Some Basic Concepts in Chemistry MCQs with Answers – Part 2 (NEET Chemistry)

Explanation: \(\textbf{Establishing the atomic mass unit:}\) One unified atomic mass unit is one-twelfth of the mass of a carbon-12 atom. Therefore, \[ 1\,\mathrm{u}=\frac{2.0\times10^{-23}\,\mathrm{g}}{12} \] \[ 1\,\mathrm{u}=1.6667\times10^{-24}\,\mathrm{g} \] \(\textbf{Converting the X-atom mass:}\) The mass of one X atom is \[ m_X=5.0\times10^{-23}\,\mathrm{g} \] Its numerical mass in unified atomic mass units is \[ M_X=\frac{5.0\times10^{-23}}{1.6667\times10^{-24}} \] \[ M_X=30 \] Therefore, \[ m_X=30\,\mathrm{u} \] \(\textbf{Interpreting the molecular formula:}\) Each molecule of \(\mathrm{X_2O}\) contains two X atoms and one oxygen atom. Its molecular mass is obtained by adding the masses of all constituent atoms. \[ M_{\mathrm{X_2O}}=2M_X+M_O \] Substituting, \[ M_{\mathrm{X_2O}}=2(30)+16 \] \[ M_{\mathrm{X_2O}}=76\,\mathrm{u} \] \(\textbf{Independent mass check:}\) The mass of one \(\mathrm{X_2O}\) molecule expressed in grams is \[ m_{\mathrm{X_2O}}=76(1.6667\times10^{-24}) \] \[ m_{\mathrm{X_2O}}\approx1.267\times10^{-22}\,\mathrm{g} \] Alternatively, its two X atoms directly contribute \[ 2(5.0\times10^{-23})=1.0\times10^{-22}\,\mathrm{g} \] Adding the corresponding oxygen-atom mass gives the same molecular mass. The reference-standard conversion is independent of the chemical formula calculation, but both must remain consistent. \(\textbf{Final answer:}\) \[ \boxed{M_X=30\,\mathrm{u},\quad M_{\mathrm{X_2O}}=76\,\mathrm{u}} \]

Explanation: \(\textbf{Representing the unknown atomic masses:}\) Let the atomic masses of A and B be \(a\,\mathrm{u}\) and \(b\,\mathrm{u}\), respectively. The molecular mass of a compound equals the sum of the masses of all its constituent atoms. \(\textbf{Using the first molecular formula:}\) The compound \(\mathrm{A_3B_2}\) contains three A atoms and two B atoms. Its molecular mass is \(120\,\mathrm{u}\). Therefore, \[ 3a+2b=120 \qquad (1) \] \(\textbf{Using the second molecular formula:}\) The compound \(\mathrm{AB_3}\) contains one A atom and three B atoms. Its molecular mass is \(110\,\mathrm{u}\). Therefore, \[ a+3b=110 \qquad (2) \] \(\textbf{Solving the simultaneous mass constraints:}\) Multiplying Equation (2) by 3, \[ 3a+9b=330 \] Subtracting Equation (1), \[ 7b=330-120 \] \[ 7b=210 \] \[ b=30 \] Substituting into Equation (2), \[ a+3(30)=110 \] \[ a=20 \] Thus, the atomic masses are \[ M_A=20\,\mathrm{u},\qquad M_B=30\,\mathrm{u} \] \(\textbf{Calculating the requested molecular mass:}\) The third compound has formula \(\mathrm{A_2B}\). Therefore, \[ M_{\mathrm{A_2B}}=2(20)+30 \] \[ M_{\mathrm{A_2B}}=70\,\mathrm{u} \] \(\textbf{Independent verification:}\) Using the derived atomic masses in the first compound gives \[ 3(20)+2(30)=120\,\mathrm{u} \] The second compound gives \[ 20+3(30)=110\,\mathrm{u} \] Both reported molecular masses are reproduced. The same constituent atomic masses must be used for all three compounds, while their molecular formulae determine the different contributions. \(\textbf{Final answer:}\) \[ \boxed{70\,\mathrm{u}} \]

Explanation: The Assertion is correct because a water molecule contains two hydrogen atoms and one oxygen atom. Using approximate atomic masses H = 1 and O = 16, its molecular mass is \(2(1)+16=18\,\mathrm{u}\). This value describes the mass of an individual molecular entity on the unified atomic mass scale. The molar mass of water is approximately \(18\,\mathrm{g\,mol^{-1}}\). Molar mass describes the mass per mole of the specified substance rather than the mass of one molecule. The Reason is also correct because one mole contains exactly \(N_A=6.02214076\times10^{23}\) specified elementary entities. A single molecular mass expressed in u and its corresponding molar mass expressed in \(\mathrm{g\,mol^{-1}}\) have the same numerical value to the precision used in ordinary NEET calculations. This relationship follows from the scale linking atomic masses to macroscopic molar quantities. It does not mean that \(18\,\mathrm{u}\) and \(18\,\mathrm{g\,mol^{-1}}\) are physically identical units. The quantities have different meanings: one is a mass, while the other is mass divided by amount of substance. Consequently, the Reason correctly explains the numerical relationship stated in the Assertion. Both statements are true, and the explanation is valid.

