Class 12 Chemistry MCQs | Chapter 1: Solutions – Part 2

Explanation: \( \textbf{Given:} \) Mass of glucose \( = 18,\text{g} \) Molar mass of glucose \( = 180,\text{g mol}^{-1} \) Mass of water \( = 18,\text{g} \) Molar mass of water \( = 18,\text{g mol}^{-1} \) \( \textbf{Required:} \) Mole fraction of glucose \( \textbf{Relevant formula:} \) \[ x_{\text{glucose}} = \frac{n_{\text{glucose}}}{n_{\text{glucose}} + n_{\text{water}}} \] \( \textbf{Why this formula applies:} \) Mole fraction is defined as moles of a component divided by total moles in the solution. \( \textbf{Identify known values:} \) \[ n_{\text{glucose}} = \frac{18}{180} = 0.10,\text{mol} \] \[ n_{\text{water}} = \frac{18}{18} = 1.0,\text{mol} \] \( \textbf{Substitution:} \) \[ x_{\text{glucose}} = \frac{0.10}{0.10 + 1.0} \] \( \textbf{Intermediate simplification:} \) \[ x_{\text{glucose}} = \frac{0.10}{1.10} \] \( \textbf{Final simplification:} \) \[ x_{\text{glucose}} = 0.091 \] \( \textbf{Unit:} \) Mole fraction is unitless. \( \textbf{Final Answer:} \) Mole fraction of glucose \( = 0.091 \)

Explanation: \( \textbf{Given:} \) Moles of solute \( = 0.25,\text{mol} \) Moles of solvent \( = 0.75,\text{mol} \) \( \textbf{Required:} \) Mole fraction of solvent \( \textbf{Relevant formula:} \) \[ x_{\text{solvent}} = \frac{n_{\text{solvent}}}{n_{\text{solute}} + n_{\text{solvent}}} \] \( \textbf{Why this formula applies:} \) Mole fraction is the ratio of moles of the chosen component to total moles present. \( \textbf{Substitution:} \) \[ x_{\text{solvent}} = \frac{0.75}{0.25 + 0.75} \] \( \textbf{Intermediate simplification:} \) \[ x_{\text{solvent}} = \frac{0.75}{1.00} \] \( \textbf{Final simplification:} \) \[ x_{\text{solvent}} = 0.75 \] \( \textbf{Unit:} \) Mole fraction has no unit. \( \textbf{Final Answer:} \) Mole fraction of solvent \( = 0.75 \)

Explanation: \( \textbf{Given:} \) Mole fraction of solute \( = 0.25 \) \( \textbf{Required:} \) Ratio of moles of solute to moles of solvent \( \textbf{Relevant formula:} \) \[ x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \] \( \textbf{Why this formula applies:} \) The given mole fraction directly connects the moles of solute with total moles of both components. \( \textbf{Let:} \) \[ n_{\text{solute}} = n \] \[ n_{\text{solvent}} = m \] \( \textbf{Substitution:} \) \[ \frac{n}{n+m} = 0.25 = \frac{1}{4} \] \( \textbf{Cross interpretation:} \) If the solute part is 1 out of total 4 parts, then solvent part must be 3 out of total 4 parts. \( \textbf{Therefore:} \) \[ n:m = 1:3 \] \( \textbf{Unit:} \) A mole ratio is unitless. \( \textbf{Final Answer:} \) Ratio of moles of solute to moles of solvent \( = 1:3 \)

Explanation: If the mole fractions of two components are equal, each contributes the same fraction of the total moles. That means the numbers of moles of the two components are equal, though their masses or volumes need not be equal.

