Class 12 Chemistry MCQs | Chapter 1: Solutions – Part 5

Explanation: \( \textbf{Given:} \) Depression in freezing point, \( \Delta T_f = 0.93,\text{K} \) \[ K_f = 1.86,\text{K kg mol}^{-1} \] \( \textbf{Required:} \) Molality, \( m \) \( \textbf{Relevant formula:} \) \[ \Delta T_f = K_f m \] \( \textbf{Why this formula applies:} \) For a dilute solution, freezing point depression is directly proportional to molality. \( \textbf{Rearrange the formula:} \) \[ m = \frac{\Delta T_f}{K_f} \] \( \textbf{Substitution:} \) \[ m = \frac{0.93}{1.86} \] \( \textbf{Intermediate simplification:} \) \[ m = 0.50 \] \( \textbf{Final simplification:} \) \[ m = 0.50,\text{mol kg}^{-1} \] \( \textbf{Unit:} \) Molality is expressed in \( \text{mol kg}^{-1} \). \( \textbf{Final Answer:} \) Molality \( = 0.50,\text{mol kg}^{-1} \)

Explanation: \( \textbf{Given:} \) Molar mass of solute, \( M_2 = 90,\text{g mol}^{-1} \) Mass of solvent \( = 200,\text{g} = 0.200,\text{kg} \) Depression in freezing point, \( \Delta T_f = 0.60,\text{K} \) \[ K_f = 3.0,\text{K kg mol}^{-1} \] \( \textbf{Required:} \) Mass of solute \( \textbf{Relevant formulas:} \) \[ \Delta T_f = K_f m \] \[ m = \frac{n_2}{W_1} \] \( \textbf{Why these formulas apply:} \) The freezing point depression gives molality, and molality gives moles of solute in the given mass of solvent. \( \textbf{First find molality:} \) \[ m = \frac{0.60}{3.0} = 0.20,\text{mol kg}^{-1} \] \( \textbf{Now find moles of solute:} \) \[ n_2 = mW_1 = 0.20 \times 0.200 = 0.040,\text{mol} \] \( \textbf{Convert moles into mass:} \) \[ w_2 = n_2 M_2 = 0.040 \times 90 \] \( \textbf{Final simplification:} \) \[ w_2 = 3.6,\text{g} \] \( \textbf{Unit:} \) Mass is expressed in \( \text{g} \). \( \textbf{Final Answer:} \) Mass of solute \( = 3.6,\text{g} \)

Explanation: The relations are \( \Delta T_b = K_b m \) and \( \Delta T_f = K_f m \). Both depend directly on molality, but the proportionality constants \( K_b \) and \( K_f \) are different and characteristic of the solvent.

Explanation: \( \textbf{Given:} \) Freezing point of pure solvent, \( T_f^0 = 15.0^\circ\text{C} \) Depression in freezing point, \( \Delta T_f = 2.5,\text{K} \) \( \textbf{Required:} \) Freezing point of solution, \( T_f \) \( \textbf{Relevant formula:} \) \[ \Delta T_f = T_f^0 - T_f \] \( \textbf{Why this formula applies:} \) Depression of freezing point is the amount by which the solution freezes below the pure solvent. \( \textbf{Substitution:} \) \[ 2.5 = 15.0 - T_f \] \( \textbf{Intermediate simplification:} \) \[ T_f = 15.0 - 2.5 \] \( \textbf{Final simplification:} \) \[ T_f = 12.5^\circ\text{C} \] \( \textbf{Unit:} \) A temperature difference of \( 2.5,\text{K} \) is numerically equal to \( 2.5^\circ\text{C} \). \( \textbf{Final Answer:} \) Freezing point of the solution \( = 12.5^\circ\text{C} \)

Explanation: For the same molality, the solvent with larger \( K_f \) shows greater \( \Delta T_f\), so its solution has lower freezing point.

Explanation: Dissolved solute particles disturb the equilibrium between pure solid solvent and the liquid phase. Because of this, the system must be cooled further before the solvent can freeze, so the freezing point falls.

Explanation: Adding salt to ice-water lowers the freezing point of water. This allows the mixture to remain liquid at temperatures below \( 0^\circ\text{C} \), producing a colder medium that helps freeze the ice cream mixture more effectively.

Explanation: In osmosis, the membrane permits passage of solvent but not solute. The solvent therefore tends to move from the more dilute side, where solvent concentration is higher, toward the more concentrated side.

