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SubjectsChemistry
मोलर द्रव्यमान का निर्धारण
In this Class 12 Chemistry topic from Chapter 01: Solutions, students learn how the molar mass of a solute can be determined from the measurable properties of a solution. The topic connects mass, moles, concentration, and colligative properties such as relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. Students also practise selecting suitable formulas, interpreting experimental data, and recognising how observed results can indicate association or dissociation of solute particles.
Correct answer: C. 600 mL
Explanation: Dilution does not change the amount of solute, so M1V1 = M2V2. Using consistent volume units, 0.2 × 200 = 0.05 × V2, giving V2 = 800 mL as the final solution volume. The water added is only the increase in volume: 800 − 200 = 600 mL. Thus option C is correct; 800 mL is the final volume, not added water.
Correct answer: B. 100 g mol−1
Explanation: Molarity is moles per litre, so first convert 250 mL to 0.250 L. The number of moles is n = M × V = 0.05 × 0.250 = 0.0125 mol. Molar mass is mass divided by amount: 1.25/0.0125 = 100 g mol−1. Hence option B is correct. The other values arise from an arithmetic error or from treating millilitres as litres.
Correct answer: A. Molar mass
Explanation: For a dilute solution of a non-volatile solute, relative lowering of vapour pressure gives the solute mole fraction. If the masses of solute and solvent and the solvent molar mass are known, the solute moles can be calculated from that mole fraction. Dividing the known solute mass by its calculated moles gives molar mass. This is why the method is useful for molecular-mass determination.
Correct answer: A. Because it is related to moles and mole fraction of solute
Explanation: For a dilute solution with a non-volatile solute, relative lowering of vapour pressure is approximately the solute mole fraction. Mole fraction contains the number of solute moles, n = w/M, where w is mass and M is molar mass. Measuring the lowering therefore permits calculation of n and then M, provided the solution is sufficiently dilute and assumptions are valid.
Correct answer: A. Molar mass of solute
Explanation: For a dilute solution, π = nRT/V can be rearranged to find the amount of solute, and the amount in grams divided by moles gives molar mass. This method is especially useful for macromolecules because osmotic pressure can be measured at low concentration and ordinary temperature. Colour, container height and odour are not obtained from this relation.
Correct answer: A. Because osmotic pressure can be measured even at low concentration
Explanation: For dilute solutions, osmotic pressure is proportional to concentration: π = nRT/V. Measuring π therefore allows the number of solute moles to be calculated, after which molar mass follows from mass divided by moles. The method is valuable for large biomolecules because their solutions may be very dilute and heating methods could decompose them. It still requires a solvent and calculation.
Correct answer: A. Molar mass of solute
Explanation: A measured colligative change can be related to the number of solute moles through equations such as ΔTf = iKf m or π = cRT. If the solute mass and solution data are known, its moles can be calculated and molar mass follows from mass divided by moles. Colour and container shape are irrelevant.
Correct answer: A. Osmotic pressure
Explanation: Large biomolecules such as proteins are usually studied in very dilute solutions. Osmotic pressure remains measurable at such low concentrations and is proportional to molar concentration through π = CRT. It also avoids heating, which could decompose a biomolecule; colour, density, and magnetism are not the standard colligative method here.
Correct answer: A. Measurement is possible even at low concentration
Explanation: Macromolecules such as polymers and biomolecules may decompose on heating, so boiling- or freezing-point methods are inconvenient. Their solutions are often dilute, yet osmotic pressure is measurable and is directly proportional to concentration at a fixed temperature: π = CRT. Thus the osmotic method can determine molar mass without requiring a large temperature change or heating the macromolecule.
Correct answer: C. Osmotic pressure method
Explanation: Large biomolecules such as proteins and polymers are usually studied in very dilute solutions. Their boiling-point elevation or freezing-point depression can be extremely small and difficult to measure accurately. Osmotic pressure remains measurable at low concentration and can be determined near room temperature, so it is particularly useful for calculating the molar mass of such high-molar-mass substances.
Correct answer: A. Osmotic pressure
Explanation: For a macromolecule, the number of moles in a given mass is very small, so changes in vapour pressure, boiling point or freezing point may be too small to measure reliably. Osmotic pressure can be appreciable even in a very dilute solution and is measured near ordinary temperatures without heating or freezing the sample. Consequently it is preferred for protein and polymer molar-mass determination.
Correct answer: B. 200 g mol⁻¹
Explanation: For a non-electrolyte, the freezing-point depression is ΔT_f = K_f m, where m is molality. Thus, m = 0.186/1.86 = 0.10 mol kg⁻¹. The solvent mass is 500 g = 0.500 kg, so moles of solute = 0.10 × 0.500 = 0.050 mol. Therefore, molar mass = mass/moles = 10/0.050 = 200 g mol⁻¹. Hence, option B is correct.
Correct answer: A. Osmotic pressure
Explanation: Osmotic pressure is especially suitable for finding the molar mass of proteins and other macromolecules. Such substances generally form very dilute solutions, so the changes in boiling point or freezing point may be too small to measure accurately. Osmotic pressure can be measured at or near room temperature, avoiding heating that might decompose a protein. Since π = CRT for a dilute solution, the molar concentration and hence molar mass can be calculated from the measured pressure.
