Some Basic Concepts in Chemistry MCQs with Answers – Part 6 (NEET Chemistry)

Explanation: \(\textbf{Separating the pure reactant from the limestone sample:}\) The 25.0 g sample is not entirely calcium carbonate. Only the stated 80.0% contributes to the decomposition reaction. \[ m_{\mathrm{CaCO_3,initial}}=25.0\left(\frac{80.0}{100}\right) \] \[ m_{\mathrm{CaCO_3,initial}}=20.0\,\text{g} \] \(\textbf{Converting pure carbonate mass into amount:}\) \[ M_{\mathrm{CaCO_3}}=40+12+3(16) \] \[ M_{\mathrm{CaCO_3}}=100\,\mathrm{g\,mol^{-1}} \] Therefore, \[ n_{\mathrm{CaCO_3,initial}}=\frac{20.0}{100} \] \[ n_{\mathrm{CaCO_3,initial}}=0.200\,\mathrm{mol} \] \(\textbf{Applying the actual decomposition percentage:}\) Only 60.0% of the calcium carbonate decomposes. \[ n_{\mathrm{CaCO_3,decomposed}}=0.200\left(\frac{60.0}{100}\right) \] \[ n_{\mathrm{CaCO_3,decomposed}}=0.120\,\mathrm{mol} \] The amount remaining is \[ n_{\mathrm{CaCO_3,remaining}}=0.200-0.120 \] \[ n_{\mathrm{CaCO_3,remaining}}=0.080\,\mathrm{mol} \] Its mass is \[ m_{\mathrm{CaCO_3,remaining}}=0.080(100) \] \[ m_{\mathrm{CaCO_3,remaining}}=8.0\,\text{g} \] \(\textbf{Calculating carbon dioxide actually produced:}\) One mole of decomposed calcium carbonate produces one mole of carbon dioxide. \[ n_{\mathrm{CO_2,produced}}=0.120\,\mathrm{mol} \] \[ m_{\mathrm{CO_2,produced}}=0.120(44) \] \[ m_{\mathrm{CO_2,produced}}=5.28\,\text{g} \] \(\textbf{Applying the separate gas-recovery condition:}\) The apparatus collects only 75.0% of the carbon dioxide produced. \[ m_{\mathrm{CO_2,collected}}=5.28\left(\frac{75.0}{100}\right) \] \[ m_{\mathrm{CO_2,collected}}=3.96\,\text{g} \] \(\textbf{Distinguishing three separate quantities:}\) The original limestone purity determines how much reactive carbonate was available. The decomposition percentage determines how much carbonate actually reacted. The collection efficiency determines how much of the resulting gas was recovered. Gas not collected is still gas that was chemically produced. Consequently, the 75.0% collection factor must not be applied when determining unreacted calcium carbonate. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{CO_2,collected}}=3.96\,\text{g},\quad m_{\mathrm{CaCO_3,remaining}}=8.0\,\text{g}} \]

Explanation: \(\textbf{Identifying the original limiting reagent:}\) The reaction requires three moles of hydrogen per mole of nitrogen. The hydrogen required to consume all the nitrogen is \[ n_{\mathrm{H_2,required}}=3(0.40) \] \[ n_{\mathrm{H_2,required}}=1.20\,\mathrm{mol} \] Only \(0.90\,\mathrm{mol}\) hydrogen is supplied. Therefore, hydrogen is the limiting reagent for the original mixture. \(\textbf{Calculating the theoretical ammonia maximum:}\) Three moles of hydrogen can produce two moles of ammonia. \[ n_{\mathrm{NH_3,theoretical}}=0.90\left(\frac23\right) \] \[ n_{\mathrm{NH_3,theoretical}}=0.60\,\mathrm{mol} \] \(\textbf{Applying the stated reaction conversion into product:}\) The actual ammonia formation is 80.0% of this theoretical maximum. \[ n_{\mathrm{NH_3,formed}}=0.60\left(\frac{80.0}{100}\right) \] \[ n_{\mathrm{NH_3,formed}}=0.48\,\mathrm{mol} \] \(\textbf{Recovering nitrogen consumed during actual reaction:}\) One mole of nitrogen forms two moles of ammonia. \[ n_{\mathrm{N_2,consumed}}=0.48\left(\frac12\right) \] \[ n_{\mathrm{N_2,consumed}}=0.24\,\mathrm{mol} \] Therefore, \[ n_{\mathrm{N_2,remaining}}=0.40-0.24 \] \[ n_{\mathrm{N_2,remaining}}=0.16\,\mathrm{mol} \] \(\textbf{Applying the product-recovery efficiency:}\) Only 75.0% of the \(0.48\,\mathrm{mol}\) ammonia actually formed is recovered. \[ n_{\mathrm{NH_3,recovered}}=0.48\left(\frac{75.0}{100}\right) \] \[ n_{\mathrm{NH_3,recovered}}=0.36\,\mathrm{mol} \] \(\textbf{Calculating recovered yield relative to the original maximum:}\) \[ \%\mathrm{recovered\ yield}=\frac{0.36}{0.60}\times100 \] \[ \%\mathrm{recovered\ yield}=60.0\% \] \(\textbf{Checking the hydrogen inventory:}\) Formation of \(0.48\,\mathrm{mol}\) ammonia consumes \[ n_{\mathrm{H_2,consumed}}=0.48\left(\frac32\right) \] \[ n_{\mathrm{H_2,consumed}}=0.72\,\mathrm{mol} \] Hydrogen remaining after reaction is \(0.90-0.72=0.18\,\mathrm{mol}\). This is consistent with incomplete reaction before product separation. The product lost during recovery was already formed and therefore cannot be counted as unreacted nitrogen. Reaction extent and product recovery describe two different stages. \(\textbf{Final answer:}\) \[ \boxed{\%\mathrm{recovered\ yield}=60.0\%,\quad n_{\mathrm{N_2,remaining}}=0.16\,\mathrm{mol}} \]

