Solutions MCQs with Answers – Part 1 (NEET Chemistry)

Explanation: Molality describes the amount of dissolved solute relative to the mass of the solvent used to prepare the solution. Its defining relation is: \[ m=\frac{n_{\text{solute}}}{w_{\text{solvent}}(\mathrm{kg})} \] The numerator represents the number of moles of solute, whereas the denominator represents the mass of solvent expressed in kilograms. Consequently, the commonly used unit of molality is \(\mathrm{mol\,kg^{-1}}\). The distinction between solvent mass and total solution mass is essential because the dissolved solute also contributes to the mass of the solution. Using the total solution mass would therefore produce a different concentration ratio. Using the solution volume in litres would instead define molarity, not molality. Similarly, a ratio involving grams of solute without converting them into moles does not represent molality. The defining feature of molality is its mass-of-solvent basis, which also makes it independent of temperature when the composition remains unchanged. Thus, the correct expression is moles of solute divided by kilograms of solvent.

Explanation: \(\textbf{Given:}\) Mass of glucose \(=18\,\mathrm g\). Mass of final solution \(=198\,\mathrm g\). Molar mass of glucose \(=180\,\mathrm{g\,mol^{-1}}\). \(\textbf{Identify the solvent mass:}\) The given \(198\,\mathrm g\) represents the total solution mass, not the mass of water. \[ w_{\text{water}}=198-18=180\,\mathrm g \] \[ w_{\text{water}}=0.180\,\mathrm{kg} \] \(\textbf{Calculate solute moles:}\) \[ n_{\text{glucose}}=\frac{18}{180}=0.100\,\mathrm{mol} \] \(\textbf{Apply the molality relation:}\) \[ m=\frac{n_{\text{glucose}}}{w_{\text{water}}(\mathrm{kg})} \] \[ m=\frac{0.100}{0.180} \] \[ m=0.5556\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Consistency check:}\) The solvent mass is smaller than the total solution mass, so using \(0.198\,\mathrm{kg}\) as the denominator would underestimate the molality. The correct denominator is \(0.180\,\mathrm{kg}\), and the resulting unit is moles per kilogram. \(\textbf{Final answer:}\) \[ \boxed{m\approx0.556\,\mathrm{mol\,kg^{-1}}} \]

Explanation: \(\textbf{Evaluate Statement I:}\) Statement I is correct because molality is determined by solute moles and solvent mass. When the sealed solution is heated without evaporation, precipitation, decomposition or any other compositional change, neither quantity changes. The defining relation remains: \[ m=\frac{n_{\text{solute}}}{w_{\text{solvent}}(\mathrm{kg})} \] Neither the numerator nor the denominator directly involves solution volume. Heating may change the density and volume of the solution, but such changes do not alter its molality when the composition is preserved. \(\textbf{Evaluate Statement II:}\) Statement II is incorrect because thermal expansion does not increase the amount of solute or decrease the mass of solvent. Even when the solution volume increases, molality remains fixed under the stated conditions. The misconception arises from confusing molality with molarity, whose denominator is solution volume. If the volume increases while solute moles remain constant, molarity decreases. Molality, however, continues to describe the same number of solute moles per kilogram of solvent. \(\textbf{Conclusion:}\) Statement I is correct, whereas Statement II is incorrect, because volume expansion alone does not change a mass-based concentration.

Explanation: \(\textbf{Given:}\) Total solution mass \(=530\,\mathrm g\). Molality \(=1.00\,\mathrm{mol\,kg^{-1}}\). Molar mass of urea \(=60\,\mathrm{g\,mol^{-1}}\). \(\textbf{Introduce the unknown composition:}\) Let the mass of urea be \(x\,\mathrm g\). The mass of water must then be \(530-x\) grams because the solution contains only urea and water. \[ w_{\text{water}}=\frac{530-x}{1000}\,\mathrm{kg} \] \(\textbf{Express the solute amount:}\) \[ n_{\text{urea}}=\frac{x}{60}\,\mathrm{mol} \] \(\textbf{Apply the specified molality:}\) \[ 1.00=\frac{x/60}{(530-x)/1000} \] Rearranging: \[ 60(530-x)=1000x \] \[ 31800-60x=1000x \] \[ 1060x=31800 \] \[ x=30.0\,\mathrm g \] \(\textbf{Verify the reconstructed solution:}\) The remaining mass is water: \[ w_{\text{water}}=530-30=500\,\mathrm g=0.500\,\mathrm{kg} \] \[ n_{\text{urea}}=\frac{30}{60}=0.500\,\mathrm{mol} \] \[ m=\frac{0.500}{0.500}=1.00\,\mathrm{mol\,kg^{-1}} \] The reconstructed composition satisfies both the mass balance and the required molality. \(\textbf{Final answer:}\) \[ \boxed{w_{\text{urea}}=30.0\,\mathrm g} \]

