6: Molar Mass Determination

The freezing-point relation is ΔT_f = K_f m. Substitution gives m = 0.186/1.86 = 0.100 mol kg⁻¹. Convert the mass of water to kilograms: 500 g = 0.500 kg. Therefore, moles of solute = molality × kilograms of solvent = 0.100 × 0.500 = 0.050 mol. The molar mass is M = 5.0 g/0.050 mol = 100 g mol⁻¹. Consequently, option C is correct. This calculation assumes a dilute solution and no abnormal association or dissociation of solute particles.

For a non-dissociating solute, \(\Delta T_b=K_bm\), so \(m=0.52/0.52=1\,\mathrm{mol\,kg^{-1}}\). The solvent mass is \(0.100\,\mathrm{kg}\), giving moles of solute equal to \(1\times0.100=0.100\,\mathrm{mol}\). Thus, \(M=2/0.100=20\,\mathrm{g\,mol^{-1}}\), making option A correct. Converting grams of solvent to kilograms is essential.

Osmotic pressure is the most suitable property for polymers and other macromolecules because it can be measured accurately even in very dilute solutions and near room temperature. Since \(\pi=CRT\), the measured pressure gives the molar concentration, from which molar mass can be calculated. The temperature changes in boiling-point elevation and freezing-point depression are often extremely small for such solutes.

A very dilute solution of a non-volatile solute is preferred because the osmotic-pressure equation \(\pi=CRT\) is most reliable near ideal-dilute conditions. Dilution reduces solute–solute interactions and non-ideal effects, while non-volatility prevents complications from loss of solute into the vapour phase. Concentrated or saturated solutions may deviate substantially from ideal behaviour.

Use \(\Delta T_f=K_fm\) and \(m=w_B\times1000/(M_Bw_A)\). Rearranging gives \(M_B=K_fw_B\times1000/(\Delta T_fw_A)\). Substitution yields \(M_B=(2.0\times1.2\times1000)/(0.40\times60)=2400/24=100\,\mathrm{g\,mol^{-1}}\). Therefore, option C is correct. The factor 1000 is required because the solvent mass is supplied in grams but molality uses kilograms.

For a solution containing a dissociating or associating solute, the boiling-point elevation is \(\Delta T_b=iK_bm\). Rearranging, \(i=\Delta T_b/(K_bm)\). Substituting the data gives \(i=0.104/(0.52\times0.1)=0.104/0.052=2\). Therefore, option B is correct. A value of 2 indicates that the effective number of solute particles is twice the ideal undissociated count under the assumed conditions.

Osmotic pressure is preferred for proteins and other macromolecules because even a very dilute solution can produce a measurable pressure, whereas boiling-point and freezing-point changes may be too small to measure accurately. The relation \(\pi=CRT\) connects osmotic pressure with molar concentration, allowing the molar mass to be calculated from the amount and volume of solution. This also permits measurements near room temperature.

For a dilute solution, osmotic pressure is proportional to the number of dissolved solute particles. If solute molecules associate or some solute remains undissolved, the actual number of particles in solution becomes smaller than expected. The measured osmotic pressure is therefore lower, and calculation gives an apparently larger molar mass. Complete dissociation would produce the opposite trend.

For a non-electrolyte, freezing-point depression is ΔTf = Kf m. Therefore, the molality is 0.50/4.0 = 0.125 mol kg⁻¹. The solvent mass is 200 g = 0.200 kg, so the amount of solute is 0.125 × 0.200 = 0.025 mol. Its molar mass is consequently 2/0.025 = 80 g mol⁻¹. Converting grams of solvent to kilograms is essential.

Use the dilute-solution osmotic-pressure relation π = wRT/(MV), so M = wRT/(πV). The solution volume is 250 mL = 0.250 L. Substitution gives M = (0.25 × 0.082 × 300)/(0.246 × 0.250), which is approximately 100 g mol⁻¹. Pressure must be in atm and volume in litres because the supplied gas constant uses L atm units.

Acetic acid associates in benzene, commonly forming dimers. Association lowers the number of independent solute particles, so the van’t Hoff factor is less than one and the observed boiling-point elevation is smaller than the ideal value. If the ideal equation is used without correction, the calculation gives a molar mass larger than the true value. Thus, option A is correct.