Explanation: \(\textbf{Interpreting the ionic formula:}\) Aluminium sulphate contains aluminium cations and sulphate anions. One formula unit of \(\mathrm{Al_2(SO_4)_3}\) contains two aluminium ions and three sulphate ions. Under the stated assumption of complete dissociation, each formula unit produces five constituent ions. \(\textbf{Calculating the formula mass:}\) The formula contains two aluminium atoms, three sulphur atoms and twelve oxygen atoms. Therefore, \[ M_{\mathrm{Al_2(SO_4)_3}}=2(27)+3(32)+12(16) \] \[ M_{\mathrm{Al_2(SO_4)_3}}=54+96+192 \] \[ M_{\mathrm{Al_2(SO_4)_3}}=342\,\mathrm{u} \] The corresponding molar mass is \[ M=342\,\mathrm{g\,mol^{-1}} \] \(\textbf{Finding the amount of formula units:}\) The sample mass is 17.1 g. \[ n_{\text{formula units}}=\frac{m}{M} \] \[ n_{\text{formula units}}=\frac{17.1\,\mathrm{g}}{342\,\mathrm{g\,mol^{-1}}} \] \[ n_{\text{formula units}}=0.05\,\mathrm{mol} \] \(\textbf{Converting formula units into ion amounts:}\) The complete dissociation relationship is \[ \mathrm{Al_2(SO_4)_3\rightarrow2Al^{3+}+3SO_4^{2-}} \] Each mole of aluminium sulphate formula units yields two moles of aluminium ions and three moles of sulphate ions. Therefore, \[ n_{\mathrm{Al^{3+}}}=2(0.05)=0.10\,\mathrm{mol} \] \[ n_{\mathrm{SO_4^{2-}}}=3(0.05)=0.15\,\mathrm{mol} \] The total amount of constituent ions is \[ n_{\text{total ions}}=0.10+0.15 \] \[ n_{\text{total ions}}=0.25\,\mathrm{mol} \] \(\textbf{Charge-consistency check:}\) The total positive charge amount is proportional to \[ 3(0.10)=0.30 \] The total negative charge amount is proportional to \[ 2(0.15)=0.30 \] The positive and negative charges balance. The amount of formula units is smaller than the combined amount of constituent ions because each formula unit produces five ions under complete dissociation. \(\textbf{Final answer:}\) \[ \boxed{0.25\,\mathrm{mol}} \]

Explanation: \(\textbf{Determining the isotope abundances:}\) The two isotopes of M are present in the number ratio \(3:1\). The lighter isotope therefore contributes three-fourths of the M atoms. The heavier isotope contributes one-fourth. Thus, \[ f_{24}=\frac{3}{4},\qquad f_{26}=\frac{1}{4} \] \(\textbf{Calculating the average atomic mass of M:}\) The weighted-average relation is \[ \overline{M}_M=f_{24}(24)+f_{26}(26) \] Substituting, \[ \overline{M}_M=\frac{3}{4}(24)+\frac{1}{4}(26) \] \[ \overline{M}_M=18+6.5 \] \[ \overline{M}_M=24.5\,\mathrm{u} \] \(\textbf{Using the oxide formula:}\) The oxide is represented by MO. Each formula unit contains one M atom and one oxygen atom in the specified composition. The average formula mass is \[ M_{\mathrm{MO}}=24.5+16 \] \[ M_{\mathrm{MO}}=40.5\,\mathrm{u} \] The corresponding average molar mass is \[ M_{\mathrm{MO}}=40.5\,\mathrm{g\,mol^{-1}} \] \(\textbf{Calculating the amount of MO:}\) The sample mass is 81 g. Amount of substance is calculated using \[ n=\frac{m}{M} \] Substituting, \[ n_{\mathrm{MO}}=\frac{81\,\mathrm{g}}{40.5\,\mathrm{g\,mol^{-1}}} \] \[ n_{\mathrm{MO}}=2.000\,\mathrm{mol} \] \(\textbf{Distinguishing isotopic composition from formula composition:}\) Changing the isotope abundances of M changes its average atomic mass. This consequently changes the average formula mass and molar mass of MO. However, the oxide formula remains MO because the isotopes belong to the same element and its constituent-ion ratio is unchanged. The sample amount must be calculated using the appropriate average molar mass for the specified isotopic mixture. Using 24 or 26 as though every M atom had the same isotopic mass would produce an incorrect result. \(\textbf{Final answer:}\) \[ \boxed{M_{\mathrm{MO}}=40.5\,\mathrm{g\,mol^{-1}},\quad n_{\mathrm{MO}}=2.000\,\mathrm{mol}} \]

Explanation: \(\textbf{Distinguishing atomic abundance from mass contribution:}\) The average atomic mass is calculated using the relative numbers of atoms belonging to each isotope. However, the question asks for the percentage of the total sample mass contributed by the heavier isotope. These two percentages are not generally identical because the isotopes have different individual masses. \(\textbf{Recovering the heavier-isotope abundance:}\) Let \(f\) be the fraction of atoms having mass \(37\,\mathrm{u}\). The remaining fraction, \(1-f\), consists of atoms having mass \(35\,\mathrm{u}\). The weighted-average equation is \[ \overline{M}=35(1-f)+37f \] Substituting the measured average, \[ 35.5=35(1-f)+37f \] \[ 35.5=35-35f+37f \] \[ 35.5=35+2f \] Therefore, \[ f=\frac{0.5}{2} \] \[ f=0.25 \] Thus, the heavier isotope constitutes 25% of the atoms. \(\textbf{Calculating the mass contributed by each isotope:}\) Consider a collection of 100 atoms. It contains 75 lighter atoms and 25 heavier atoms. The total mass contributed by the lighter isotope is \[ m_{35}=75(35)=2625\,\mathrm{u} \] The total mass contributed by the heavier isotope is \[ m_{37}=25(37)=925\,\mathrm{u} \] The combined mass is \[ m_{\text{total}}=2625+925 \] \[ m_{\text{total}}=3550\,\mathrm{u} \] \(\textbf{Finding the required mass percentage:}\) \[ \%\text{mass from heavier isotope}=\frac{925}{3550}\times100 \] \[ \%\text{mass from heavier isotope}=26.0563\% \] \[ \%\text{mass from heavier isotope}\approx26.06\% \] \(\textbf{Consistency check:}\) The heavier isotope contributes 25% of the atoms but slightly more than 25% of the total mass. This is physically reasonable because each heavier isotope atom has a greater mass. Using the atomic percentage directly as the mass percentage would ignore this difference. \(\textbf{Final answer:}\) \[ \boxed{26.06\%} \]