Explanation: \(\textbf{Compare the initial solute amounts:}\) The two solutes have different molar masses, so their unequal masses do not necessarily represent unequal numbers of moles. \[ n_P=\frac{30}{60}=0.500\,\mathrm{mol} \] \[ n_Q=\frac{90}{180}=0.500\,\mathrm{mol} \] Both preparations initially contain the same amount of solute and the same mass of solvent. Therefore, their initial molalities are equal. \[ m_{P,\text{initial}}=m_{Q,\text{initial}}=\frac{0.500}{0.300} \] \[ m_{\text{initial}}=1.667\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Account for the selective dilution:}\) Only solution P receives additional water. The number of moles of urea remains \(0.500\,\mathrm{mol}\). \[ w_{P,\text{water}}=300+150=450\,\mathrm g \] \[ m_P=\frac{0.500}{0.450}=1.111\,\mathrm{mol\,kg^{-1}} \] Solution Q remains unchanged. \[ m_Q=\frac{0.500}{0.300}=1.667\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Compare the final concentrations:}\) \[ \frac{m_P}{m_Q}=\frac{0.500/0.450}{0.500/0.300} \] The common solute amount cancels. \[ \frac{m_P}{m_Q}=\frac{300}{450}=\frac23 \] \(\textbf{Physical interpretation:}\) The initial equality arises from equal solute moles per kilogram of water, not from equal solute masses. Adding water only to P decreases its molality while Q retains its original concentration. \(\textbf{Final answer:}\) \[ \boxed{\frac{m_P}{m_Q}=\frac23} \]

Explanation: \(\textbf{Identify the concentration basis:}\) Both dissolved substances share the same \(500\,\mathrm g\) of water as solvent. However, the molality of an individual solute counts only the moles of that particular solute. The combined molality counts the total analytical moles of both dissolved solutes per kilogram of the common solvent. \(\textbf{Calculate the separate solute amounts:}\) \[ n_{\text{urea}}=\frac{30}{60}=0.500\,\mathrm{mol} \] \[ n_{\text{glucose}}=\frac{18}{180}=0.100\,\mathrm{mol} \] The solvent mass is: \[ w_{\text{water}}=0.500\,\mathrm{kg} \] \(\textbf{Determine urea molality:}\) \[ m_{\text{urea}}=\frac{0.500}{0.500} \] \[ m_{\text{urea}}=1.00\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Determine the combined molality:}\) \[ n_{\text{total solutes}}=0.500+0.100=0.600\,\mathrm{mol} \] \[ m_{\text{combined}}=\frac{0.600}{0.500} \] \[ m_{\text{combined}}=1.20\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Interpret the difference:}\) Glucose contributes additional dissolved moles but does not change the number of urea moles. The mass of the two solutes must not be included in the solvent-mass denominator. For these nonreacting molecular solutes, their individual analytical molalities add to give the combined value. \(\textbf{Final answer:}\) \[ \boxed{m_{\text{urea}}=1.00\,m,\qquad m_{\text{combined}}=1.20\,m} \]