Explanation: Statement I is correct because the oxygen masses must be compared against the same mass of iron. In FeO, 56 mass units of iron combine with 16 mass units of oxygen. In \(\mathrm{Fe_2O_3}\), 112 mass units of iron combine with 48 mass units of oxygen. For the same 56 mass units of iron, the second oxide therefore contains 24 mass units of oxygen. The ratio of oxygen masses is \[ 16:24=2:3 \] This is a simple whole-number ratio. Statement II is correct because the displayed equation conserves both iron and oxygen atoms. Four FeO formula units contain four iron atoms and four oxygen atoms. One additional oxygen molecule contributes two more oxygen atoms. Two \(\mathrm{Fe_2O_3}\) formula units contain four iron atoms and six oxygen atoms. Thus, the equation is balanced. Statement III is incorrect because the two oxides have different iron mass fractions and different formula molar masses. FeO has molar mass 72 g mol\(^{-1}\), whereas \(\mathrm{Fe_2O_3}\) has molar mass 160 g mol\(^{-1}\). Equal sample masses therefore correspond to different amounts of formula units. Counting the iron atoms within those formula units does not produce equal amounts. Statement IV is correct because the same two elements form distinct compounds with different fixed combining mass ratios. When the mass of one element is held constant, the masses of the other occur in a simple whole-number ratio. That observation is the central feature of the law of multiple proportions. The balanced oxidation equation is consistent with the distinct compositions but does not make the two oxides chemically identical. Therefore, Statements I, II and IV are correct.

Explanation: \(\textbf{Representing the unknown coefficients:}\) Write the equation as \[ a\mathrm{X_3Y_2}+b\mathrm{O_2}\rightarrow c\mathrm{X_2O_3}+d\mathrm{YO_2} \] Every coefficient must satisfy the conservation of X, Y and O atoms. \(\textbf{Balancing element X:}\) The reactant contains three X atoms per formula unit. The corresponding oxide contains two X atoms per formula unit. Therefore, \[ 3a=2c \] \(\textbf{Balancing element Y:}\) The reactant contains two Y atoms per formula unit. The Y oxide contains one Y atom per formula unit. Hence, \[ 2a=d \] \(\textbf{Choosing the smallest coefficients compatible with X and Y:}\) Taking \(a=2\) gives \[ c=3,\qquad d=4 \] The provisional equation becomes \[ \mathrm{2X_3Y_2+bO_2\rightarrow3X_2O_3+4YO_2} \] \(\textbf{Counting product oxygen atoms:}\) Three \(\mathrm{X_2O_3}\) formula units contain nine oxygen atoms. Four \(\mathrm{YO_2}\) formula units contain eight oxygen atoms. Thus, \[ N_{O,\mathrm{products}}=3(3)+4(2) \] \[ N_{O,\mathrm{products}}=17 \] The required number of oxygen molecules on this coefficient basis is \[ b=\frac{17}{2} \] \(\textbf{Removing the fractional coefficient:}\) Multiplying all coefficients by two gives \[ \mathrm{4X_3Y_2+17O_2\rightarrow6X_2O_3+8YO_2} \] Therefore, \[ a:b:c:d=4:17:6:8 \] \(\textbf{Calculating the oxygen requirement:}\) Four moles of the binary compound require seventeen moles of oxygen. \[ n_{\mathrm{O_2}}=0.040\left(\frac{17}{4}\right) \] \[ n_{\mathrm{O_2}}=0.170\,\mathrm{mol} \] \(\textbf{Checking atom conservation:}\) The reactant side contains twelve X atoms, eight Y atoms and thirty-four oxygen atoms per four formula units of the compound. The product side contains \(6(2)=12\) X atoms and eight Y atoms. Its oxygen count is \(6(3)+8(2)=34\). All elemental inventories agree. The odd total number of product oxygen atoms on the provisional coefficient basis explains why fractional oxygen coefficients initially arise. A common multiplier converts the valid fractional equation into the required smallest whole-number equation. \(\textbf{Final answer:}\) \[ \boxed{a:b:c:d=4:17:6:8,\quad n_{\mathrm{O_2}}=0.170\,\mathrm{mol}} \]

