6: Molar Mass Determination

In the standard colligative-property formula, w₁ denotes the mass of the solvent and w₂ denotes the mass of the solute. In this question, 100 g is the solvent mass, so w₁ = 100 g, while 2 g is the solute mass represented by w₂. Correct identification of these symbols is essential for obtaining the correct molar mass.

In the standard colligative-property formula, w₁ denotes the mass of the solvent and w₂ denotes the mass of the solute. Here, the solute is the substance that is dissolved, and its stated mass is 2 g. Therefore, w₂ = 2 g and represents the mass of solute, not the solvent, volume, or boiling point. Correctly identifying w₁ and w₂ is essential before substituting values into a molar-mass equation.

Begin with the osmotic-pressure equation πV = (w/M)RT, where w is the solute mass and M is its molar mass. Multiplying by M and dividing by πV gives MπV = wRT, so M = wRT/(πV). Temperature must be expressed in kelvin and units must remain consistent.

To convert Celsius temperature to kelvin, add approximately 273 to the Celsius value. Thus, 27°C + 273 = 300 K. Using the exact conversion gives 27 + 273.15 = 300.15 K, which is approximately 300 K. This absolute-temperature value is used in gas-law and osmotic-pressure calculations.

Normal colligative behaviour is observed when solute particles neither dissociate into smaller particles nor associate to form larger particles in the solution. Glucose is a covalent, nonelectrolyte solute and generally remains as individual molecules. Sodium chloride and calcium chloride dissociate into ions, while acetic acid can associate in benzene. Therefore, glucose gives the normal result.

Sodium chloride is an electrolyte. In a suitable solution it dissociates into Na⁺ and Cl⁻ ions, so one formula unit produces more solute particles than expected for an undissociated substance. Colligative effects therefore become larger. If the ordinary formula is used without correcting for the van’t Hoff factor, the calculated or observed molar mass is lower than the true value.

Acetic acid molecules can form hydrogen-bonded dimers in a nonpolar solvent such as benzene. Association decreases the effective number of solute particles in the solution. Since colligative properties depend on the number of particles, the measured effect is smaller than expected for separate molecules. Using the ordinary formula then gives a higher observed molar mass; this corresponds to a van’t Hoff factor less than one.

Molality is moles of solute per kilogram of solvent. The solvent mass is 100 g = 0.1 kg. Therefore, moles of solute = 0.1 mol kg⁻¹ × 0.1 kg = 0.01 mol. The molar mass is mass divided by moles: M = 1 g ÷ 0.01 mol = 100 g mol⁻¹. Hence, option A is correct.

Molar mass is calculated using M = mass ÷ amount in moles. Here the mass is 2 g and the amount is 0.02 mol. Thus, M = 2 g ÷ 0.02 mol = 100 g mol⁻¹. Option A would result from treating 0.02 as 0.1, and option B would involve an incorrect decimal calculation. Therefore, option C is the only correct answer.

The depression in freezing point is given by \(\Delta T_f=iK_fm\), where \(i\) is the van’t Hoff factor and \(m\) is molality. Because the solute is a nonelectrolyte, it neither ionizes nor dissociates appreciably, so \(i=1\). Therefore, \(m=\Delta T_f/K_f=0.186/1.86=0.10\,mol\,kg^{-1}\). Hence, option B is correct. The units also confirm the result because K divided by K kg mol⁻¹ gives mol kg⁻¹.

For elevation of boiling point, the relation is \(\Delta T_b=iK_bm\). Since the solute is a nonelectrolyte, its van’t Hoff factor is \(i=1\). Substituting the given values gives \(m=\Delta T_b/K_b=0.052/0.52=0.10\,mol\,kg^{-1}\). Thus, option B is correct. The unit \(K\,kg\,mol^{-1}\) for \(K_b\) ensures that the calculated concentration is expressed as mol kg⁻¹, the unit of molality.

Molar mass is defined as the mass of a substance divided by the amount of that substance in moles. Substituting the given values, M = 0.5 g ÷ 0.005 mol. Since 0.005 × 100 = 0.5, the result is 100 g mol⁻¹. Therefore, option C is correct. Careful handling of the decimal places prevents choosing the smaller distractors.

K_f is the cryoscopic constant, also called the freezing-point depression constant. Its value depends on the nature of the pure solvent, because it reflects how that solvent’s freezing point changes when particles are dissolved in it. Different solvents have different K_f values. It is not a property of the solute, container, or thermometer.