Explanation: \(\textbf{Identify the conserved quantity:}\) Evaporation removes water but leaves the dissolved solute amount unchanged under the specified conditions. Each final solution must therefore satisfy: \[ m_{\text{final}}=\frac{n_{\text{solute}}}{w_{\text{water, final}}(\mathrm{kg})} \] For a final molality of \(2.00\,\mathrm{mol\,kg^{-1}}\): \[ w_{\text{water, final}}=\frac{n_{\text{solute}}}{2.00} \] \(\textbf{Sample P:}\) \[ n_P=\frac{12}{60}=0.200\,\mathrm{mol} \] \[ w_{P,\text{final}}=\frac{0.200}{2.00}=0.100\,\mathrm{kg}=100\,\mathrm g \] \[ w_{P,\text{evaporated}}=200-100=100\,\mathrm g \] \(\textbf{Sample Q:}\) \[ n_Q=\frac{36}{180}=0.200\,\mathrm{mol} \] \[ w_{Q,\text{final}}=100\,\mathrm g \] \[ w_{Q,\text{evaporated}}=160-100=60\,\mathrm g \] \(\textbf{Sample R:}\) \[ n_R=\frac{18}{180}=0.100\,\mathrm{mol} \] \[ w_{R,\text{final}}=50\,\mathrm g \] \[ w_{R,\text{evaporated}}=80-50=30\,\mathrm g \] \(\textbf{Sample S:}\) \[ n_S=\frac{30}{60}=0.500\,\mathrm{mol} \] \[ w_{S,\text{final}}=250\,\mathrm g \] \[ w_{S,\text{evaporated}}=300-250=50\,\mathrm g \] \(\textbf{Compare the required changes:}\) The four required evaporation masses are \(100\), \(60\), \(30\) and \(50\,\mathrm g\), respectively. Sample P requires the greatest removal of water. The largest initial solvent mass alone does not determine the answer because the final permitted solvent mass also depends on solute moles. Hence, Sample P is correct.

Explanation: \(\textbf{Initial composition:}\) The original solution contains \(1.00\,\mathrm{mol}\) urea and \(0.400\,\mathrm{kg}\) water. \[ m_0=\frac{1.00}{0.400}=2.50\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Effect of withdrawing a homogeneous portion:}\) Removing half of the homogeneous solution removes half the solute and half the solvent. It does not remove half the solute while retaining all the water. The composition of the remaining portion is: \[ n_{\text{urea, remaining}}=0.500\,\mathrm{mol} \] \[ w_{\text{water, remaining}}=200\,\mathrm g \] Its molality immediately after withdrawal is still: \[ m_{\text{after withdrawal}}=\frac{0.500}{0.200}=2.50\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Effect of adding pure water:}\) The additional \(100\,\mathrm g\) of water increases only the solvent mass. \[ w_{\text{water, final}}=200+100=300\,\mathrm g \] \[ m_f=\frac{0.500}{0.300}=\frac53\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Determine the requested ratio:}\) \[ \frac{m_f}{m_0}=\frac{5/3}{5/2} \] \[ \frac{m_f}{m_0}=\frac23 \] \(\textbf{Check the physical direction:}\) Proportional withdrawal alone preserves molality, whereas subsequent solvent addition decreases it. Therefore, the final-to-initial ratio must be less than unity. \(\textbf{Final answer:}\) \[ \boxed{\frac{m_f}{m_0}=\frac23} \]

Explanation: \(\textbf{Establish the common starting point:}\) Both solutions initially contain the same solute amount and solvent mass. \[ m_0=\frac{0.500}{0.500}=1.00\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Process P — Add dissolved solute:}\) Additional sucrose increases the number of solute moles while leaving the mass of water unchanged. \[ n_P=0.500+0.500=1.000\,\mathrm{mol} \] \[ w_P=0.500\,\mathrm{kg} \] \[ m_{P,\text{intermediate}}=\frac{1.000}{0.500}=2.00\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Process Q — Remove solvent:}\) Water evaporation decreases the denominator but retains the original solute amount. \[ n_Q=0.500\,\mathrm{mol} \] \[ w_Q=0.500-0.250=0.250\,\mathrm{kg} \] \[ m_{Q,\text{intermediate}}=\frac{0.500}{0.250}=2.00\,\mathrm{mol\,kg^{-1}} \] Both intermediate solutions therefore have identical molalities, although their absolute compositions are different. \(\textbf{Apply the common final treatment:}\) Adding \(0.250\,\mathrm{kg}\) water to P gives: \[ m_{P,\text{final}}=\frac{1.000}{0.500+0.250} \] \[ m_{P,\text{final}}=\frac43\,\mathrm{mol\,kg^{-1}} \] Adding the same water mass to Q gives: \[ m_{Q,\text{final}}=\frac{0.500}{0.250+0.250} \] \[ m_{Q,\text{final}}=1.00\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Interpret the unequal response:}\) Equal intermediate molalities represent equal solute-to-solvent ratios, not equal absolute amounts of solute and solvent. Consequently, adding an identical mass of water does not produce an identical proportional dilution. \(\textbf{Final answer:}\) \[ \boxed{\frac{m_P}{m_Q}=\frac43} \]