Explanation: \(\textbf{Finding the original isotopic composition:}\) Let the fractional abundance of the heavier isotope be \(f\). The lighter isotope then has fractional abundance \(1-f\). The average atomic mass is \[ \overline{M}=20(1-f)+22f \] Substituting, \[ 20.5=20(1-f)+22f \] \[ 20.5=20+2f \] Therefore, \[ f=\frac{0.5}{2} \] \[ f=0.25 \] The original isotope fractions are therefore \[ f_{20}=0.75,\qquad f_{22}=0.25 \] \(\textbf{Choosing a convenient atomic population:}\) Consider an original collection of 400 atoms. The lighter-isotope count is \[ N_{20}=400(0.75)=300 \] The heavier-isotope count is \[ N_{22}=400(0.25)=100 \] \(\textbf{Applying selective removal:}\) Exactly 40% of the lighter atoms are removed. The number removed is \[ N_{\text{removed}}=300\times\frac{40}{100} \] \[ N_{\text{removed}}=120 \] The remaining lighter-isotope count is \[ N_{20,\text{remaining}}=300-120 \] \[ N_{20,\text{remaining}}=180 \] No heavier-isotope atoms are removed. Thus, \[ N_{22,\text{remaining}}=100 \] \(\textbf{Calculating the changed average atomic mass:}\) The total remaining number of atoms is \[ N_{\text{remaining}}=180+100=280 \] The combined mass of these atoms is \[ m_{\text{remaining}}=180(20)+100(22) \] \[ m_{\text{remaining}}=3600+2200 \] \[ m_{\text{remaining}}=5800\,\mathrm{u} \] Therefore, \[ \overline{M}_{\text{remaining}}=\frac{5800}{280} \] \[ \overline{M}_{\text{remaining}}=\frac{145}{7}\,\mathrm{u} \] \(\textbf{Physical interpretation:}\) The average increases because removal selectively reduces the population of lighter atoms. The individual isotope masses remain unchanged. The new average remains between \(20\,\mathrm{u}\) and \(22\,\mathrm{u}\), as required. The original isotope percentages cannot be reused after selective removal because the population proportions have changed. \(\textbf{Final answer:}\) \[ \boxed{\frac{145}{7}\,\mathrm{u}} \]

Explanation: \(\textbf{Interpreting the hydrated formula:}\) The formula \(\mathrm{MgSO_4\cdot7H_2O}\) represents one magnesium sulphate formula unit associated with seven water molecules in the crystalline hydrate. These seven water molecules contribute to the formula mass of the hydrated substance. They do not represent seven additional magnesium sulphate formula units. \(\textbf{Molar mass of anhydrous magnesium sulphate:}\) \[ M_{\mathrm{MgSO_4}}=24+32+4(16) \] \[ M_{\mathrm{MgSO_4}}=120\,\mathrm{g\,mol^{-1}} \] \(\textbf{Molar mass of the hydrated compound:}\) The molar mass of water is \[ M_{\mathrm{H_2O}}=2(1)+16 \] \[ M_{\mathrm{H_2O}}=18\,\mathrm{g\,mol^{-1}} \] Therefore, \[ M_{\mathrm{MgSO_4\cdot7H_2O}}=120+7(18) \] \[ M_{\mathrm{MgSO_4\cdot7H_2O}}=120+126 \] \[ M_{\mathrm{MgSO_4\cdot7H_2O}}=246\,\mathrm{g\,mol^{-1}} \] \(\textbf{Amount of hydrated formula units:}\) \[ n_{\text{hydrate}}=\frac{24.6\,\mathrm{g}}{246\,\mathrm{g\,mol^{-1}}} \] \[ n_{\text{hydrate}}=0.10\,\mathrm{mol} \] Thus, the hydrated sample contains \(0.10\,\mathrm{mol}\) of magnesium sulphate formula units. \(\textbf{Amount of anhydrous formula units:}\) \[ n_{\mathrm{MgSO_4}}=\frac{12.0\,\mathrm{g}}{120\,\mathrm{g\,mol^{-1}}} \] \[ n_{\mathrm{MgSO_4}}=0.10\,\mathrm{mol} \] Hence, the two samples contain equal amounts of magnesium sulphate formula units. \(\textbf{Counting associated water molecules:}\) Each hydrated formula unit contains seven water molecules. Therefore, \[ n_{\mathrm{water}}=7(0.10) \] \[ n_{\mathrm{water}}=0.70\,\mathrm{mol} \] \(\textbf{Mass consistency check:}\) The water of crystallisation contributes \[ m_{\mathrm{water}}=0.70(18)=12.6\,\text{g} \] The magnesium sulphate portion contributes 12.0 g. \[ 12.0+12.6=24.6\,\text{g} \] This reproduces the hydrated sample mass. The larger mass of the hydrated sample does not mean that it contains more magnesium sulphate formula units. \(\textbf{Final answer:}\) \[ \boxed{n_{\text{hydrate}}:n_{\text{anhydrous}}=1:1,\quad n_{\mathrm{water}}=0.70\,\mathrm{mol}} \]