Explanation: \(\textbf{Identifying the limiting-reagent changeover:}\) The equation requires one mole of sodium sulphate per mole of barium chloride. The fixed barium chloride amount is \(0.120\,\mathrm{mol}\). Therefore, sodium sulphate is limiting below \(0.120\,\mathrm{mol}\). At \(0.120\,\mathrm{mol}\), the reactants are stoichiometrically equivalent. Beyond that amount, barium chloride becomes limiting. This explains the change from a rising straight line to a plateau. \(\textbf{Calculating precipitate formation in the first trial:}\) The sodium sulphate amount is \(0.080\,\mathrm{mol}\). Since it is limiting, \[ n_{\mathrm{BaSO_4,formed},1}=0.080\,\mathrm{mol} \] The molar mass of barium sulphate is \[ M_{\mathrm{BaSO_4}}=137+32+4(16) \] \[ M_{\mathrm{BaSO_4}}=233\,\mathrm{g\,mol^{-1}} \] The mass chemically formed is \[ m_{\mathrm{formed},1}=0.080(233) \] \[ m_{\mathrm{formed},1}=18.64\,\text{g} \] \(\textbf{Applying the recovery percentage in the first trial:}\) \[ m_{\mathrm{recovered},1}=18.64\left(\frac{80.0}{100}\right) \] \[ m_{\mathrm{recovered},1}=14.912\,\text{g} \] Thus, the reported recovered mass is approximately 14.91 g. \(\textbf{Evaluating the second trial after the plateau:}\) The sodium sulphate amount is \(0.160\,\mathrm{mol}\). Only \(0.120\,\mathrm{mol}\) barium chloride is available. Therefore, \[ n_{\mathrm{BaSO_4,formed},2}=0.120\,\mathrm{mol} \] The mass chemically formed is \[ m_{\mathrm{formed},2}=0.120(233) \] \[ m_{\mathrm{formed},2}=27.96\,\text{g} \] \(\textbf{Applying the same recovery percentage:}\) \[ m_{\mathrm{recovered},2}=27.96\left(\frac{80.0}{100}\right) \] \[ m_{\mathrm{recovered},2}=22.368\,\text{g} \] The recovered mass is therefore approximately 22.37 g. \(\textbf{Interpreting the graph correctly:}\) The plateau represents exhaustion of barium chloride, not a change in the balanced reaction. Because the recovery fraction is constant, it multiplies the entire theoretical product-mass graph without changing the location of the plateau. An 80.0% recovery does not mean that only 80.0% of the limiting reagent reacted. The missing precipitate represents separation loss after chemical formation. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{recovered},1}=14.91\,\text{g},\quad m_{\mathrm{recovered},2}=22.37\,\text{g}} \]

Explanation: \(\textbf{Recovering the solids after thermal decomposition:}\) The amount of magnesium hydroxide decomposed is \[ n_{\mathrm{decomposed}}=0.500\left(\frac{40.0}{100}\right) \] \[ n_{\mathrm{decomposed}}=0.200\,\mathrm{mol} \] The decomposition produces the same mole amount of magnesium oxide. Therefore, \[ n_{\mathrm{MgO,initial\ for\ acid}}=0.200\,\mathrm{mol} \] The magnesium hydroxide remaining after heating is \[ n_{\mathrm{Mg(OH)_2,initial\ for\ acid}}=0.500-0.200 \] \[ n_{\mathrm{Mg(OH)_2,initial\ for\ acid}}=0.300\,\mathrm{mol} \] \(\textbf{Comparing the two acid-consumption coefficients:}\) Both solids consume two moles of HCl per mole of formula units. Both produce one mole of magnesium chloride per mole of solid consumed. Thus, regardless of which component reacts, \[ n_{\mathrm{MgCl_2}}=\frac{n_{\mathrm{HCl,consumed}}}{2} \] Using the stated complete HCl consumption, \[ n_{\mathrm{MgCl_2}}=\frac{0.400}{2} \] \[ n_{\mathrm{MgCl_2}}=0.200\,\mathrm{mol} \] \(\textbf{Testing whether the residual solid composition is unique:}\) Let \(x\) be the amount of MgO consumed and \(y\) the amount of magnesium hydroxide consumed. The acid inventory gives \[ 2x+2y=0.400 \] \[ x+y=0.200 \] The available solids impose \[ 0\leq x\leq0.200 \] and \[ 0\leq y\leq0.300 \] These conditions admit multiple possibilities. For example, consuming \(0.200\,\mathrm{mol}\) MgO and no hydroxide satisfies the acid balance. So does consuming \(0.100\,\mathrm{mol}\) of each solid. Both possibilities form \(0.200\,\mathrm{mol}\) magnesium chloride. However, they leave different individual quantities of MgO and magnesium hydroxide. \(\textbf{Identifying what the measurements determine:}\) The total amount of magnesium-containing solid formula units consumed is fixed. The precise allocation of that consumption between the two solids is not fixed. The acid measurement supplies only one independent constraint for the two component consumptions. An additional selective measurement or explicit preferential-reaction condition would be required to determine the residual component amounts. Therefore, Option C is necessarily correct.

Explanation: The starting sample contains two different carbonates. Its total mass and total formula-unit amount are sufficient to reconstruct the initial component amounts, because the carbonates have different molar masses. However, knowing their initial proportions does not automatically determine which component reacts to what extent when the supplied acid is insufficient for complete conversion of both. The two balanced reactions have the same HCl-to-carbonate mole ratio. Every mole of either carbonate consumes two moles of HCl. Every mole of either carbonate also produces one mole of carbon dioxide. Since all \(0.300\,\mathrm{mol}\) HCl reacts, the combined amount of carbonate consumed must be \[ n_{\mathrm{carbonate,consumed}}=\frac{0.300}{2} \] \[ n_{\mathrm{carbonate,consumed}}=0.150\,\mathrm{mol} \] The carbon dioxide amount consequently is \[ n_{\mathrm{CO_2}}=0.150\,\mathrm{mol} \] This result is independent of how the acid consumption is distributed between calcium carbonate and magnesium carbonate. By contrast, the amount of calcium chloride formed depends specifically on the calcium carbonate consumed. Likewise, the magnesium chloride amount depends on the magnesium carbonate consumed. Different allowed allocations of the \(0.150\,\mathrm{mol}\) total carbonate consumption can leave different residual masses of each solid. The observations therefore establish the total carbon dioxide amount but do not uniquely establish the individual carbonate residues or individual metal chloride products. The key distinction is between a total stoichiometric quantity shared by two reactions and component-specific quantities that require an additional independent measurement. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{CO_2}}=0.150\,\mathrm{mol}} \]