Explanation: \(\textbf{Identify conserved solute amount during mixing:}\) Both solutions contain the same nonreacting dissolved solute. Their urea moles can therefore be added directly. \[ n_{\text{total}}=0.800+0.400=1.200\,\mathrm{mol} \] \(\textbf{Combine the solvent masses:}\) The total mass of water is also additive. \[ w_{\text{water, mixed}}=400+600=1000\,\mathrm g \] \[ w_{\text{water, mixed}}=1.000\,\mathrm{kg} \] The initial molality of the mixture is: \[ m_{\text{mixed}}=\frac{1.200}{1.000}=1.200\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Work backward from the required molality:}\) Evaporation changes the solvent mass but not the number of urea moles. For the final solution: \[ m_{\text{final}}=\frac{n_{\text{total}}}{w_{\text{water, final}}} \] \[ 2.00=\frac{1.200}{w_{\text{water, final}}} \] \[ w_{\text{water, final}}=\frac{1.200}{2.00}=0.600\,\mathrm{kg} \] \(\textbf{Calculate the solvent loss:}\) \[ w_{\text{evaporated}}=1.000-0.600 \] \[ w_{\text{evaporated}}=0.400\,\mathrm{kg} \] \[ w_{\text{evaporated}}=400\,\mathrm g \] \(\textbf{Check the final concentration:}\) \[ m_{\text{check}}=\frac{1.200}{0.600}=2.00\,\mathrm{mol\,kg^{-1}} \] The final solvent mass is smaller than the initial mixed solvent mass, consistent with the required concentration increase. Averaging the two original molalities directly would not correctly account for their unequal solvent masses. \(\textbf{Final answer:}\) \[ \boxed{w_{\text{evaporated}}=400\,\mathrm g} \]

Explanation: \(\textbf{Identify the unknown composition:}\) Let the initial mass of urea be \(x\,\mathrm g\). The initial mass of water is therefore \(600-x\) grams. After evaporation, the urea mass remains unchanged while the water mass becomes \(550-x\) grams. \(\textbf{Express the initial molality:}\) The number of moles of urea is: \[ n_{\text{urea}}=\frac{x}{60} \] Hence: \[ m_0=\frac{x/60}{(600-x)/1000} \] \(\textbf{Use the observed concentration change:}\) A \(10\%\) increase means: \[ m_f=1.10m_0 \] Because solute moles remain constant, the molality ratio depends only on the reciprocal ratio of solvent masses. \[ \frac{m_f}{m_0}=\frac{600-x}{550-x} \] Substituting the observed increase: \[ 1.10=\frac{600-x}{550-x} \] \[ 1.10(550-x)=600-x \] \[ 605-1.10x=600-x \] \[ 0.10x=5 \] \[ x=50\,\mathrm g \] \(\textbf{Reconstruct the initial molality:}\) Initial urea moles: \[ n_{\text{urea}}=\frac{50}{60}=0.8333\,\mathrm{mol} \] Initial water mass: \[ w_{\text{water}}=600-50=550\,\mathrm g=0.550\,\mathrm{kg} \] \[ m_0=\frac{0.8333}{0.550} \] \[ m_0=1.515\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Independent verification:}\) After evaporating \(50\,\mathrm g\) water, the solvent mass becomes \(0.500\,\mathrm{kg}\). \[ m_f=\frac{0.8333}{0.500}=1.6667\,\mathrm{mol\,kg^{-1}} \] \[ \frac{1.6667}{1.515}\approx1.10 \] The calculated increase agrees with the stated \(10\%\). \(\textbf{Final answer:}\) \[ \boxed{m_0\approx1.52\,\mathrm{mol\,kg^{-1}}} \]