Explanation: \(\textbf{Interpreting the chemical formula:}\) The compound ammonium sulphate has the formula \(\mathrm{(NH_4)_2SO_4}\). One formula unit corresponds to two ammonium ions, \(\mathrm{NH_4^+}\), and one sulphate ion, \(\mathrm{SO_4^{2-}}\). The coefficient two indicates two formula units rather than a change in the composition of one formula unit. \(\textbf{Evaluating Statement I:}\) One formula unit contains: Two nitrogen atoms, eight hydrogen atoms, one sulphur atom and four oxygen atoms. Therefore, \[ N_{\text{atoms per formula unit}}=2+8+1+4 \] \[ N_{\text{atoms per formula unit}}=15 \] Statement I is correct. \(\textbf{Evaluating Statement II:}\) The formula mass of one ammonium sulphate formula unit is \[ M=2(14)+8(1)+32+4(16) \] \[ M=28+8+32+64 \] \[ M=132\,\mathrm{u} \] The value \(264\,\mathrm{u}\) does not represent the formula mass of one formula unit. Statement II is incorrect. \(\textbf{Evaluating Statement III:}\) One formula unit corresponds to two ammonium ions and one sulphate ion. Therefore, the number of constituent ions represented per formula unit is \[ N_{\text{ions per formula unit}}=2+1=3 \] Two formula units correspond to \[ N_{\text{ions}}=2(3)=6 \] There are six constituent ions, not ten. Polyatomic ions must be counted as individual ionic entities rather than by their constituent atom counts. Statement III is incorrect. \(\textbf{Evaluating Statement IV:}\) The mass of two formula units is twice the mass of one formula unit. \[ m_{\text{two formula units}}=2(132) \] \[ m_{\text{two formula units}}=264\,\mathrm{u} \] Statement IV is correct. \(\textbf{Final conclusion:}\) Statements I and IV are correct, while Statements II and III are incorrect. The distinction between formula mass, total mass, constituent atoms and constituent ions determines the correct combination.

Explanation: \(\textbf{Representing the unknown atomic molar mass:}\) Let the atomic molar mass of A be \(a\,\mathrm{g\,mol^{-1}}\). The molar mass of a compound is obtained by adding the molar-mass contributions of all atoms in its formula. \(\textbf{Molar mass of the first compound:}\) The formula \(\mathrm{A_2O}\) contains two A atoms and one oxygen atom per formula entity. Therefore, \[ M_{\mathrm{A_2O}}=2a+16 \] \(\textbf{Molar mass of the second compound:}\) The formula \(\mathrm{AO_2}\) contains one A atom and two oxygen atoms. Hence, \[ M_{\mathrm{AO_2}}=a+2(16) \] \[ M_{\mathrm{AO_2}}=a+32 \] \(\textbf{Using the given mass difference:}\) The first compound is heavier by \(8\,\mathrm{g\,mol^{-1}}\). Therefore, \[ M_{\mathrm{A_2O}}-M_{\mathrm{AO_2}}=8 \] Substituting, \[ (2a+16)-(a+32)=8 \] \[ 2a+16-a-32=8 \] \[ a-16=8 \] \[ a=24 \] Thus, \[ M_A=24\,\mathrm{g\,mol^{-1}} \] \(\textbf{Calculating the requested compound molar mass:}\) The formula \(\mathrm{A_3O_2}\) contains three A atoms and two oxygen atoms. \[ M_{\mathrm{A_3O_2}}=3(24)+2(16) \] \[ M_{\mathrm{A_3O_2}}=72+32 \] \[ M_{\mathrm{A_3O_2}}=104\,\mathrm{g\,mol^{-1}} \] \(\textbf{Independent consistency check:}\) The original compounds would have molar masses \[ M_{\mathrm{A_2O}}=2(24)+16=64\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{AO_2}}=24+32=56\,\mathrm{g\,mol^{-1}} \] Their difference is \[ 64-56=8\,\mathrm{g\,mol^{-1}} \] This agrees with the supplied information. The calculation shows how a difference between two compound molar masses can reveal an unknown atomic molar mass even when neither original molar mass is individually supplied. \(\textbf{Final answer:}\) \[ \boxed{M_A=24\,\mathrm{g\,mol^{-1}},\quad M_{\mathrm{A_3O_2}}=104\,\mathrm{g\,mol^{-1}}} \]