Explanation: \( \textbf{Given:} \) Mass of ethanol \( = 46,\text{g} \) Molar mass of ethanol \( = 46,\text{g mol}^{-1} \) Mass of water \( = 72,\text{g} \) Molar mass of water \( = 18,\text{g mol}^{-1} \) \( \textbf{Required:} \) Mole fraction of ethanol \( \textbf{Relevant formula:} \) \[ x_{\text{ethanol}} = \frac{n_{\text{ethanol}}}{n_{\text{ethanol}} + n_{\text{water}}} \] \( \textbf{Why this formula applies:} \) Mole fraction is based on moles, so masses must first be converted into moles. \( \textbf{Identify known values:} \) \[ n_{\text{ethanol}} = \frac{46}{46} = 1.0,\text{mol} \] \[ n_{\text{water}} = \frac{72}{18} = 4.0,\text{mol} \] \( \textbf{Substitution:} \) \[ x_{\text{ethanol}} = \frac{1.0}{1.0 + 4.0} \] \( \textbf{Intermediate simplification:} \) \[ x_{\text{ethanol}} = \frac{1.0}{5.0} \] \( \textbf{Final simplification:} \) \[ x_{\text{ethanol}} = 0.200 \] \( \textbf{Unit:} \) Mole fraction is unitless. \( \textbf{Final Answer:} \) Mole fraction of ethanol \( = 0.200 \)

Explanation: Mole fraction is a ratio of moles to total moles, so it has no unit. Molarity expresses moles of solute per litre of solution, so its unit is \( \text{mol L}^{-1} \).

Explanation: \( \textbf{Given:} \) Moles of benzene \( = 3,\text{mol} \) Moles of toluene \( = 2,\text{mol} \) \( \textbf{Required:} \) Mole fraction of benzene \( \textbf{Relevant formula:} \) \[ x_{\text{benzene}} = \frac{n_{\text{benzene}}}{n_{\text{benzene}} + n_{\text{toluene}}} \] \( \textbf{Why this formula applies:} \) The mole fraction of a component is its share in the total number of moles. \( \textbf{Substitution:} \) \[ x_{\text{benzene}} = \frac{3}{3+2} \] \( \textbf{Intermediate simplification:} \) \[ x_{\text{benzene}} = \frac{3}{5} \] \( \textbf{Final simplification:} \) \[ x_{\text{benzene}} = 0.60 \] \( \textbf{Unit:} \) Mole fraction has no unit. \( \textbf{Final Answer:} \) Mole fraction of benzene \( = 0.60 \)

Explanation: \( \textbf{Given:} \) Mole fraction of solvent, \( x_1 = 0.82 \) \( \textbf{Required:} \) Mole fraction of solute, \( x_2 \) \( \textbf{Relevant principle:} \) For a binary solution, \[ x_1 + x_2 = 1 \] \( \textbf{Why this principle applies:} \) A binary solution contains only two components, so the sum of their mole fractions must be 1. \( \textbf{Substitution:} \) \[ x_2 = 1 - 0.82 \] \( \textbf{Intermediate simplification:} \) \[ x_2 = 0.18 \] \( \textbf{Final simplification:} \) No further simplification is needed. \( \textbf{Unit:} \) Mole fraction is unitless. \( \textbf{Final Answer:} \) \[ x_2 = 0.18 \]

Explanation: \( \textbf{Given:} \) Total moles of solution \( = 2.0,\text{mol} \) Mole fraction of component \( A \), \( x_A = 0.25 \) \( \textbf{Required:} \) Number of moles of component \( A \) \( \textbf{Relevant formula:} \) \[ x_A = \frac{n_A}{n_{\text{total}}} \] \( \textbf{Why this formula applies:} \) Mole fraction is defined as the ratio of moles of the chosen component to the total moles in the solution. \( \textbf{Substitution:} \) \[ 0.25 = \frac{n_A}{2.0} \] \( \textbf{Intermediate simplification:} \) \[ n_A = 0.25 \times 2.0 \] \( \textbf{Final simplification:} \) \[ n_A = 0.50,\text{mol} \] \( \textbf{Unit:} \) Amount of substance is expressed in mol. \( \textbf{Final Answer:} \) Number of moles of component \( A = 0.50,\text{mol} \)

Explanation: For equal masses, the substance with smaller molar mass contains more moles because \( n = \frac{m}{M} \). Since urea has lower molar mass than glucose, the same mass of urea corresponds to a larger number of moles.