Explanation: \(\textbf{Determine the initial solute amount:}\) \[ n_{\text{urea}}=\frac{12}{60}=0.200\,\mathrm{mol} \] \(\textbf{Calculate the original molarity:}\) Molarity is defined using the final solution volume rather than the solvent mass. \[ M=\frac{n_{\text{solute}}}{V_{\text{solution}}(\mathrm L)} \] \[ M_0=\frac{0.200}{0.200}=1.000\,\mathrm{mol\,L^{-1}} \] \(\textbf{Apply the temperature-induced volume change:}\) The number of moles of urea remains constant. The new volume is \(0.220\,\mathrm L\). \[ M_f=\frac{0.200}{0.220} \] \[ M_f=0.9091\,\mathrm{mol\,L^{-1}} \] Molarity decreases because the same solute amount occupies a greater solution volume. \(\textbf{Examine molality separately:}\) Molality depends on the mass of solvent: \[ m=\frac{n_{\text{solute}}}{w_{\text{solvent}}(\mathrm{kg})} \] Heating does not change either quantity under the stated conditions. Therefore, the molality remains unchanged even though the molarity decreases. \(\textbf{Conclusion:}\) The new molarity is approximately \(0.909\,\mathrm M\), while the molality remains at its original value. The distinction arises because molarity is volume-based and molality is solvent-mass-based.

Explanation: \(\textbf{Choose a convenient composition basis:}\) A molality of \(1.00\,\mathrm{mol\,kg^{-1}}\) means that \(1.00\,\mathrm{mol}\) of X is present per kilogram of water. Consider exactly \(1.000\,\mathrm{kg}\) of water. \[ n_X=1.00\,\mathrm{mol} \] \(\textbf{Determine the total solution mass:}\) \[ w_X=1.00\times100=100\,\mathrm g \] \[ w_{\text{solution}}=1000+100=1100\,\mathrm g \] \(\textbf{Use density to reconstruct the initial volume:}\) \[ \rho=\frac{w_{\text{solution}}}{V_{\text{solution}}} \] \[ V_0=\frac{1100\,\mathrm g}{1.10\,\mathrm{g\,mL^{-1}}} \] \[ V_0=1000\,\mathrm{mL}=1.000\,\mathrm L \] \(\textbf{Determine the initial molarity:}\) \[ M_0=\frac{1.00}{1.000}=1.00\,\mathrm M \] \(\textbf{Account for thermal expansion:}\) The final volume is \(105\%\) of the original volume. \[ V_f=1.05(1.000)=1.050\,\mathrm L \] The amount of X is unchanged. \[ M_f=\frac{1.00}{1.050} \] \[ M_f=0.95238\,\mathrm M \] \(\textbf{Check the mass-based concentration:}\) No solute or solvent is lost. The solution still contains \(1.00\,\mathrm{mol}\) X per \(1.000\,\mathrm{kg}\) water. \[ m_f=\frac{1.00}{1.000}=1.00\,\mathrm{mol\,kg^{-1}} \] The reduction in molarity is consistent with the volume increase. It would be incorrect to use the initial density to calculate the final volume after expansion without accounting for the changed condition. \(\textbf{Final answer:}\) \[ \boxed{m_f=1.00\,\mathrm{mol\,kg^{-1}},\qquad M_f\approx0.952\,\mathrm{mol\,L^{-1}}} \]

Explanation: \(\textbf{Find total glucose moles:}\) The solutions contain the same nonreacting solute, so the individual solute amounts are additive. \[ n_P=M_PV_P=2.00(0.100)=0.200\,\mathrm{mol} \] \[ n_Q=M_QV_Q=1.00(0.100)=0.100\,\mathrm{mol} \] \[ n_{\text{total}}=0.200+0.100=0.300\,\mathrm{mol} \] \(\textbf{Use the measured final volume for molarity:}\) The final volume is explicitly \(198\,\mathrm{mL}\), not \(200\,\mathrm{mL}\). \[ V_f=0.198\,\mathrm L \] \[ M_f=\frac{0.300}{0.198} \] \[ M_f=1.51515\,\mathrm{mol\,L^{-1}} \] \(\textbf{Reconstruct the solution masses from density:}\) \[ w_P=\rho_PV_P=1.16(100)=116\,\mathrm g \] \[ w_Q=\rho_QV_Q=1.08(100)=108\,\mathrm g \] \[ w_{\text{total solution}}=116+108=224\,\mathrm g \] \(\textbf{Separate solute mass from solvent mass:}\) \[ w_{\text{glucose}}=0.300(180)=54.0\,\mathrm g \] \[ w_{\text{water}}=224-54=170\,\mathrm g \] \[ w_{\text{water}}=0.170\,\mathrm{kg} \] \(\textbf{Calculate the final molality:}\) \[ m_f=\frac{0.300}{0.170} \] \[ m_f=1.7647\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Check the distinction between the two calculations:}\) Molarity requires the actually measured final volume, whereas molality requires the conserved total mass of water. The final solution volume need not equal the sum of the two measured starting volumes. Using \(200\,\mathrm{mL}\) for molarity or \(224\,\mathrm g\) as the solvent mass would introduce different errors. \(\textbf{Final answer:}\) \[ \boxed{M_f\approx1.52\,\mathrm M,\qquad m_f\approx1.76\,\mathrm{mol\,kg^{-1}}} \]