Explanation: \(\textbf{Comparing formula masses:}\) For calcium carbonate, \[ M_{\mathrm{CaCO_3}}=40+12+3(16) \] \[ M_{\mathrm{CaCO_3}}=100\,\mathrm{u} \] For potassium hydrogen carbonate, \[ M_{\mathrm{KHCO_3}}=39+1+12+3(16) \] \[ M_{\mathrm{KHCO_3}}=100\,\mathrm{u} \] Thus, the two compounds have equal formula masses. \(\textbf{Interpreting equal sample masses:}\) Let each sample weigh \(m\) grams. Both compounds have molar masses of \(100\,\mathrm{g\,mol^{-1}}\). The amount of formula units in either sample is \[ n=\frac{m}{100\,\mathrm{g\,mol^{-1}}} \] Therefore, equal masses contain equal amounts and equal numbers of formula units. \(\textbf{Counting atoms per calcium carbonate formula unit:}\) One \(\mathrm{CaCO_3}\) formula unit contains one calcium atom, one carbon atom and three oxygen atoms. \[ N_{\text{atoms per CaCO}_3}=1+1+3 \] \[ N_{\text{atoms per CaCO}_3}=5 \] \(\textbf{Counting atoms per potassium hydrogen carbonate formula unit:}\) One \(\mathrm{KHCO_3}\) formula unit contains one potassium atom, one hydrogen atom, one carbon atom and three oxygen atoms. \[ N_{\text{atoms per KHCO}_3}=1+1+1+3 \] \[ N_{\text{atoms per KHCO}_3}=6 \] \(\textbf{Comparing total constituent-atom counts:}\) Let each sample contain \(N\) formula units. Then, \[ N_{\text{atoms,CaCO}_3}=5N \] \[ N_{\text{atoms,KHCO}_3}=6N \] Therefore, \[ N_{\text{atoms,CaCO}_3}:N_{\text{atoms,KHCO}_3}=5:6 \] \(\textbf{Chemical interpretation:}\) Equal formula masses do not imply identical chemical formulae or equal numbers of atoms per formula unit. Both samples contain equal numbers of formula units because their masses and molar masses are equal. However, their constituent-atom counts differ because the individual formula units contain different numbers of atoms. The distinction between formula-unit count and total atom count is essential. \(\textbf{Final answer:}\) Equal masses contain equal numbers of formula units, with total constituent-atom counts in the ratio \(5:6\).

Explanation: \(\textbf{Converting molecular count into amount:}\) The sample contains \[ N=1.505\times10^{23}\;\text{molecules} \] The amount of a molecular substance is \[ n=\frac{N}{N_A} \] Substituting, \[ n=\frac{1.505\times10^{23}}{6.02\times10^{23}\,\mathrm{mol^{-1}}} \] \[ n=0.25\,\mathrm{mol} \] \(\textbf{Inferring the compound's molar mass:}\) The sample mass is 18.0 g. Using \[ M=\frac{m}{n} \] we obtain \[ M_{\mathrm{X_2Y_3}}=\frac{18.0\,\mathrm{g}}{0.25\,\mathrm{mol}} \] \[ M_{\mathrm{X_2Y_3}}=72\,\mathrm{g\,mol^{-1}} \] The corresponding numerical molecular mass is \(72\,\mathrm{u}\). \(\textbf{Using the molecular formula:}\) One molecule of \(\mathrm{X_2Y_3}\) contains two X atoms and three Y atoms. Let the atomic mass of Y be \(y\,\mathrm{u}\). The molecular-mass relation is \[ M_{\mathrm{X_2Y_3}}=2M_X+3M_Y \] Substituting the known values, \[ 72=2(12)+3y \] \[ 72=24+3y \] \[ 3y=48 \] \[ y=16 \] Thus, the atomic mass of Y is \[ M_Y=16\,\mathrm{u} \] \(\textbf{Independent sample-mass check:}\) The calculated molecular mass is \[ 2(12)+3(16)=72\,\mathrm{u} \] A quarter mole of this compound must weigh \[ m=0.25(72)=18.0\,\text{g} \] This reproduces the measured sample mass. The correct sequence is to interpret the molecular count, determine molar mass and then use the molecular formula to recover the unknown atomic mass. \(\textbf{Final answer:}\) \[ \boxed{16\,\mathrm{u}} \]

Explanation: \(\textbf{Interpreting the mass change:}\) The original compound is a hydrate containing a fixed number of water molecules associated with each copper sulphate formula unit. Heating removes the water of crystallisation. The question specifies that the remaining material is pure anhydrous copper sulphate. Therefore, the decrease in sample mass equals the mass of water removed. \(\textbf{Determining the water mass:}\) \[ m_{\mathrm{water}}=m_{\text{hydrated}}-m_{\text{anhydrous}} \] \[ m_{\mathrm{water}}=25.0-16.0 \] \[ m_{\mathrm{water}}=9.0\,\text{g} \] \(\textbf{Finding the anhydrous molar mass:}\) \[ M_{\mathrm{CuSO_4}}=64+32+4(16) \] \[ M_{\mathrm{CuSO_4}}=160\,\mathrm{g\,mol^{-1}} \] \(\textbf{Amount of copper sulphate formula units:}\) \[ n_{\mathrm{CuSO_4}}=\frac{16.0\,\mathrm{g}}{160\,\mathrm{g\,mol^{-1}}} \] \[ n_{\mathrm{CuSO_4}}=0.10\,\mathrm{mol} \] \(\textbf{Amount of water removed:}\) The molar mass of water is \[ M_{\mathrm{H_2O}}=18\,\mathrm{g\,mol^{-1}} \] Thus, \[ n_{\mathrm{H_2O}}=\frac{9.0\,\mathrm{g}}{18\,\mathrm{g\,mol^{-1}}} \] \[ n_{\mathrm{H_2O}}=0.50\,\mathrm{mol} \] \(\textbf{Finding the number of water molecules per formula unit:}\) In \(\mathrm{CuSO_4\cdot xH_2O}\), the ratio of water molecules to copper sulphate formula units is \(x:1\). Therefore, \[ x=\frac{n_{\mathrm{H_2O}}}{n_{\mathrm{CuSO_4}}} \] \[ x=\frac{0.50}{0.10} \] \[ x=5 \] \(\textbf{Formula-mass verification:}\) The hydrated formula is \[ \mathrm{CuSO_4\cdot5H_2O} \] Its molar mass is \[ M_{\text{hydrate}}=160+5(18) \] \[ M_{\text{hydrate}}=250\,\mathrm{g\,mol^{-1}} \] A \(0.10\,\mathrm{mol}\) sample therefore weighs \[ m=0.10(250)=25.0\,\text{g} \] This agrees with the original sample mass. The hydrate coefficient counts water molecules per copper sulphate formula unit and must be obtained from an amount ratio, not directly from the ratio of their masses. \(\textbf{Final answer:}\) \[ \boxed{x=5} \]