Correct answer: C. 80 g mol⁻¹
Explanation: For a non-dissociated solute, the van’t Hoff factor is i = 1, so ΔT_f = K_f m. Therefore, the molality is m = 0.186/1.86 = 0.10 mol kg⁻¹. The solvent mass is 250 g = 0.250 kg, so moles of solute = 0.10 × 0.250 = 0.025 mol. Hence, molar mass = mass/moles = 2/0.025 = 80 g mol⁻¹. Therefore, option C is correct.
Correct answer: B. 100 g mol⁻¹
Explanation: For a non-dissociated solute, i = 1, so ΔT_b = K_bm. The molality is m = 0.052/0.52 = 0.10 mol kg⁻¹. The solvent mass is 500 g = 0.500 kg, so the number of solute moles is 0.10 × 0.500 = 0.050 mol. Therefore, molar mass = 5 g/0.050 mol = 100 g mol⁻¹. Hence, option B is correct.
Correct answer: A. Because dilute solutions behave nearly ideally
Explanation: For a dilute solution, osmotic pressure is described accurately by π = iCRT, and interactions among dissolved particles are relatively small. This makes the solution approach ideal behaviour and gives a reliable relation between osmotic pressure, concentration, and molar mass. In concentrated solutions, solute–solute and solute–solvent interactions, activity effects, and possible association or dissociation can cause deviations and reduce the accuracy of the calculated molar mass.
Correct answer: D. Osmotic pressure
Explanation: Osmotic pressure is preferred for macromolecules such as proteins because it can be measured at room temperature and does not require heating or cooling that might decompose the solute. For a dilute solution, π = cRT = (w/MV)RT, so the molar mass M can be calculated from the measured osmotic pressure. The other colligative changes are often extremely small for very high-molar-mass solutes.
Correct answer: C. Osmotic-pressure method
Explanation: Polymers have extremely large molar masses, so a given mass contains very few molecules and produces very small changes in boiling point, freezing point, or vapour pressure. Osmotic pressure is the preferred method because it can be measured for dilute polymer solutions at ordinary temperature and is sufficiently useful even when the pressure is small. Using π = cRT and c = w/(MV), the polymer molar mass can be calculated.
Correct answer: A. π = CRT
Explanation: For a dilute solution, osmotic pressure follows the van’t Hoff equation π = CRT, where C is the molar concentration, R is the gas constant, and T is the absolute temperature. Since C can be written using solute mass, molar mass, and solution volume, the equation allows calculation of the unknown molar mass. The other equations describe different colligative properties.
Correct answer: A. Osmotic pressure
Explanation: Osmotic pressure is the most suitable colligative property for finding the molar mass of proteins, polymers, and other macromolecules. Such substances are usually studied in very dilute solutions, where boiling-point elevation and freezing-point depression may be extremely small and difficult to measure accurately. Osmotic pressure can be measured at room temperature, and the relation Π = CRT allows the molar mass to be calculated reliably.
Correct answer: A. Molality
Explanation: The boiling-point elevation relation is ΔTᵦ = Kᵦm for a non-electrolyte, or ΔTᵦ = iKᵦm when the van’t Hoff factor is included. Since ΔTᵦ is measured experimentally and Kᵦ is known for the solvent, molality is calculated first using m = ΔTᵦ/Kᵦ. The solvent mass then gives moles of solute, and the measured solute mass finally gives molar mass.
Correct answer: A. The small amount of solute can still be determined from a measurable colligative effect
Explanation: A very dilute solution contains relatively few solute particles, so its solute mole fraction is small, but the resulting colligative effect is not necessarily zero. Measurements such as vapour-pressure lowering, boiling-point elevation, freezing-point depression, or osmotic pressure can be used to estimate the amount of solute. Combining the measured solute amount in moles with its known mass gives molar mass through M = mass/moles.
Correct answer: A. M = wRT/(πV)
Explanation: For a dilute solution, the osmotic-pressure equation is πV = nRT. If w grams of solute having molar mass M are dissolved, the number of moles is n = w/M. Substitution gives πV = wRT/M, and rearrangement gives M = wRT/(πV). Thus option A is correct. Temperature must be expressed in kelvin, and the units of pressure and volume must agree with the chosen gas constant.
Correct answer: A. M_B = K_b w_B × 1000/(ΔT_b w_A)
Explanation: Boiling-point elevation is given by ΔTb = Kb m. If wB grams of solute with molar mass MB are dissolved in wA grams of solvent, then molality m = (wB/MB)/(wA/1000). Substitution gives ΔTb = Kb wB × 1000/(MB wA). Rearranging yields MB = Kb wB × 1000/(ΔTb wA), which is option A. Here wA is solvent mass and wB is solute mass.
Correct answer: A. M = (K_f w₂ × 1000)/(ΔT_f w₁)
Explanation: Freezing-point depression follows ΔT_f = K_f m, where molality m is the number of moles of solute per kilogram of solvent. If w₂ grams of solute have molar mass M, the moles are w₂/M. Since w₁ grams of solvent equal w₁/1000 kilograms, substitution gives ΔT_f = K_f(w₂/M)(1000/w₁). Rearranging produces M = K_f w₂ × 1000/(ΔT_f w₁). Thus option A is correct, and the factor 1000 converts grams of solvent into kilograms.