Explanation: \(\textbf{Recovering the amount decomposed from carbon dioxide:}\) Two moles of sodium hydrogen carbonate produce one mole of carbon dioxide. Therefore, \[ n_{\mathrm{NaHCO_3,decomposed}}=2(0.100) \] \[ n_{\mathrm{NaHCO_3,decomposed}}=0.200\,\mathrm{mol} \] \(\textbf{Determining the corresponding water loss:}\) The equation forms equal mole amounts of carbon dioxide and water. Thus, \[ n_{\mathrm{H_2O,formed}}=0.100\,\mathrm{mol} \] The combined escaped gas mass is \[ m_{\mathrm{gases}}=0.100(44)+0.100(18) \] \[ m_{\mathrm{gases}}=4.4+1.8 \] \[ m_{\mathrm{gases}}=6.2\,\text{g} \] \(\textbf{Checking the observed solid residue:}\) If only these gases escape, the final solid mass must be \[ m_{\mathrm{solid,final}}=20.0-6.2 \] \[ m_{\mathrm{solid,final}}=13.8\,\text{g} \] This agrees with the reported residue. However, the residue mass supplies no independent information about how much undecomposed sodium hydrogen carbonate or inert solid remains. \(\textbf{Testing alternative starting compositions:}\) The molar mass of sodium hydrogen carbonate is \[ M_{\mathrm{NaHCO_3}}=23+1+12+3(16) \] \[ M_{\mathrm{NaHCO_3}}=84\,\mathrm{g\,mol^{-1}} \] The amount decomposed weighs \[ m_{\mathrm{NaHCO_3,decomposed}}=0.200(84) \] \[ m_{\mathrm{NaHCO_3,decomposed}}=16.8\,\text{g} \] Therefore, the original sample must have contained at least 16.8 g sodium hydrogen carbonate. One possible starting composition is 16.8 g sodium hydrogen carbonate and 3.2 g inert material. Another is 18.0 g sodium hydrogen carbonate and 2.0 g inert material. In either case, decomposition of 16.8 g sodium hydrogen carbonate produces the same 6.2 g of escaping gases. The remaining solid mass is 13.8 g in both cases. Yet the two samples have different original sodium hydrogen carbonate percentages. \(\textbf{Identifying the missing independent condition:}\) The amount decomposed is fixed by the gas measurement. The amount originally present is not, because the problem does not state complete decomposition. The residue measurement is merely the original mass minus the same measured gas loss. It cannot determine a second unknown composition quantity. Therefore, the original purity and residual sodium hydrogen carbonate amount are underdetermined. Option D is correct.

Explanation: \(\textbf{Using the selective absorption sequence:}\) Stage I removes only carbon dioxide already present in the original mixture. The unabsorbed CO passes into Stage II. Therefore, carbon dioxide produced in Stage II can be used to determine the original CO amount independently of the original carbon dioxide content. \(\textbf{Recovering the original CO amount:}\) Stage II produces 8.8 g carbon dioxide. \[ n_{\mathrm{CO_2,II}}=\frac{8.8}{44} \] \[ n_{\mathrm{CO_2,II}}=0.200\,\mathrm{mol} \] The balanced oxidation equation has a \(1:1\) mole relationship between CO consumed and carbon dioxide formed. Therefore, \[ n_{\mathrm{CO,initial}}=0.200\,\mathrm{mol} \] \(\textbf{Determining the original CO mass:}\) \[ M_{\mathrm{CO}}=12+16 \] \[ M_{\mathrm{CO}}=28\,\mathrm{g\,mol^{-1}} \] Thus, \[ m_{\mathrm{CO,initial}}=0.200(28) \] \[ m_{\mathrm{CO,initial}}=5.6\,\text{g} \] \(\textbf{Recovering the original carbon dioxide mass:}\) The original gas mixture weighs 14.4 g. Consequently, \[ m_{\mathrm{CO_2,initial}}=14.4-5.6 \] \[ m_{\mathrm{CO_2,initial}}=8.8\,\text{g} \] All of this carbon dioxide is absorbed in Stage I. \(\textbf{Determining the original carbon dioxide amount:}\) \[ n_{\mathrm{CO_2,initial}}=\frac{8.8}{44} \] \[ n_{\mathrm{CO_2,initial}}=0.200\,\mathrm{mol} \] \(\textbf{Calculating the original CO mole fraction:}\) The original combined molecular amount is \[ n_{\mathrm{total}}=0.200+0.200 \] \[ n_{\mathrm{total}}=0.400\,\mathrm{mol} \] Therefore, \[ X_{\mathrm{CO}}=\frac{0.200}{0.400} \] \[ X_{\mathrm{CO}}=\frac12 \] \(\textbf{Checking the options against the recovered amounts:}\) The calculated CO mole fraction is \(\frac12\), and the original carbon dioxide mass is 8.8 g. These values correspond to Option D. The Stage II carbon dioxide is newly formed from CO and is not part of the original mixture mass. However, its mass happens to equal the original carbon dioxide mass in this particular example. The two amounts must still be distinguished by their physical origin. \(\textbf{Final answer:}\) \[ \boxed{X_{\mathrm{CO}}=\frac12,\quad m_{\mathrm{CO_2,absorbed}}=8.8\,\text{g}} \]