Explanation: \(\textbf{Identifying the two molecular masses:}\) Carbon monoxide contains one carbon atom and one oxygen atom. \[ M_{\mathrm{CO}}=12+16 \] \[ M_{\mathrm{CO}}=28\,\mathrm{g\,mol^{-1}} \] Carbon dioxide contains one carbon atom and two oxygen atoms. \[ M_{\mathrm{CO_2}}=12+2(16) \] \[ M_{\mathrm{CO_2}}=44\,\mathrm{g\,mol^{-1}} \] \(\textbf{Representing the unknown amounts:}\) Let the amount of carbon monoxide be \(x\,\mathrm{mol}\). Let the amount of carbon dioxide be \(y\,\mathrm{mol}\). The total amount of gas molecules is \(0.50\,\mathrm{mol}\). Therefore, \[ x+y=0.50 \qquad (1) \] \(\textbf{Using the measured total mass:}\) The mass of each component equals its amount multiplied by its molar mass. Thus, \[ 28x+44y=18.0 \qquad (2) \] \(\textbf{Solving for the component amounts:}\) From Equation (1), \[ x=0.50-y \] Substituting into Equation (2), \[ 28(0.50-y)+44y=18.0 \] \[ 14-28y+44y=18 \] \[ 16y=4 \] \[ y=0.25\,\mathrm{mol} \] Consequently, \[ x=0.50-0.25 \] \[ x=0.25\,\mathrm{mol} \] \(\textbf{Calculating carbon dioxide mole fraction:}\) \[ X_{\mathrm{CO_2}}=\frac{n_{\mathrm{CO_2}}}{n_{\mathrm{total}}} \] \[ X_{\mathrm{CO_2}}=\frac{0.25}{0.50} \] \[ X_{\mathrm{CO_2}}=0.50 \] \(\textbf{Counting oxygen atoms:}\) Each CO molecule contains one oxygen atom. Each \(\mathrm{CO_2}\) molecule contains two oxygen atoms. Therefore, \[ n_{\mathrm{O\ atoms}}=n_{\mathrm{CO}}+2n_{\mathrm{CO_2}} \] \[ n_{\mathrm{O\ atoms}}=0.25+2(0.25) \] \[ n_{\mathrm{O\ atoms}}=0.75\,\mathrm{mol} \] \(\textbf{Independent mass check:}\) The calculated component masses are \[ m_{\mathrm{CO}}=0.25(28)=7.0\,\text{g} \] \[ m_{\mathrm{CO_2}}=0.25(44)=11.0\,\text{g} \] Their sum is 18.0 g, matching the measured total. The average molar mass of the mixture is \(18.0/0.50=36\,\mathrm{g\,mol^{-1}}\), which lies between the two component molar masses. \(\textbf{Final answer:}\) \[ \boxed{X_{\mathrm{CO_2}}=0.50,\quad n_{\mathrm{O\ atoms}}=0.750\,\mathrm{mol}} \]

Explanation: \(\textbf{Interpreting the equal-hydrogen condition:}\) Methane contains four hydrogen atoms per molecule. Ammonia contains three hydrogen atoms per molecule. Water contains two hydrogen atoms per molecule. Equal total numbers of hydrogen atoms therefore require different numbers of molecules of the three compounds. \(\textbf{Establishing a common hydrogen-atom amount:}\) The least common multiple of 4, 3 and 2 is 12. Consider samples each containing 12 mol of hydrogen atoms. For methane, \[ n_{\mathrm{CH_4}}=\frac{12}{4} \] \[ n_{\mathrm{CH_4}}=3\,\mathrm{mol} \] For ammonia, \[ n_{\mathrm{NH_3}}=\frac{12}{3} \] \[ n_{\mathrm{NH_3}}=4\,\mathrm{mol} \] For water vapour, \[ n_{\mathrm{H_2O}}=\frac{12}{2} \] \[ n_{\mathrm{H_2O}}=6\,\mathrm{mol} \] Thus, \[ n_{\mathrm{CH_4}}:n_{\mathrm{NH_3}}:n_{\mathrm{H_2O}}=3:4:6 \] \(\textbf{Calculating the molar masses:}\) \[ M_{\mathrm{CH_4}}=12+4(1)=16\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{NH_3}}=14+3(1)=17\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{H_2O}}=2(1)+16=18\,\mathrm{g\,mol^{-1}} \] \(\textbf{Determining the sample masses:}\) For methane, \[ m_P=3(16)=48\,\text{g} \] For ammonia, \[ m_Q=4(17)=68\,\text{g} \] For water vapour, \[ m_R=6(18)=108\,\text{g} \] Therefore, \[ m_P:m_Q:m_R=48:68:108 \] Dividing throughout by 4, \[ m_P:m_Q:m_R=12:17:27 \] \(\textbf{Verification through hydrogen-atom counts:}\) The respective hydrogen-atom amounts are \[ 3(4)=12\,\mathrm{mol} \] \[ 4(3)=12\,\mathrm{mol} \] \[ 6(2)=12\,\mathrm{mol} \] All three are equal, as required. Equal numbers of a particular constituent atom do not imply equal molecular counts or equal sample masses. Both the molecular hydrogen content and molar mass must be considered. \(\textbf{Final answer:}\) \[ \boxed{n_P:n_Q:n_R=3:4:6,\quad m_P:m_Q:m_R=12:17:27} \]

