Some Basic Concepts of Chemistry MCQs with Answers – Part 5 (Class 11 Chemistry)

Explanation: \( \textbf{Mass of solute:} \) \(2.65\,g\). \( \textbf{Molar mass:} \) \(106\,g\,mol^{-1}\). \( \textbf{Moles of solute:} \) \[ n=\frac{2.65\,g}{106\,g\,mol^{-1}}=0.0250\,mol \] \( \textbf{Solution volume:} \) \[ 250\,mL=0.250\,L \] \( \textbf{Molarity:} \) \[ M=\frac{0.0250\,mol}{0.250\,L} \] \( \textbf{Calculation:} \) \[ M=0.100\,M \] \( \textbf{Final answer:} \) the solution concentration is \(0.100\,M\). The final volume of the solution, not the mass of water used, is the denominator for molarity.

Explanation: \( \textbf{Initial concentration:} \) \(M_1=0.500\,M\). \( \textbf{Initial volume:} \) \(V_1=10.0\,mL\). \( \textbf{Final volume:} \) \(V_2=100\,mL\). \( \textbf{Dilution relation:} \) \[ M_1V_1=M_2V_2 \] \( \textbf{Substitution:} \) \[ (0.500)(10.0)=M_2(100) \] \( \textbf{Solve:} \) \[ M_2=\frac{0.500\times10.0}{100}=0.0500\,M \] \( \textbf{Final answer:} \) the final concentration is \(0.0500\,M\). The volume increases ten times, so the molarity becomes one-tenth of its original value.

Explanation: \( \textbf{Moles of \(\mathrm{HCl}\):} \) \[ n=(0.100\,mol\,L^{-1})(0.0400\,L)=0.00400\,mol \] \( \textbf{Reaction ratio:} \) \[ 1\,mol\ \mathrm{Ba(OH)_2}:2\,mol\ \mathrm{HCl} \] \( \textbf{Moles of \(\mathrm{Ba(OH)_2}\) required:} \) \[ n_{\mathrm{Ba(OH)_2}}=\frac{0.00400}{2}=0.00200\,mol \] \( \textbf{Base concentration:} \) \(0.100\,mol\,L^{-1}\). \( \textbf{Volume of base:} \) \[ V=\frac{0.00200\,mol}{0.100\,mol\,L^{-1}}=0.0200\,L \] \( \textbf{Convert to millilitres:} \) \[ 0.0200\,L=20.0\,mL \] \( \textbf{Final answer:} \) \(20.0\,mL\) of \(\mathrm{Ba(OH)_2}\) is required. One mole of \(\mathrm{Ba(OH)_2}\) supplies the stoichiometric equivalent needed for two moles of \(\mathrm{HCl}\) in this reaction.

Explanation: The assertion is true because \(100\,mL\) diluted to \(1.0\,L\) means the final volume is \(10\) times the initial volume. Since no solute is added or removed during dilution, the moles of solute remain the same. The concentration therefore becomes one-tenth of the original value. The reason states this dilution principle correctly and explains the numerical change. The calculation can also be shown as \((1.0)(100)=M_2(1000)\), giving \(M_2=0.10\,M\).

Explanation: \( \textbf{Moles in original solution:} \) \[ n=\frac{4.00\,g}{40.0\,g\,mol^{-1}}=0.100\,mol \] \( \textbf{Original volume:} \) \[ 500\,mL=0.500\,L \] \( \textbf{Original molarity:} \) \[ M=\frac{0.100\,mol}{0.500\,L}=0.200\,M \] \( \textbf{Aliquot volume:} \) \[ 20.0\,mL=0.0200\,L \] \( \textbf{Moles in aliquot:} \) \[ n=MV=(0.200\,mol\,L^{-1})(0.0200\,L) \] \( \textbf{Calculation:} \) \[ n=0.00400\,mol \] \( \textbf{Final answer:} \) the \(20.0\,mL\) aliquot contains \(0.00400\,mol\) of \(\mathrm{NaOH}\). A measured portion of a uniform solution contains the same concentration as the original solution, but fewer total moles because its volume is smaller.