Explanation: \(\textbf{Recovering the initial oxide amount:}\) The molar mass of iron(III) oxide is \[ M_{\mathrm{Fe_2O_3}}=2(56)+3(16) \] \[ M_{\mathrm{Fe_2O_3}}=160\,\mathrm{g\,mol^{-1}} \] Therefore, \[ n_{\mathrm{Fe_2O_3,initial}}=\frac{16.0}{160} \] \[ n_{\mathrm{Fe_2O_3,initial}}=0.100\,\mathrm{mol} \] \(\textbf{Checking the first-stage limiting reagent:}\) Complete reduction of this oxide would require \[ n_{\mathrm{CO,required}}=3(0.100) \] \[ n_{\mathrm{CO,required}}=0.300\,\mathrm{mol} \] Only \(0.180\,\mathrm{mol}\) CO is available. Thus, CO is limiting. \(\textbf{Calculating oxide actually reduced:}\) Three moles of CO reduce one mole of iron(III) oxide. \[ n_{\mathrm{Fe_2O_3,consumed}}=\frac{0.180}{3} \] \[ n_{\mathrm{Fe_2O_3,consumed}}=0.060\,\mathrm{mol} \] The oxide remaining is \[ n_{\mathrm{Fe_2O_3,remaining}}=0.100-0.060 \] \[ n_{\mathrm{Fe_2O_3,remaining}}=0.040\,\mathrm{mol} \] Its mass is \[ m_{\mathrm{Fe_2O_3,remaining}}=0.040(160) \] \[ m_{\mathrm{Fe_2O_3,remaining}}=6.4\,\text{g} \] \(\textbf{Determining the iron formed:}\) One mole of iron(III) oxide produces two moles of iron. Therefore, \[ n_{\mathrm{Fe,formed}}=2(0.060) \] \[ n_{\mathrm{Fe,formed}}=0.120\,\mathrm{mol} \] \(\textbf{Applying the selective second-stage reaction:}\) The problem specifies that only elemental iron generates hydrogen under the given conditions. The iron-to-hydrogen mole ratio is \(1:1\). Therefore, \[ n_{\mathrm{H_2}}=0.120\,\mathrm{mol} \] \(\textbf{Checking the first-stage solid mass:}\) The iron produced weighs \[ m_{\mathrm{Fe}}=0.120(56) \] \[ m_{\mathrm{Fe}}=6.72\,\text{g} \] Together with the 6.4 g unreacted oxide, the intermediate solids weigh \[ m_{\mathrm{solids}}=6.72+6.4 \] \[ m_{\mathrm{solids}}=13.12\,\text{g} \] The decrease from the original 16.0 g oxide mass is 2.88 g. The oxygen removed from \(0.060\,\mathrm{mol}\) iron(III) oxide is \[ m_{O,\mathrm{removed}}=3(0.060)(16) \] \[ m_{O,\mathrm{removed}}=2.88\,\text{g} \] This agrees with the solid mass change. The second-stage hydrogen measures the elemental iron produced, not the combined iron atoms present in both intermediate solids. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{H_2}}=0.120\,\mathrm{mol},\quad m_{\mathrm{Fe_2O_3,remaining}}=6.4\,\text{g}} \]

Explanation: Statement I is correct because the balanced equation consumes three gaseous molecules and forms two for each full stoichiometric reaction unit. That unit forms two moles of C while decreasing the total gas amount by one mole. Therefore, formation of one mole of C corresponds to a decrease of half a mole in the total gaseous amount. Statement II is correct because the reacting volume ratios at common temperature and pressure follow the balanced coefficients. Two volumes of A react with one volume of B and produce two volumes of C. The specified 4.0 L A and 2.0 L B are present in the exact required ratio. They therefore form 4.0 L C at the stated reference conditions. Statement III is incorrect because the number of gas molecules is not the same as the total mass. The reaction rearranges conserved atoms into different molecular groupings. Although the total molecular amount decreases, all atoms and their masses remain within a closed system. Statement IV is correct because an inert gas contributes to total pressure but does not enter the displayed chemical equation. At constant temperature and fixed volume, adding inert gas increases the total gaseous amount and therefore increases the total pressure. However, it does not change the mole amounts of A or B initially available for reaction. The identity of the limiting reagent remains determined by their stoichiometric comparison. The statements collectively distinguish molecular-count changes, measured gas volumes, conservation of mass and the contribution of inert gases. Statements I, II and IV are correct.

Explanation: \(\textbf{Converting the products into a constraint on original oxides:}\) Each oxide formula unit contains one metal atom. Both displayed reactions form one metal chloride formula unit per original oxide formula unit. Therefore, the measured combined chloride amount equals the combined oxide amount. Let \(x\) be the original CaO amount and \(y\) the original MgO amount. \[ x+y=0.120 \qquad (1) \] \(\textbf{Constructing the original mass equation:}\) The molar masses are \[ M_{\mathrm{CaO}}=40+16=56\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{MgO}}=24+16=40\,\mathrm{g\,mol^{-1}} \] Thus, \[ 56x+40y=5.6 \qquad (2) \] \(\textbf{Solving the mixture composition:}\) From Equation (1), \[ x=0.120-y \] Substituting, \[ 56(0.120-y)+40y=5.6 \] \[ 6.72-56y+40y=5.6 \] \[ 16y=1.12 \] \[ y=0.070\,\mathrm{mol} \] Therefore, \[ x=0.120-0.070 \] \[ x=0.050\,\mathrm{mol} \] \(\textbf{Calculating the original magnesium oxide mass:}\) \[ m_{\mathrm{MgO}}=0.070(40) \] \[ m_{\mathrm{MgO}}=2.8\,\text{g} \] \(\textbf{Recovering both anhydrous chloride amounts:}\) The calcium chloride amount is \(0.050\,\mathrm{mol}\). The magnesium chloride amount is \(0.070\,\mathrm{mol}\). Their molar masses are \[ M_{\mathrm{CaCl_2}}=40+2(35.5)=111\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{MgCl_2}}=24+2(35.5)=95\,\mathrm{g\,mol^{-1}} \] \(\textbf{Calculating the combined chloride mass:}\) \[ m_{\mathrm{chlorides}}=0.050(111)+0.070(95) \] \[ m_{\mathrm{chlorides}}=5.55+6.65 \] \[ m_{\mathrm{chlorides}}=12.20\,\text{g} \] \(\textbf{Independent elemental-mass check:}\) The original two oxides contain \(0.120\,\mathrm{mol}\) oxygen atoms. The oxygen mass removed from the oxide portion is \[ m_O=0.120(16) \] \[ m_O=1.92\,\text{g} \] The chloride products incorporate \[ n_{\mathrm{Cl,atoms}}=2(0.120) \] \[ n_{\mathrm{Cl,atoms}}=0.240\,\mathrm{mol} \] chlorine atoms. Their mass is \[ m_{\mathrm{Cl}}=0.240(35.5) \] \[ m_{\mathrm{Cl}}=8.52\,\text{g} \] The predicted chloride mass is therefore \[ m_{\mathrm{chlorides}}=5.6-1.92+8.52 \] \[ m_{\mathrm{chlorides}}=12.20\,\text{g} \] This matches the direct calculation. \(\textbf{Checking the options:}\) The calculated pair is 2.8 g magnesium oxide and 12.2 g chlorides, corresponding to Option A. The measured total chloride amount and the independently checked mass balance both support this pair. The correctly aligned numerical result is shown below. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{MgO}}=2.8\,\text{g},\quad m_{\mathrm{chlorides}}=12.20\,\text{g}} \]