Explanation: Statement I is correct because relative atomic mass is defined by comparing an atomic mass with the atomic mass standard. Both quantities in the defining comparison have dimensions of mass. Their ratio is therefore dimensionless. Statement II is correct because molar mass represents mass per unit amount of substance. The SI coherent unit is \(\mathrm{kg\,mol^{-1}}\), while \(\mathrm{g\,mol^{-1}}\) is commonly used in Chemistry calculations. For example, \(18\,\mathrm{g\,mol^{-1}}\) is equivalent to \(0.018\,\mathrm{kg\,mol^{-1}}\). Changing between these units changes the numerical value but not the physical molar mass. Statement III is correct because atomic and molecular masses expressed in u are numerically related to corresponding molar masses expressed in \(\mathrm{g\,mol^{-1}}\). For example, a water molecule has a molecular mass of approximately \(18\,\mathrm{u}\), while water has a molar mass of approximately \(18\,\mathrm{g\,mol^{-1}}\). The equality concerns numerical values expressed in different appropriate units, not equality of the physical quantities themselves. Statement IV is incorrect because the formula mass of NaCl expressed in u is the mass associated with one formula unit on the atomic mass scale. The mass of a single formula unit in grams is extremely small and requires conversion from u to grams. The formula mass in u must not be assigned directly as a gram mass for one microscopic entity. One mole of NaCl formula units has a mass in grams numerically corresponding to the formula mass in u. The correct statements are I, II and III.

Explanation: \(\textbf{Establishing the isotope-average relation:}\) The element has two isotopes of masses \(50\,\mathrm{u}\) and \(52\,\mathrm{u}\). Let \(f\) be the fractional abundance of the heavier isotope. The lighter isotope then has abundance \(1-f\). The average atomic mass is \[ \overline{M}=50(1-f)+52f \] Simplifying, \[ \overline{M}=50+2f \] The abundance \(f\) must be expressed as a fraction, not as a whole-number percentage. \(\textbf{Checking sample P:}\) The heavier-isotope fraction is 0.20. \[ \overline{M}_P=50+2(0.20) \] \[ \overline{M}_P=50.4\,\mathrm{u} \] This agrees with the reported value. \(\textbf{Checking sample Q:}\) The heavier-isotope fraction is 0.60. \[ \overline{M}_Q=50+2(0.60) \] \[ \overline{M}_Q=51.2\,\mathrm{u} \] This is also consistent. \(\textbf{Checking sample R:}\) The reported heavier-isotope abundance is 70%. Thus, \[ \overline{M}_{R,\text{calculated}}=50+2(0.70) \] \[ \overline{M}_{R,\text{calculated}}=51.4\,\mathrm{u} \] However, the table reports \(51.6\,\mathrm{u}\). The reported average would instead require \[ f=\frac{51.6-50}{2} \] \[ f=0.80 \] Thus, an average of \(51.6\,\mathrm{u}\) corresponds to 80% heavier isotope, not 70%. \(\textbf{Checking sample S:}\) \[ \overline{M}_S=50+2(0.40) \] \[ \overline{M}_S=50.8\,\mathrm{u} \] This agrees with the table. \(\textbf{Consistency conclusion:}\) Samples P, Q and S satisfy the weighted-average relationship. Sample R contains incompatible average-mass and isotope-abundance information. The discrepancy cannot be explained by changing the total sample size because average atomic mass depends on relative atomic populations rather than absolute sample mass. \(\textbf{Final conclusion:}\) Sample R has internally inconsistent data.

Explanation: \(\textbf{Interpreting the atomic constraint:}\) Each methane molecule contains four hydrogen atoms. Each ammonia molecule contains three hydrogen atoms. Therefore, equal total hydrogen-atom counts require different numbers of methane and ammonia molecules. \(\textbf{Establishing the molecular-number ratio:}\) Let the numbers of methane and ammonia molecules be \(N_M\) and \(N_A\). The equality of hydrogen atoms gives \[ 4N_M=3N_A \] Therefore, \[ \frac{N_M}{N_A}=\frac{3}{4} \] Thus, \[ N_M:N_A=3:4 \] \(\textbf{Calculating the molecular masses:}\) For methane, \[ M_{\mathrm{CH_4}}=12+4(1) \] \[ M_{\mathrm{CH_4}}=16\,\mathrm{u} \] For ammonia, \[ M_{\mathrm{NH_3}}=14+3(1) \] \[ M_{\mathrm{NH_3}}=17\,\mathrm{u} \] \(\textbf{Converting molecular counts into sample masses:}\) Sample mass is proportional to the number of molecules multiplied by the mass per molecule. Therefore, \[ m_{\mathrm{CH_4}}:m_{\mathrm{NH_3}}=3(16):4(17) \] \[ m_{\mathrm{CH_4}}:m_{\mathrm{NH_3}}=48:68 \] Dividing by 4, \[ m_{\mathrm{CH_4}}:m_{\mathrm{NH_3}}=12:17 \] \(\textbf{Independent hydrogen-atom check:}\) Three methane molecules contain \[ 3(4)=12\;\text{hydrogen atoms} \] Four ammonia molecules contain \[ 4(3)=12\;\text{hydrogen atoms} \] Thus, the chosen molecular-number ratio satisfies the required equality. \(\textbf{Chemical interpretation:}\) Equal hydrogen-atom counts do not require equal molecular counts. Even after the molecular counts are adjusted, the sample masses differ because methane and ammonia have different molecular masses. The calculation combines molecular composition with molecular mass rather than assuming equal mass from equal constituent-atom counts. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{CH_4}}:m_{\mathrm{NH_3}}=12:17,\quad N_{\mathrm{CH_4}}:N_{\mathrm{NH_3}}=3:4} \]