Explanation: \( \textbf{Given:} \) Moles of solvent \( = 0.80,\text{mol} \) Mole fraction of solute \( = 0.20 \) \( \textbf{Required:} \) Moles of solute \( \textbf{Relevant formula:} \) \[ x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \] \( \textbf{Why this formula applies:} \) The mole fraction of solute is its mole share in the total amount of solution. \( \textbf{Let:} \) Moles of solute \( = n \) \( \textbf{Substitution:} \) \[ \frac{n}{n + 0.80} = 0.20 \] \( \textbf{Intermediate simplification:} \) \[ n = 0.20(n + 0.80) \] \[ n = 0.20n + 0.16 \] \[ 0.80n = 0.16 \] \( \textbf{Final simplification:} \) \[ n = \frac{0.16}{0.80} = 0.20,\text{mol} \] \( \textbf{Unit:} \) Amount of substance is written in mol. \( \textbf{Final Answer:} \) Number of moles of solute \( = 0.20,\text{mol} \)

Explanation: Mole fraction expresses the fraction of total moles contributed by a component. Therefore, a larger mole fraction means that component contributes a larger portion of the total number of moles.

Explanation: Mole fraction depends only on the ratio of moles of the components. If temperature changes but the actual numbers of moles remain the same, the mole fractions do not change.

Explanation: Molarity is defined as the number of moles of solute present in one litre of solution. The volume used is total solution volume, not solvent mass or solute volume.

Explanation: Molarity measures moles of solute per litre of solution. Therefore, its standard unit is \( \text{mol L}^{-1} \).

Explanation: Molarity depends on the volume of the solution. Since volume generally changes with temperature, the molarity also changes even when the amount of solute remains the same.

Explanation: Molarity is defined as the number of moles of solute present per litre of the final solution. The denominator is total solution volume, not the volume of solvent taken initially.

Explanation: \( \textbf{Given:} \) Moles of solute \( = 0.50,\text{mol} \) Volume of solution \( = 250,\text{mL} \) \( \textbf{Required:} \) Molarity of the solution \( \textbf{Relevant formula:} \) \[ M = \frac{\text{moles of solute}}{\text{volume of solution in litre}} \] \( \textbf{Why this formula applies:} \) Molarity gives the number of moles of solute present in one litre of solution. \( \textbf{Identify known values:} \) \[ 250,\text{mL} = 0.250,\text{L} \] \( \textbf{Substitution:} \) \[ M = \frac{0.50}{0.250} \] \( \textbf{Intermediate simplification:} \) \[ M = 2.0 \] \( \textbf{Final simplification:} \) \[ M = 2.0,\text{mol L}^{-1} \] \( \textbf{Unit:} \) Molarity is expressed in \( \text{mol L}^{-1} \). \( \textbf{Final Answer:} \) Molarity \( = 2.0,\text{mol L}^{-1} \)

Explanation: During dilution, the number of moles of solute stays constant, but the total volume of solution increases. Since molarity is moles per litre of solution, its value decreases.

Explanation: \( \textbf{Given:} \) Molarity \( = 0.20,\text{mol L}^{-1} \) Volume of solution \( = 500,\text{mL} \) \( \textbf{Required:} \) Number of moles of solute \( \textbf{Relevant formula:} \) \[ M = \frac{n}{V} \] \( \textbf{Why this formula applies:} \) Molarity directly relates moles of solute to volume of solution in litre. \( \textbf{Identify known values:} \) \[ 500,\text{mL} = 0.500,\text{L} \] \( \textbf{Rearrange the formula:} \) \[ n = MV \] \( \textbf{Substitution:} \) \[ n = 0.20 \times 0.500 \] \( \textbf{Intermediate simplification:} \) \[ n = 0.100 \] \( \textbf{Final simplification:} \) \[ n = 0.10,\text{mol} \] \( \textbf{Unit:} \) Amount of substance is written in mol. \( \textbf{Final Answer:} \) Number of moles of solute \( = 0.10,\text{mol} \)