Explanation: Statement I is correct because both phases of Sample P contain the same chemical species, \(\mathrm{H_2O}\). The liquid and vapour differ in physical state, not in the chemical formula of their constituent molecules. A system containing one chemical substance can have more than one physical phase. Statement II is correct because Sample Q contains two distinct molecular substances, oxygen and ozone. Both consist exclusively of oxygen atoms, but \(\mathrm{O_2}\) and \(\mathrm{O_3}\) have different molecular compositions and are different forms of the same element. Their physical mixing does not produce a new compound. Consequently, the sample remains a mixture despite containing only one type of element. Statement III is incorrect because a uniform appearance does not establish chemical purity. Dissolved sodium chloride and water remain distinct chemical substances in a homogeneous solution. Their relative amounts can be varied without producing a new compound having a fixed composition. Statement IV is correct because all constituent molecules in Sample S have the same formula, \(\mathrm{CO_2}\). Each molecule contains one carbon atom and two oxygen atoms. That fixed molecular composition identifies carbon dioxide as a compound rather than a mixture. The important distinction is that physical uniformity, number of phases, number of elements and number of chemical substances are different classification criteria. A two-phase system need not contain two substances, and a one-phase system need not be chemically pure. Thus, Statements I, II and IV are correct.

Explanation: Statement I is correct because Dalton originally regarded atoms of the same element as identical in mass. Isotopes demonstrate that atoms belonging to one element can possess different atomic masses. This requires refinement of the original postulate rather than rejection of the entire atomic description of matter. Statement II is incorrect because ordinary chemical reactions conserve atoms of each element and their associated mass. Different isotopes may have different masses, but the existence of isotopes does not mean that mass disappears or is created during a chemical reaction. A reaction involving an isotope still follows the appropriate elemental and isotopic mass balance. Statement III is correct because atoms and molecules represent different levels of chemical description. For example, oxygen atoms may occur in \(\mathrm{O_2}\) and \(\mathrm{O_3}\) molecules. The different molecular forms do not imply that oxygen atoms cease to belong to the element oxygen. Statement IV is correct because an ordinary chemical reaction rearranges constituent atoms into new chemical combinations. The formulas of the reactants and products may therefore differ while the elemental inventories remain conserved. Such rearrangement does not require transmutation of one element into another. Dalton's framework successfully connected atom conservation with fixed chemical combination, although some original statements required revision. The discovery of isotopes particularly affects the assumption of identical masses for all atoms of one element. It does not abolish the distinction between an element, its atoms and the molecules formed from those atoms. Thus, Statements I, III and IV are correct.

Explanation: \(\textbf{Establishing a common mass reference:}\) The law of multiple proportions compares the masses of one element combining with a fixed mass of another element. The original samples contain unequal amounts of X. Therefore, their Y masses cannot be compared directly. \(\textbf{Normalising Compound P:}\) For P, \[ m_X:m_Y=6.0:4.0 \] The mass of Y corresponding to 18.0 g X is \[ m_{Y,P}=4.0\left(\frac{18.0}{6.0}\right) \] \[ m_{Y,P}=12.0\,\text{g} \] \(\textbf{Normalising Compound Q:}\) For Q, \[ m_X:m_Y=9.0:9.0 \] Using the same 18.0 g X reference, \[ m_{Y,Q}=9.0\left(\frac{18.0}{9.0}\right) \] \[ m_{Y,Q}=18.0\,\text{g} \] \(\textbf{Normalising Compound R:}\) For R, \[ m_X:m_Y=3.0:5.0 \] Therefore, \[ m_{Y,R}=5.0\left(\frac{18.0}{3.0}\right) \] \[ m_{Y,R}=30.0\,\text{g} \] \(\textbf{Constructing the combining-mass ratio:}\) The masses of Y for the same mass of X are \[ m_{Y,P}:m_{Y,Q}:m_{Y,R}=12:18:30 \] Dividing by the greatest common factor gives \[ 12:18:30=2:3:5 \] \(\textbf{Identifying the chemical law:}\) The three pure compounds contain the same pair of elements in different fixed combining proportions. The normalised masses of Y occur in a simple whole-number ratio. This is the characteristic comparison described by the law of multiple proportions. The law of definite proportions instead concerns the constant elemental mass ratio within one particular pure compound. Conservation of mass concerns the complete mass balance of a chemical transformation. Neither of those laws directly names the three-compound comparison made here. \(\textbf{Final answer:}\) \[ \boxed{m_{Y,P}:m_{Y,Q}:m_{Y,R}=2:3:5} \]