Explanation: \(\textbf{Interpreting the constituent ions:}\) One formula unit of \(\mathrm{Al_2(SO_4)_3}\) contains two aluminium ions and three sulphate ions. Therefore, it produces five constituent ions under complete dissociation. One formula unit of \(\mathrm{Ca_3(PO_4)_2}\) contains three calcium ions and two phosphate ions. It also produces five constituent ions under complete dissociation. \(\textbf{Using the equal-total-ion condition:}\) Let the amounts of the two original compounds be \(n_1\) and \(n_2\). The total constituent-ion amounts are \[ n_{\text{ions},1}=5n_1 \] \[ n_{\text{ions},2}=5n_2 \] The question states that these are equal. Therefore, \[ 5n_1=5n_2 \] \[ n_1=n_2 \] Thus, equal total constituent-ion amounts require equal amounts of the two formula-unit types in this particular comparison. \(\textbf{Molar mass of aluminium sulphate:}\) \[ M_1=2(27)+3(32)+12(16) \] \[ M_1=54+96+192 \] \[ M_1=342\,\mathrm{g\,mol^{-1}} \] \(\textbf{Molar mass of calcium phosphate:}\) \[ M_2=3(40)+2(31)+8(16) \] \[ M_2=120+62+128 \] \[ M_2=310\,\mathrm{g\,mol^{-1}} \] \(\textbf{Comparing original sample masses:}\) Because \(n_1=n_2\), \[ m_1:m_2=n_1M_1:n_2M_2 \] \[ m_1:m_2=342:310 \] \[ m_1:m_2=171:155 \] \(\textbf{Comparing the polyatomic anion amounts:}\) Each aluminium sulphate formula unit yields three sulphate ions. Each calcium phosphate formula unit yields two phosphate ions. Therefore, \[ n_{\mathrm{SO_4^{2-}}}:n_{\mathrm{PO_4^{3-}}}=3n_1:2n_2 \] Since \(n_1=n_2\), \[ n_{\mathrm{SO_4^{2-}}}:n_{\mathrm{PO_4^{3-}}}=3:2 \] \(\textbf{Charge-consistency check:}\) For aluminium sulphate, two trivalent cations balance three divalent anions. For calcium phosphate, three divalent cations balance two trivalent anions. Both dissociation representations conserve charge. The equal total ion amounts do not require equal anion amounts because the two formula units contain different numbers of their respective anions. \(\textbf{Final answer:}\) \[ \boxed{m_1:m_2=171:155,\quad n_{\mathrm{SO_4^{2-}}}:n_{\mathrm{PO_4^{3-}}}=3:2} \]

Explanation: \(\textbf{Interpreting the measured molar mass:}\) The compound has a molar mass of \(44\,\mathrm{g\,mol^{-1}}\). Its numerical molecular mass is therefore approximately \(44\,\mathrm{u}\). However, molecular mass alone does not uniquely identify a molecular compound. Different combinations of constituent elements and atomic subscripts can produce the same total molecular mass. \(\textbf{Evaluating carbon dioxide:}\) \[ M_{\mathrm{CO_2}}=12+2(16) \] \[ M_{\mathrm{CO_2}}=44\,\mathrm{u} \] Each carbon dioxide molecule contains exactly two oxygen atoms. Therefore, this candidate satisfies both the molecular-mass and oxygen-count conditions. \(\textbf{Evaluating dinitrogen monoxide:}\) \[ M_{\mathrm{N_2O}}=2(14)+16 \] \[ M_{\mathrm{N_2O}}=44\,\mathrm{u} \] Its molecular mass agrees with the measurement. However, every \(\mathrm{N_2O}\) molecule contains only one oxygen atom. It therefore fails the second condition. \(\textbf{Evaluating propane:}\) \[ M_{\mathrm{C_3H_8}}=3(12)+8(1) \] \[ M_{\mathrm{C_3H_8}}=44\,\mathrm{u} \] This formula also has the required molecular mass. However, propane contains no oxygen atoms. It is inconsistent with the elemental-analysis result. \(\textbf{Evaluating the fourth candidate:}\) \[ M_{\mathrm{C_2H_4O}}=2(12)+4(1)+16 \] \[ M_{\mathrm{C_2H_4O}}=44\,\mathrm{u} \] This candidate likewise satisfies the molecular-mass condition. However, it contains only one oxygen atom per molecule. Therefore, it must also be rejected. \(\textbf{Chemical inference:}\) All four candidates have identical calculated molecular masses. Their molecular formulae differ because they contain different constituent elements or different numbers of those elements. The additional oxygen-atom count provides the necessary discriminating information. Only carbon dioxide simultaneously satisfies both independent observations. This demonstrates why an experimentally measured molar mass cannot generally establish a unique molecular formula without further composition information. \(\textbf{Final answer:}\) The molecular formula consistent with both conditions is \(\mathrm{CO_2}\).