space volume, it contains twice the equilibrium vapour amount. Statement IV is correct. \(\textbf{Integrated interpretation:}\) Exposed liquid surface area influences evaporation kinetics. Vapour-space volume influences the amount of saturated vapour required. Neither changes the equilibrium vapour pressure of a pure liquid at fixed temperature while a liquid phase remains available to maintain saturation. Statements I, II and IV are correct.

Explanation: \(\textbf{Count the independently moving molecules in P:}\) The first macromolecular fraction contributes: \[ n_{P,1}=\frac{5.00}{1.00\times10^4} \] \[ n_{P,1}=5.00\times10^{-4}\,\mathrm{mol} \] The second fraction contributes: \[ n_{P,2}=\frac{5.00}{3.00\times10^4} \] \[ n_{P,2}=\frac53\times10^{-4}\,\mathrm{mol} \] Therefore: \[ n_{P,\text{total}}=\frac{20}{3}\times10^{-4} \] \[ n_{P,\text{total}}=6.6667\times10^{-4}\,\mathrm{mol} \] \(\textbf{Count the molecules in Q:}\) Sample Q contains one molecular species. \[ n_Q=\frac{10.0}{1.50\times10^4} \] \[ n_Q=6.6667\times10^{-4}\,\mathrm{mol} \] Thus, both samples contain the same total molecular amount. \(\textbf{Calculate osmotic pressure:}\) Both have final volume \(1.000\,\mathrm L\) and temperature \(300\,\mathrm K\). \[ \pi=\frac{nRT}{V} \] Therefore: \[ \pi_P=\pi_Q=(6.6667\times10^{-4})(0.0820)(300) \] \[ \pi_P=\pi_Q=0.0164\,\mathrm{atm} \] \(\textbf{Compare the underlying molecular distributions:}\) P contains two distinct macromolecular masses. Q contains only one stipulated molecular mass. Their total sample masses and total molecule counts nevertheless coincide. Consequently, they have equal ideal osmotic pressures. \(\textbf{Interpret what the measurement identifies:}\) The osmotic measurement determines the total number of dissolved molecules per unit volume. Combining that count with total solute mass gives an apparent number-based average molar mass. However, one pressure measurement does not independently establish whether the sample contains one molecular size or several. Different molecular-mass distributions can produce the same mass concentration and effective molecular count. An independent molecular-characterisation measurement would be required to distinguish P from Q. \(\textbf{Final answer:}\) \[ \boxed{\pi_P=\pi_Q=0.0164\,\mathrm{atm}} \]

Explanation: \(\textbf{Calculate the original osmotic pressure:}\) X remains molecularly dissolved and contributes one particle per molecule. \[ \pi_1=C_XRT \] \[ \pi_1=0.00200(0.0820)(300) \] \[ \pi_1=0.0492\,\mathrm{atm} \] \(\textbf{Convert the pressure into SI units:}\) \[ \pi_1=0.0492(1.01325\times10^5) \] \[ \pi_1=4985.19\,\mathrm{Pa} \] \(\textbf{Relate osmotic pressure to hydrostatic head:}\) At equilibrium: \[ \pi=\rho gh \] Therefore: \[ h_1=\frac{\pi_1}{\rho g} \] \[ h_1=\frac{4985.19}{1000(9.80)} \] \[ h_1\approx0.509\,\mathrm m \] \(\textbf{Count the additional ionic particles:}\) Complete NaCl dissociation gives: \[ \mathrm{NaCl\rightarrow Na^++Cl^-} \] The effective ion concentration introduced is: \[ C_{\text{NaCl,effective}}=2(0.00100) \] \[ C_{\text{NaCl,effective}}=0.00200\,\mathrm M \] \(\textbf{Reconstruct the final effective concentration:}\) The original macromolecular contribution remains \(0.00200\,\mathrm M\). Therefore: \[ C_{\text{effective},2}=0.00200+0.00200 \] \[ C_{\text{effective},2}=0.00400\,\mathrm M \] \(\textbf{Calculate the new osmotic pressure:}\) \[ \pi_2=0.00400(0.0820)(300) \] \[ \pi_2=0.0984\,\mathrm{atm} \] This is twice the initial pressure. \(\textbf{Determine the new hydrostatic height:}\) Because the liquid density and gravitational acceleration are unchanged: \[ \frac{h_2}{h_1}=\frac{\pi_2}{\pi_1}=2 \] Therefore: \[ h_2=2(0.5087) \] \[ h_2\approx1.018\,\mathrm m \] \(\textbf{Interpret the membrane and concentration conditions:}\) The added NaCl formula units are fewer in molarity than the macromolecule molecules. Nevertheless, complete dissociation makes their effective particle contribution equal to that of the original macromolecule. The osmotic pressure therefore doubles. Because the column density is fixed, the hydrostatic height required to balance that pressure also doubles. The concentration-maintaining reservoir is important because otherwise a substantial change in column volume could alter the concentrations during the measurement. \(\textbf{Final answer:}\) \[ \boxed{h_1\approx0.509\,\mathrm m,\qquad h_2\approx1.018\,\mathrm m} \]