Explanation: \(\textbf{Identify what equal molarity does and does not imply:}\) Both solutions contain \(0.500\,\mathrm{mol}\) of their respective solutes per litre. However, urea and sucrose have different molar masses. Consequently, equal molarities do not guarantee equal molalities. The solution densities must be used to determine the actual water masses. \(\textbf{Analyse solution A using a one-litre basis:}\) \[ n_A=0.500\,\mathrm{mol} \] \[ w_{\text{solution A}}=1.01(1000)=1010\,\mathrm g \] Urea mass: \[ w_{\text{urea}}=0.500(60)=30\,\mathrm g \] Water mass: \[ w_{\text{water A}}=1010-30=980\,\mathrm g \] Thus: \[ m_A=\frac{0.500}{0.980} \] \[ m_A=0.5102\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Analyse solution B:}\) \[ n_B=0.500\,\mathrm{mol} \] \[ w_{\text{solution B}}=1.07(1000)=1070\,\mathrm g \] Sucrose mass: \[ w_{\text{sucrose}}=0.500(342)=171\,\mathrm g \] Water mass: \[ w_{\text{water B}}=1070-171=899\,\mathrm g \] Initially: \[ m_B=\frac{0.500}{0.899}=0.5562\,\mathrm{mol\,kg^{-1}} \] Solution B is initially more concentrated on a molal basis, despite having the same molarity as A. \(\textbf{Determine the required final solvent mass:}\) Both one-litre samples contain \(0.500\,\mathrm{mol}\) solute. For their molalities to become equal, the solvent masses must also become equal. Therefore, solution B must ultimately contain \(980\,\mathrm g\) of water. \[ w_{\text{water to add}}=980-899 \] \[ w_{\text{water to add}}=81\,\mathrm g \] \(\textbf{Verify the final molality:}\) \[ m_{B,\text{final}}=\frac{0.500}{0.980}=0.5102\,\mathrm{mol\,kg^{-1}} \] This matches the molality of solution A. The calculation does not require assuming any particular final volume after adding water. \(\textbf{Final answer:}\) \[ \boxed{w_{\text{water to add}}=81\,\mathrm g} \]

Explanation: \(\textbf{Interpret the graph axes:}\) Molality is defined by: \[ m=\frac{n_{\text{solute}}}{w_{\text{solvent}}(\mathrm{kg})} \] Define the horizontal-axis quantity as: \[ x=\frac{1}{w_{\text{solvent}}(\mathrm{kg})} \] The relation becomes: \[ m=n_{\text{solute}}x \] This has the form of a straight line through the origin. \(\textbf{Extract the physical meaning of slope:}\) Comparing the equation with \(y=(\text{slope})x\): \[ \text{slope}=n_{\text{solute}} \] The slope therefore represents the number of moles of sucrose retained throughout the dilution series. \[ n_{\text{sucrose}}=0.240\,\mathrm{mol} \] The units also confirm this interpretation: \[ \frac{\mathrm{mol\,kg^{-1}}}{\mathrm{kg^{-1}}}=\mathrm{mol} \] \(\textbf{Calculate molality for the selected preparation:}\) Water mass: \[ w_{\text{water}}=480\,\mathrm g=0.480\,\mathrm{kg} \] \[ m=\frac{0.240}{0.480} \] \[ m=0.500\,\mathrm{mol\,kg^{-1}} \] \(\textbf{Use the independent solution-volume information:}\) Molarity is determined from the number of solute moles divided by the measured final solution volume. \[ V_{\text{solution}}=530\,\mathrm{mL}=0.530\,\mathrm L \] \[ M=\frac{0.240}{0.530} \] \[ M=0.45283\,\mathrm{mol\,L^{-1}} \] \(\textbf{Distinguish the two denominators:}\) The \(480\,\mathrm g\) value represents solvent mass and belongs in the molality calculation. The \(530\,\mathrm{mL}\) value represents final solution volume and belongs in the molarity calculation. No solution density is needed because both denominators have been supplied independently. \(\textbf{Final answer:}\) \[ \boxed{m=0.500\,\mathrm{mol\,kg^{-1}},\qquad M\approx0.453\,\mathrm{mol\,L^{-1}}} \]