Explanation: The law of definite proportions concerns the fixed elemental mass composition of a particular pure compound. Sodium chloride has the formula \(\mathrm{NaCl}\), representing a fixed sodium-to-chlorine atomic ratio of \(1:1\). Dissolving sodium chloride in water does not change that formula into a new compound containing different amounts of sodium and chlorine. The two prepared solutions contain different relative amounts of sodium chloride and water. They are therefore mixtures with different overall compositions. A mixture is not required to maintain one fixed solute-to-solvent mass ratio. The sodium chloride component nevertheless retains its characteristic elemental composition. The mass percentage of sodium chloride in the whole solution must not be confused with the mass percentage of sodium within pure sodium chloride. Adding water changes the former but not the latter. The student's conclusion applies a law governing pure compounds to the variable composition of a mixture. The distinction remains valid whether the pure compound is examined before dissolution or as a dissolved solute. Different concentrations can therefore coexist with complete consistency of the law of definite proportions. Option D correctly identifies both the classification error and the invariant chemical composition.

Explanation: \(\textbf{Determining the average atomic mass in Sample P:}\) P contains 80% lighter-isotope atoms and 20% heavier-isotope atoms. \[ \overline{M}_P=0.80(24)+0.20(26) \] \[ \overline{M}_P=19.2+5.2 \] \[ \overline{M}_P=24.4\,\mathrm{g\,mol^{-1}} \] \(\textbf{Determining the amount of Z atoms in P:}\) \[ n_P=\frac{12.2}{24.4} \] \[ n_P=0.500\,\mathrm{mol} \] The heavier-isotope amount in P is \[ n_{26,P}=0.20(0.500) \] \[ n_{26,P}=0.100\,\mathrm{mol} \] \(\textbf{Determining the average atomic mass in Sample Q:}\) Q contains 20% lighter-isotope atoms and 80% heavier-isotope atoms. \[ \overline{M}_Q=0.20(24)+0.80(26) \] \[ \overline{M}_Q=4.8+20.8 \] \[ \overline{M}_Q=25.6\,\mathrm{g\,mol^{-1}} \] \(\textbf{Determining the amount of Z atoms in Q:}\) \[ n_Q=\frac{25.6}{25.6} \] \[ n_Q=1.000\,\mathrm{mol} \] Its heavier-isotope amount is \[ n_{26,Q}=0.80(1.000) \] \[ n_{26,Q}=0.800\,\mathrm{mol} \] \(\textbf{Combining atom amounts rather than mass percentages:}\) The total amount of Z atoms is \[ n_{\mathrm{total}}=0.500+1.000 \] \[ n_{\mathrm{total}}=1.500\,\mathrm{mol} \] The total amount of heavier-isotope atoms is \[ n_{26,\mathrm{total}}=0.100+0.800 \] \[ n_{26,\mathrm{total}}=0.900\,\mathrm{mol} \] Therefore, \[ \%\,^{26}\mathrm{Z}=\frac{0.900}{1.500}\times100 \] \[ \%\,^{26}\mathrm{Z}=60\% \] \(\textbf{Calculating the final average atomic mass:}\) The combined sample mass is \[ m_{\mathrm{total}}=12.2+25.6 \] \[ m_{\mathrm{total}}=37.8\,\text{g} \] Therefore, \[ \overline{M}_Z=\frac{37.8}{1.500} \] \[ \overline{M}_Z=25.2\,\mathrm{g\,mol^{-1}} \] The corresponding average atomic mass is 25.2 u. \(\textbf{Checking the isotope-weighted average directly:}\) The lighter-isotope atom fraction is 40%. Thus, \[ \overline{m}_Z=0.40(24)+0.60(26) \] \[ \overline{m}_Z=25.2\,\mathrm{u} \] Both calculations agree. The percentages in P and Q cannot simply be averaged because the two samples contain different numbers of atoms. The correct weighting follows their atomic amounts, not their sample masses alone. \(\textbf{Final answer:}\) \[ \boxed{\%\,^{26}\mathrm{Z}=60\%,\quad\overline{m}_Z=25.2\,\mathrm{u}} \]

Explanation: \(\textbf{Separating known and unknown isotope fractions:}\) Let \(x\) be the atom-number fraction of the 10 u isotope. Let \(y\) be the atom-number fraction of the 12 u isotope. The 11 u isotope contributes the known fraction 0.30. \(\textbf{Applying the complete fraction constraint:}\) The atom-number fractions of all three isotopes must sum to one. \[ x+0.30+y=1 \] Therefore, \[ x+y=0.70 \qquad (1) \] \(\textbf{Applying the measured average atomic mass:}\) The weighted-average relation is \[ \overline{m}=10x+11(0.30)+12y \] Substituting the measured average, \[ 11=10x+3.30+12y \] Thus, \[ 10x+12y=7.70 \qquad (2) \] \(\textbf{Solving the two isotope constraints:}\) From Equation (1), \[ x=0.70-y \] Substitute into Equation (2). \[ 10(0.70-y)+12y=7.70 \] \[ 7.00-10y+12y=7.70 \] \[ 2y=0.70 \] \[ y=0.35 \] Therefore, \[ x=0.70-0.35 \] \[ x=0.35 \] \(\textbf{Converting fractions into percentages:}\) The lighter isotope has atom-number percentage \[ 100x=35\% \] The heavier isotope has atom-number percentage \[ 100y=35\% \] \(\textbf{Understanding why the result is symmetric:}\) The measured average atomic mass is exactly 11 u. The other two isotope masses lie one unit below and one unit above that value. Equal atom-number fractions of those two isotopes consequently produce equal and opposite deviations from the average. The central 11 u isotope contributes no deviation from the average. Because its percentage is independently known, the remaining 70% must be divided equally between the extreme isotopes. The result satisfies both the average-mass relation and the complete atom-fraction condition. \(\textbf{Final answer:}\) \[ \boxed{\%\,^{10}\mathrm{X}=35\%,\quad\%\,^{12}\mathrm{X}=35\%} \]