Explanation: The essential feature of a semipermeable membrane is selective passage. It allows solvent molecules to pass through but prevents solute particles from crossing, which makes osmosis possible.

Explanation: Pure solvent has the maximum escaping tendency of solvent molecules. Across a semipermeable membrane, solvent therefore moves into the solution side until equilibrium conditions are established.

Explanation: Osmotic pressure is defined as the minimum external pressure that must be applied on the solution side to just prevent the inward flow of solvent through a semipermeable membrane.

Explanation: Osmotic pressure is one of the standard colligative properties. Its magnitude depends primarily on the number of solute particles present in a given volume of solution rather than on their chemical identity.

Explanation: For a dilute solution, osmotic pressure is directly proportional to molar concentration and absolute temperature. The relation is \( \Pi = CRT \), where \( C \) is molar concentration, \( R \) is the gas constant, and \( T \) is absolute temperature.

Explanation: Isotonic solutions are those that have equal osmotic pressure at the same temperature. Because of this equality, there is no net osmosis between them through a semipermeable membrane.

Explanation: \( \textbf{Given:} \) \[ C = 0.10,\text{mol L}^{-1} \] \[ T = 300,\text{K} \] \[ R = 0.083,\text{L bar mol}^{-1}\text{K}^{-1} \] \( \textbf{Required:} \) Osmotic pressure, \( \Pi \) \( \textbf{Relevant formula:} \) \[ \Pi = CRT \] \( \textbf{Why this formula applies:} \) For a dilute solution, osmotic pressure is proportional to concentration and absolute temperature. \( \textbf{Substitution:} \) \[ \Pi = 0.10 \times 0.083 \times 300 \] \( \textbf{Intermediate simplification:} \) \[ \Pi = 0.10 \times 24.9 \] \( \textbf{Final simplification:} \) \[ \Pi = 2.49,\text{bar} \] \( \textbf{Unit:} \) \( \text{L} \) and \( \text{mol} \) cancel appropriately, leaving pressure in bar. \( \textbf{Final Answer:} \) Osmotic pressure \( = 2.49,\text{bar} \)

Explanation: A hypertonic solution has greater osmotic pressure than the solution with which it is compared. Therefore, solvent tends to move toward the hypertonic side through a semipermeable membrane.

Explanation: In a hypertonic medium, the surrounding solution has higher osmotic pressure than the fluid inside the cell. Water therefore moves out of the cell by osmosis, causing the cell to shrink.

Explanation: In reverse osmosis, an external pressure greater than the osmotic pressure is applied on the solution side. This reverses the natural direction of solvent flow and drives solvent toward the pure-solvent side.

Explanation: \( \textbf{Given:} \) Mass of polymer, \( w_2 = 1.00,\text{g} \) Volume of solution, \( V = 200,\text{mL} = 0.200,\text{L} \) Temperature, \( T = 300,\text{K} \) Osmotic pressure, \( \Pi = 0.123,\text{bar} \) \[ R = 0.083,\text{L bar mol}^{-1}\text{K}^{-1} \] \( \textbf{Required:} \) Molar mass of polymer, \( M_2 \) \( \textbf{Relevant formulas:} \) \[ \Pi = CRT \] \[ C = \frac{n}{V} \] \[ n = \frac{w_2}{M_2} \] \( \textbf{Why these formulas apply:} \) For a dilute solution, osmotic pressure is related to molar concentration. The number of moles of solute can be written in terms of its mass and molar mass. \( \textbf{Combine the relations:} \) \[ \Pi = \frac{w_2}{M_2 V}RT \] \( \textbf{Rearrange:} \) \[ M_2 = \frac{w_2RT}{\Pi V} \] \( \textbf{Substitution:} \) \[ M_2 = \frac{1.00 \times 0.083 \times 300}{0.123 \times 0.200} \] \( \textbf{Intermediate simplification:} \) Numerator \( = 24.9 \) Denominator \( = 0.0246 \) \[ M_2 = \frac{24.9}{0.0246} \] \( \textbf{Final simplification:} \) \[ M_2 \approx 1012,\text{g mol}^{-1} \] \( \textbf{Unit:} \) Molar mass is expressed in \( \text{g mol}^{-1} \). \( \textbf{Final Answer:} \) Molar mass of the polymer \( \approx 1010,\text{g mol}^{-1} \)

Explanation: From \( \Pi = CRT \), osmotic pressure is directly proportional to molar concentration at constant temperature. Therefore, the most concentrated solution has the greatest osmotic pressure.