Explanation: \(\textbf{Original solute amount:}\) The initial solution contains \(100.0\,\mathrm{mL}\) of \(0.500\,\mathrm M\) urea. Molarity connects the amount of dissolved solute with the final solution volume: \[ M=\frac{n}{V} \] Therefore: \[ n_{\text{initial}}=0.500\times0.1000 \] \[ n_{\text{initial}}=0.0500\,\mathrm{mol} \] \(\textbf{Account for the withdrawn portion:}\) Because the solution is homogeneous, the withdrawn aliquot has the same molarity as the original solution. The amount of urea discarded is: \[ n_{\text{discarded}}=0.500\times0.0400 \] \[ n_{\text{discarded}}=0.0200\,\mathrm{mol} \] Hence, the urea remaining in the flask is: \[ n_{\text{remaining}}=0.0500-0.0200 \] \[ n_{\text{remaining}}=0.0300\,\mathrm{mol} \] \(\textbf{Reconstruct the required final composition:}\) The final volume must be \(0.2000\,\mathrm L\), and the required molarity is \(0.800\,\mathrm M\). \[ n_{\text{required}}=0.800\times0.2000 \] \[ n_{\text{required}}=0.1600\,\mathrm{mol} \] The stock solution must supply the difference: \[ n_{\text{stock}}=0.1600-0.0300 \] \[ n_{\text{stock}}=0.1300\,\mathrm{mol} \] \(\textbf{Determine the stock volume:}\) \[ V_{\text{stock}}=\frac{n_{\text{stock}}}{M_{\text{stock}}} \] \[ V_{\text{stock}}=\frac{0.1300}{2.50} \] \[ V_{\text{stock}}=0.0520\,\mathrm L \] \[ V_{\text{stock}}=52.0\,\mathrm{mL} \] \(\textbf{Consistency check:}\) The withdrawn aliquot removes both urea and water in the original solution proportions. It would be incorrect to retain the original \(0.0500\,\mathrm{mol}\) when calculating the required stock addition. The calculated stock volume supplies exactly the missing \(0.1300\,\mathrm{mol}\), and the final dilution fixes the required molarity. \(\textbf{Final answer:}\) \[ \boxed{V_{\text{stock}}=52.0\,\mathrm{mL}} \]

Explanation: \(\textbf{Identify the two conservation relationships:}\) The solute is nonreacting, so its total number of moles is conserved during mixing. The total mass of the solutions is also conserved because no material is lost. However, the final volume cannot automatically be obtained by adding the separately measured initial volumes. The measured density of the final mixture must be used to determine its actual volume. \(\textbf{Calculate the total solute amount:}\) For solution P: \[ n_P=M_PV_P \] \[ n_P=1.00\times0.100=0.100\,\mathrm{mol} \] For solution Q: \[ n_Q=2.00\times0.100=0.200\,\mathrm{mol} \] Therefore: \[ n_{\text{total}}=0.100+0.200 \] \[ n_{\text{total}}=0.300\,\mathrm{mol} \] \(\textbf{Reconstruct the conserved solution mass:}\) \[ w_P=\rho_PV_P=1.04\times100=104\,\mathrm g \] \[ w_Q=\rho_QV_Q=1.12\times100=112\,\mathrm g \] \[ w_{\text{final}}=104+112=216\,\mathrm g \] \(\textbf{Use the final density:}\) \[ V_{\text{final}}=\frac{w_{\text{final}}}{\rho_{\text{final}}} \] \[ V_{\text{final}}=\frac{216}{1.10} \] \[ V_{\text{final}}=196.36\,\mathrm{mL} \] \[ V_{\text{final}}=0.19636\,\mathrm L \] \(\textbf{Calculate the final molarity:}\) \[ M_{\text{final}}=\frac{0.300}{0.19636} \] \[ M_{\text{final}}=1.5278\,\mathrm{mol\,L^{-1}} \] \(\textbf{Physical verification:}\) The measured final volume is slightly smaller than the sum of the two initial volumes. Consequently, the actual molarity is slightly greater than the \(1.500\,\mathrm M\) that would follow from assuming a final volume of \(200\,\mathrm{mL}\). The result is consistent with both solute conservation and the measured density. \(\textbf{Final answer:}\) \[ \boxed{M_{\text{final}}\approx1.528\,\mathrm M} \]