Explanation: \(\textbf{Recovering the first compound's molar mass:}\) The formula-unit amount is \(0.100\,\mathrm{mol}\). Therefore, \[ M_{\mathrm{X_2Y_3}}=\frac{11.5}{0.100} \] \[ M_{\mathrm{X_2Y_3}}=115\,\mathrm{g\,mol^{-1}} \] \(\textbf{Recovering the second compound's molar mass:}\) Similarly, \[ M_{\mathrm{XY_2}}=\frac{6.0}{0.100} \] \[ M_{\mathrm{XY_2}}=60\,\mathrm{g\,mol^{-1}} \] \(\textbf{Constructing formula-mass constraints:}\) Let the atomic masses of X and Y be \(a\) u and \(b\) u, respectively. The first molecular or formula composition gives \[ 2a+3b=115 \qquad (1) \] The second gives \[ a+2b=60 \qquad (2) \] \(\textbf{Solving for the constituent atomic masses:}\) Multiplying Equation (2) by two gives \[ 2a+4b=120 \] Subtracting Equation (1) from this result, \[ (2a+4b)-(2a+3b)=120-115 \] \[ b=5 \] Substituting \(b=5\) into Equation (2), \[ a+2(5)=60 \] \[ a=50 \] \(\textbf{Verifying both independent mass measurements:}\) For \(\mathrm{X_2Y_3}\), the formula mass calculated from these atomic masses is \[ 2(50)+3(5)=115\,\mathrm{u} \] Therefore, a \(0.100\,\mathrm{mol}\) sample weighs \[ 0.100(115)=11.5\,\text{g} \] For \(\mathrm{XY_2}\), the corresponding formula mass is \[ 50+2(5)=60\,\mathrm{u} \] Its \(0.100\,\mathrm{mol}\) sample weighs \[ 0.100(60)=6.0\,\text{g} \] Both recovered masses agree with the independently supplied sample data. The atomic masses are positive, and the solution satisfies the two formula-composition equations simultaneously. \(\textbf{Final answer:}\) \[ \boxed{m_X=50\,\mathrm{u},\quad m_Y=5\,\mathrm{u}} \]

Explanation: \(\textbf{Counting X atoms from each molecular species:}\) Every P molecule contains two X atoms. Every Q molecule contains one X atom. Therefore, \[ n_{X,P}=2(0.20) \] \[ n_{X,P}=0.40\,\mathrm{mol} \] \[ n_{X,Q}=0.30\,\mathrm{mol} \] The combined X-atom amount is \[ n_X=0.40+0.30 \] \[ n_X=0.70\,\mathrm{mol} \] \(\textbf{Counting Y atoms independently:}\) Each P molecule contains one Y atom, while each Q molecule contains two. Thus, \[ n_{Y,P}=0.20\,\mathrm{mol} \] \[ n_{Y,Q}=2(0.30)=0.60\,\mathrm{mol} \] The combined Y-atom amount is \[ n_Y=0.20+0.60 \] \[ n_Y=0.80\,\mathrm{mol} \] \(\textbf{Determining the constituent-atom fraction:}\) \[ n_{\mathrm{atoms,total}}=0.70+0.80 \] \[ n_{\mathrm{atoms,total}}=1.50\,\mathrm{mol} \] Therefore, \[ f_X=\frac{0.70}{1.50} \] \[ f_X=\frac7{15} \] \(\textbf{Calculating the mass contributed by X:}\) The atomic molar mass of X is \(20\,\mathrm{g\,mol^{-1}}\). \[ m_X=0.70(20) \] \[ m_X=14.0\,\text{g} \] \(\textbf{Calculating the mass contributed by Y:}\) The atomic molar mass of Y is \(10\,\mathrm{g\,mol^{-1}}\). \[ m_Y=0.80(10) \] \[ m_Y=8.0\,\text{g} \] The mixture mass is \[ m_{\mathrm{total}}=14.0+8.0 \] \[ m_{\mathrm{total}}=22.0\,\text{g} \] \(\textbf{Determining X mass percentage:}\) \[ \%\mathrm{X}=\frac{14.0}{22.0}\times100 \] \[ \%\mathrm{X}\approx63.64\% \] \(\textbf{Checking the offered choices:}\) The X-atom fraction is \(\frac7{15}\), but the X mass percentage is approximately 63.64%. The calculated pair therefore combines an X-atom fraction of \(\frac7{15}\) with an X mass percentage of approximately 63.64%. The atom fraction and mass percentage are not interchangeable because X and Y have different atomic masses. The complete quantitative result is given below. \(\textbf{Final answer:}\) \[ \boxed{f_X=\frac7{15},\quad\%\mathrm{X}\approx63.64\%} \]