Explanation: \( \textbf{Moles of \(\mathrm{Na_2SO_4}\):} \) \[ n=(0.100\,mol\,L^{-1})(0.0250\,L)=0.00250\,mol \] \( \textbf{Reaction ratio:} \) \[ 1\,mol\ \mathrm{Na_2SO_4}\rightarrow1\,mol\ \mathrm{BaSO_4} \] \( \textbf{Moles of \(\mathrm{BaSO_4}\):} \) \[ n_{\mathrm{BaSO_4}}=0.00250\,mol \] \( \textbf{Given mass of precipitate:} \) \(0.583\,g\). \( \textbf{Molar mass relation:} \) \[ M=\frac{m}{n} \] \( \textbf{Substitution:} \) \[ M=\frac{0.583\,g}{0.00250\,mol}=233\,g\,mol^{-1} \] \( \textbf{Final answer:} \) the molar mass used is \(233\,g\,mol^{-1}\). Working backward still follows the same route: solution moles, equation ratio, then mass-to-mole relation.

Explanation: Molarity is defined as \(M=\frac{n}{V}\). Rearranging this relation gives \(n=MV\). Since \(M\) has unit \(mol\,L^{-1}\), the volume should be in \(L\) so that \(L^{-1}\times L\) cancels. The expression \(\frac{V}{M}\) would give units of \(\frac{L}{mol\,L^{-1}}\), which is not moles. Checking units is a reliable way to repair a wrong rearrangement.

Explanation: A known solution volume and molarity first give moles of solute through \(n=MV\). The balanced equation then converts moles of the known reacting substance into moles of the required product. Product moles are finally converted into product mass using molar mass. Skipping the mole-ratio step would ignore the balanced equation. The correct route keeps each conversion tied to its own meaning: concentration, stoichiometric ratio, and molar mass.

Explanation: \( \textbf{Moles of \(\mathrm{HCl}\):} \) \[ n_{\mathrm{HCl}}=(0.150\,mol\,L^{-1})(0.0300\,L)=0.00450\,mol \] \( \textbf{Moles of \(\mathrm{NaOH}\):} \) \[ n_{\mathrm{NaOH}}=(0.100\,mol\,L^{-1})(0.0200\,L)=0.00200\,mol \] \( \textbf{Reaction ratio:} \) \[ 1\,mol\ \mathrm{HCl}:1\,mol\ \mathrm{NaOH} \] \( \textbf{Acid neutralised:} \) \[ 0.00200\,mol \] \( \textbf{Acid left:} \) \[ 0.00450\,mol-0.00200\,mol=0.00250\,mol \] \( \textbf{Final answer:} \) \(0.00250\,mol\) of \(\mathrm{HCl}\) remains. The comparison must be made using moles, because different volumes and molarities are both involved.

Explanation: \( \textbf{Volume of \(\mathrm{Pb(NO_3)_2}\):} \) \[ 40.0\,mL=0.0400\,L \] \( \textbf{Molarity of \(\mathrm{Pb(NO_3)_2}\):} \) \(0.250\,mol\,L^{-1}\). \( \textbf{Moles of \(\mathrm{Pb(NO_3)_2}\):} \) \[ n=(0.250)(0.0400)=0.0100\,mol \] \( \textbf{Balanced equation:} \) \[ \mathrm{Pb(NO_3)_2+2KI\rightarrow PbI_2+2KNO_3} \] \( \textbf{Mole ratio:} \) \[ 1\,mol\ \mathrm{Pb(NO_3)_2}\rightarrow1\,mol\ \mathrm{PbI_2} \] \( \textbf{Moles of precipitate:} \) \[ n_{\mathrm{PbI_2}}=0.0100\,mol \] \( \textbf{Mass of \(\mathrm{PbI_2}\):} \) \[ m=(0.0100\,mol)(461\,g\,mol^{-1})=4.61\,g \] \( \textbf{Final answer:} \) \(4.61\,g\) of \(\mathrm{PbI_2}\) forms. The coefficient \(2\) belongs to \(\mathrm{KI}\), so it does not double the mole amount of \(\mathrm{PbI_2}\).