Explanation: \(\textbf{Statement I:}\) Statement I is correct. Every representative portion of a homogeneous solution contains the same amount of solute per unit solution volume. Removing an aliquot therefore removes solute and solution volume in the same proportion. The molarity of the remaining homogeneous solution stays unchanged. \(\textbf{Statement II:}\) Statement II is incorrect. Liquid volumes are not universally additive when different liquids or solutions are mixed. Intermolecular interactions may cause contraction or expansion of the final solution volume. Therefore, adding \(100\,\mathrm{mL}\) of water to \(100\,\mathrm{mL}\) of solution does not guarantee a final volume of exactly \(200\,\mathrm{mL}\). \(\textbf{Statement III:}\) Statement III is correct. For dilution without solute loss: \[ M_1V_1=M_2V_2 \] Consequently: \[ M_2=M_1\left(\frac{20}{100}\right)=\frac{M_1}{5} \] A volumetric flask fixes the final solution volume rather than merely specifying the volume of water added. \(\textbf{Statement IV:}\) Statement IV is incorrect. The arithmetic mean of two molarities applies to equal initial solution volumes only when the final volume equals their sum and the solute is conserved. If the final volume differs from that sum, the total moles must instead be divided by the actual final volume. Thus, the validity of the arithmetic-mean result depends on the volume condition. \(\textbf{Conclusion:}\) Statements I and III are correct, whereas Statements II and IV are incorrect.

Explanation: \(\textbf{Initial urea inventory:}\) The original solution contains: \[ n_0=M_0V_0 \] \[ n_0=0.800\times0.250 \] \[ n_0=0.200\,\mathrm{mol} \] \(\textbf{First withdrawal-and-replacement operation:}\) Each withdrawal removes \(50.0\,\mathrm{mL}\) from a total of \(250\,\mathrm{mL}\). Hence, the fraction of dissolved urea retained is: \[ f=\frac{250-50}{250}=\frac45 \] The replacement water introduces no additional urea. The amount of urea after the first operation is: \[ n_1=0.200\times\frac45 \] \[ n_1=0.160\,\mathrm{mol} \] Because the volume is restored to \(0.250\,\mathrm L\): \[ M_1=\frac{0.160}{0.250} \] \[ M_1=0.640\,\mathrm M \] \(\textbf{Second withdrawal-and-replacement operation:}\) The second aliquot must be evaluated using the reduced concentration, not the original concentration. The same fraction \(\frac45\) of the then-present urea remains. \[ n_2=0.160\times\frac45 \] \[ n_2=0.128\,\mathrm{mol} \] The corresponding molarity is: \[ M_2=\frac{0.128}{0.250} \] \[ M_2=0.512\,\mathrm M \] \(\textbf{Reconstruct the missing solute:}\) To restore \(0.800\,\mathrm M\) in \(250\,\mathrm{mL}\), the flask must again contain \(0.200\,\mathrm{mol}\) of urea. \[ n_{\text{additional}}=0.200-0.128 \] \[ n_{\text{additional}}=0.072\,\mathrm{mol} \] The required urea mass is: \[ w_{\text{urea}}=n_{\text{additional}}\times60 \] \[ w_{\text{urea}}=0.072\times60 \] \[ w_{\text{urea}}=4.32\,\mathrm g \] \(\textbf{Final-volume check:}\) Dissolving solid urea may change the solution volume. Therefore, the final volume must be adjusted to exactly \(250\,\mathrm{mL}\), as stated. The calculation would not be guaranteed if the urea were simply added without restoring the final calibrated volume. \(\textbf{Final answer:}\) \[ \boxed{w_{\text{urea}}=4.32\,\mathrm g} \]