K_b is the ebullioscopic constant, or boiling-point elevation constant, of a solvent. It depends on the solvent’s nature and relates the elevation in boiling point to the molality of dissolved particles. Thus, a different solvent generally has a different K_b value. The constant is not determined by the solute’s colour or by outside air.

Molality is defined as the number of moles of solute present in one kilogram of solvent. Therefore, when the solvent mass is supplied in grams, it must be divided by 1000 and expressed in kilograms before substitution. For example, 250 g of solvent equals 0.250 kg. The other options represent volume, length, or pressure and cannot be used as the denominator of molality. Hence, option A is correct.

The number of moles is calculated by dividing the mass of a substance by its molar mass. Thus, n = w/M, where w is the mass, usually in grams, and M is the molar mass in grams per mole. The units confirm the result: g divided by g mol⁻¹ gives mol. Multiplication, addition, or the reciprocal expression does not produce the number of moles. Therefore, option A is correct.

The first step in any molar-mass calculation is to inspect and standardise the units. Mass may need to be expressed in grams, solvent mass in kilograms for molality, volume in litres for molarity, and temperature in kelvin when gas or osmotic-pressure equations are used. Applying a correct formula to inconsistent units gives a wrong result. After unit conversion, identify the relevant relation, substitute the values, calculate, and check whether the final unit is g mol⁻¹.

Osmotic pressure is preferred for macromolecules because it can be measured using very dilute solutions and usually at or near room temperature. The other colligative changes, such as boiling-point elevation and freezing-point depression, become extremely small for high-molar-mass solutes and are therefore less accurate. The relation π = cRT is especially useful for calculating molar mass. Hence, option A is correct.

Osmotic pressure is the preferred colligative property for polymers because it can be measured accurately in very dilute solutions. For a polymer, the changes in boiling point and freezing point are usually extremely small, whereas osmotic pressure remains measurable. The relation π = CRT connects osmotic pressure with concentration and can be used to determine molar mass. Therefore, option D is correct.

Osmotic pressure is most useful for determining the molar mass of polymers because even a very dilute polymer solution produces a measurable osmotic pressure. The changes in boiling point and freezing point are generally too small for reliable measurement when the solute has a very large molar mass. Using π = CRT, the concentration and hence the molar mass can be determined. Thus, option D is correct.

The number of moles of solute is n = concentration × volume. Thus, n = 0.01 mol L⁻¹ × 2 L = 0.020 mol. The molar mass is mass divided by number of moles: M = 3.6 g/0.020 mol = 180 g mol⁻¹. Therefore, option C is correct. The litre units cancel correctly in the mole calculation. Choosing 360 g mol⁻¹ would mean incorrectly using only 0.010 mol, which would correspond to 1 L rather than the given 2 L of solution.

Convert the volume to litres: 250 mL = 0.250 L. The amount of solute is n = molarity × volume = 0.04 mol L⁻¹ × 0.250 L = 0.010 mol. Therefore, the molar mass is M = mass/n = 1.2 g/0.010 mol = 120 g mol⁻¹. Hence, option C is correct. The statement that the solute is non-dissociated confirms that no van’t Hoff correction is needed; however, the calculation here uses the stated molarity directly.

Boiling-point elevation is given by ΔTb = Kb m, and molality is defined as moles of solute per kilogram of solvent. Therefore, the solvent mass must be converted from grams to kilograms before substitution. Since 1 kg = 1000 g, 100 g equals 100/1000 = 0.1 kg. Hence, option A is correct. Using 0.01 kg would represent only 10 g, not 100 g.

Molality is defined as moles of solute per kilogram of solvent. Thus, the number of solute moles is n = m × mass of solvent = 0.1 mol kg⁻¹ × 0.2 kg = 0.02 mol. The molar mass is then M = mass/n = 2 g/0.02 mol = 100 g mol⁻¹. Therefore, option B is correct. The solvent mass is already supplied in kilograms, so no further conversion is needed.

Molar mass is the mass of a substance divided by the amount of substance in moles: M = m/n. Substituting the given values gives M = 7.2 g ÷ 0.04 mol = 180 g mol⁻¹. The unit is grams per mole because grams are divided by moles. Therefore, option C is correct. The fact that the solute is in a solution does not change this basic mass–mole relationship.