Explanation: \( \textbf{Moles of \(\mathrm{Na_2CO_3}\):} \) \[ n=(0.100\,mol\,L^{-1})(0.0500\,L)=0.00500\,mol \] \( \textbf{Reaction ratio:} \) \[ 1\,mol\ \mathrm{Na_2CO_3}:2\,mol\ \mathrm{HCl} \] \( \textbf{Moles of \(\mathrm{HCl}\) required:} \) \[ n_{\mathrm{HCl}}=2(0.00500\,mol)=0.0100\,mol \] \( \textbf{Acid molarity:} \) \[ 0.200\,mol\,L^{-1} \] \( \textbf{Find acid volume:} \) \[ V=\frac{n}{M}=\frac{0.0100\,mol}{0.200\,mol\,L^{-1}} \] \( \textbf{Calculation:} \) \[ V=0.0500\,L=50.0\,mL \] \( \textbf{Final answer:} \) \(50.0\,mL\) of \(\mathrm{HCl}\) is required. The acid concentration is twice the carbonate concentration, but the equation also requires two acid moles per carbonate mole.

Explanation: \( \textbf{Mass of \(\mathrm{Na_2CO_3}\):} \) \(1.06\,g\). \( \textbf{Molar mass:} \) \(106\,g\,mol^{-1}\). \( \textbf{Total moles in \(100\,mL\):} \) \[ n=\frac{1.06\,g}{106\,g\,mol^{-1}}=0.0100\,mol \] \( \textbf{Aliquot fraction:} \) \[ \frac{25.0\,mL}{100\,mL}=0.250 \] \( \textbf{Moles in aliquot:} \) \[ 0.250(0.0100\,mol)=0.00250\,mol \] \( \textbf{Alternative route:} \) the solution is \(0.100\,M\), and \(25.0\,mL=0.0250\,L\). \[ n=(0.100)(0.0250)=0.00250\,mol \] \( \textbf{Final answer:} \) the portion contains \(0.00250\,mol\). A measured aliquot contains the same concentration as the original solution, but only a fraction of the total solute.

Explanation: \( \textbf{Stock concentration:} \) \(M_1=2.00\,M\). \( \textbf{Required concentration:} \) \(M_2=0.200\,M\). \( \textbf{Final solution volume:} \) \(V_2=250\,mL\). \( \textbf{Dilution relation:} \) \[ M_1V_1=M_2V_2 \] \( \textbf{Rearrange:} \) \[ V_1=\frac{M_2V_2}{M_1} \] \( \textbf{Substitution:} \) \[ V_1=\frac{(0.200)(250\,mL)}{2.00} \] \( \textbf{Calculation:} \) \[ V_1=25.0\,mL \] \( \textbf{Final answer:} \) \(25.0\,mL\) of stock solution is needed. The stock solution is ten times more concentrated than the final solution, so one-tenth of the final volume is taken.

Explanation: \( \textbf{Moles of solute:} \) \(0.0200\,mol\). \( \textbf{Volume of solution:} \) \[ 250\,mL=0.250\,L \] \( \textbf{Molarity:} \) \[ M=\frac{0.0200\,mol}{0.250\,L}=0.0800\,mol\,L^{-1} \] Row P is correct because it gives \(0.0800\,mol\,L^{-1}\). Row Q is also correct because \(M\) is commonly used for \(mol\,L^{-1}\). Row R is the same value in scientific notation. \( \textbf{Final answer:} \) Row S is incorrect because it fails to divide by the solution volume. The amount of solute and the concentration are not the same unless the solution volume is exactly \(1\,L\).

Explanation: \( \textbf{Initial moles of \(\mathrm{HCl}\):} \) \[ n=(0.120\,mol\,L^{-1})(0.0250\,L)=0.00300\,mol \] \( \textbf{After dilution:} \) total moles remain \(0.00300\,mol\). \( \textbf{Final diluted volume:} \) \[ 100\,mL=0.100\,L \] \( \textbf{Diluted molarity:} \) \[ M=\frac{0.00300\,mol}{0.100\,L}=0.0300\,M \] \( \textbf{Aliquot volume:} \) \[ 20.0\,mL=0.0200\,L \] \( \textbf{Moles in aliquot:} \) \[ n=(0.0300\,mol\,L^{-1})(0.0200\,L)=0.000600\,mol \] \( \textbf{Final answer:} \) the portion contains \(0.000600\,mol\) of \(\mathrm{HCl}\). Dilution changes concentration first, and the aliquot then takes only part of the diluted solution.

Explanation: The formula \(\mathrm{AlCl_3}\) contains three chloride ions per formula unit. If \(\mathrm{AlCl_3}\) dissociates completely, each mole of \(\mathrm{AlCl_3}\) produces \(3\,mol\) of \(\mathrm{Cl^-}\). Therefore, a \(0.100\,M\) \(\mathrm{AlCl_3}\) solution gives chloride ion concentration \(3\times0.100\,M=0.300\,M\). The aluminium ion concentration would be \(0.100\,M\). The subscript in the formula becomes the mole multiplier for the corresponding ion.

Explanation: \( \textbf{Initial \(\mathrm{CaCl_2}\) concentration:} \) \(0.150\,M\). \( \textbf{Initial volume:} \) \(200\,mL\). \( \textbf{Final volume:} \) \(600\,mL\). \( \textbf{Diluted \(\mathrm{CaCl_2}\) concentration:} \) \[ M_2=\frac{M_1V_1}{V_2}=\frac{(0.150)(200)}{600}=0.0500\,M \] \( \textbf{Formula relation:} \) \[ 1\,mol\ \mathrm{CaCl_2}\rightarrow2\,mol\ \mathrm{Cl^-} \] \( \textbf{Chloride concentration:} \) \[ [\mathrm{Cl^-}]=2(0.0500\,M)=0.100\,M \] \( \textbf{Final answer:} \) the chloride ion concentration is \(0.100\,M\). Dilution is applied to the compound concentration first, and the ion ratio is then read from the formula.

Explanation: \( \textbf{Moles of \(\mathrm{Cl^-}\) from \(\mathrm{NaCl}\):} \) \[ (0.100\,mol\,L^{-1})(0.100\,L)(1)=0.0100\,mol \] \( \textbf{Moles of \(\mathrm{Cl^-}\) from \(\mathrm{CaCl_2}\):} \) \[ (0.100\,mol\,L^{-1})(0.100\,L)(2)=0.0200\,mol \] \( \textbf{Total chloride moles:} \) \[ 0.0100\,mol+0.0200\,mol=0.0300\,mol \] \( \textbf{Total volume:} \) \[ 100\,mL+100\,mL=200\,mL=0.200\,L \] \( \textbf{Final chloride concentration:} \) \[ [\mathrm{Cl^-}]=\frac{0.0300\,mol}{0.200\,L}=0.150\,M \] \( \textbf{Final answer:} \) the final \(\mathrm{Cl^-}\) concentration is \(0.150\,M\). The two salts have the same molarity, but \(\mathrm{CaCl_2}\) contributes twice as many chloride ions per formula unit.

Explanation: Molarity is moles of solute per litre of solution. Equal volumes of solutions with the same molarity contain equal moles of their respective solutes. However, mass is calculated from \(m=nM\), so it depends on the molar mass of each solute. For example, \(1\,mol\) of \(\mathrm{NaCl}\) and \(1\,mol\) of glucose do not have the same mass. Equal molarity compares amount of substance, not gram mass directly.

Explanation: \( \textbf{Mass of solute:} \) \(9.80\,g\). \( \textbf{Molar mass of \(\mathrm{H_2SO_4}\):} \) \(98.0\,g\,mol^{-1}\). \( \textbf{Moles of solute:} \) \[ n=\frac{9.80\,g}{98.0\,g\,mol^{-1}}=0.100\,mol \] \( \textbf{Solution volume:} \) \[ 500\,mL=0.500\,L \] \( \textbf{Molarity:} \) \[ M=\frac{0.100\,mol}{0.500\,L} \] \( \textbf{Calculation:} \) \[ M=0.200\,M \] \( \textbf{Final answer:} \) the solution is \(0.200\,M\). The molar mass converts grams to moles, and the final solution volume supplies the litre denominator.