First calculate the amount of solute: n = 1.0/100 = 0.010 mol. The solvent mass is 100 g = 0.100 kg, so the molality is m = 0.010/0.100 = 0.10 mol kg⁻¹. For an ideal nonelectrolyte, ΔTf = Kf m = 1.86 × 0.10 = 0.186 K. The solvent mass must be converted to kilograms when calculating molality.

For boiling-point elevation, ΔTb = Kb m. Thus, the molality is 0.40/0.80 = 0.50 mol kg⁻¹. The solvent mass is 50 g = 0.050 kg, so the moles of solute are 0.50 × 0.050 = 0.025 mol. Therefore, the molar mass is 2.0/0.025 = 80 g mol⁻¹. Correct conversion of solvent mass to kilograms is essential in this calculation.

Osmotic pressure is the most useful colligative property for finding the molar mass of polymers. A polymer has a very large molar mass, so a dilute solution contains very few solute particles and produces extremely small changes in vapour pressure, boiling point, or freezing point. Osmotic pressure can still be measured accurately at low concentration and is related to concentration by π = CRT for a non-electrolyte. It also avoids heating or freezing the sample.

At very low concentration, changes in boiling point, freezing point, and vapour pressure are usually extremely small, so accurate measurement becomes difficult. Osmotic pressure, however, can be measured appreciably even for dilute solutions and is proportional to concentration through π = CRT. This makes the osmotic-pressure method particularly valuable for estimating the molar masses of polymers and other macromolecules, whose solutions must generally be kept dilute.

The freezing-point depression is ΔTf = iKf m. Since the observed depression is 1.5 times the non-dissociated value, the van’t Hoff factor is i = 1.5. For dissociation of one solute unit into two ions, i = 1 + α, where α is the fraction dissociated. Hence α = i − 1 = 1.5 − 1 = 0.5, or 50%. A 75% dissociation would produce i = 1.75, so it cannot be correct.

For boiling-point elevation, the ratio of the observed value to the value expected for a non-associated solute is the van’t Hoff factor i. Thus, i = 0.75. When monomer molecules associate to form dimers, the relation is i = 1 − α/2, where α is the degree of association. Substitution gives 0.75 = 1 − α/2, so α/2 = 0.25 and α = 0.50, or 50%. Association reduces the number of particles and therefore makes i less than one.

For a non-electrolyte, ΔTf = Kf m, so the molality is m = 0.30/1.5 = 0.20 mol kg⁻¹. The solvent mass is 90 g = 0.090 kg, so the amount of solute is n = 0.20 × 0.090 = 0.018 mol. Its molar mass is therefore M = mass/n = 1.8/0.018 = 100 g mol⁻¹. Converting grams of solvent to kilograms is essential because the cryoscopic constant uses molality.

Osmotic pressure is preferred for proteins and other macromolecules because it can be measured in very dilute solutions. Their high molar masses mean that the number of solute particles is small, making changes in boiling point, freezing point, or vapour pressure too tiny for reliable measurement. Osmotic pressure remains measurable and follows π = CRT for a dilute non-electrolyte solution, allowing concentration and molar mass to be calculated without heating or freezing the sample.

The van’t Hoff factor for an abnormal molar mass is related by \(i=M_{\text{normal}}/M_{\text{observed}}\). Hence, \(M_{\text{observed}}=200/0.8=250\,\mathrm{g\,mol^{-1}}\). Because \(i<1\), the solute particles have associated, so fewer effective particles are present and the observed molar mass becomes larger than the normal value. Therefore, option D is correct.

For a nonelectrolyte, \(\Delta T_f=K_fm\), where molality \(m=(w/M)/(W/1000)\). Rearranging gives \(M=K_fw\times1000/(\Delta T_fW)\). Substitution gives \(M=(5.0\times4.0\times1000)/(0.50\times400)=100\,\mathrm{g\,mol^{-1}}\). Thus, option C is correct. The solvent mass must be converted from grams to kilograms when calculating molality.

For a dilute nonelectrolyte solution, \(\pi V=nRT=(w/M)RT\). Therefore, \(M=wRT/(\pi V)\). Substituting the given values, \(M=(2.0\times0.082\times300)/(0.492\times1.0)=49.2/0.492=100\,\mathrm{g\,mol^{-1}}\). Hence, option B is correct. The volume must be expressed in litres because the given gas constant uses litres and atmospheres.

Osmotic pressure is especially suitable for determining the molar mass of proteins and other macromolecules. A protein solution is usually very dilute, so its boiling-point elevation and freezing-point depression are extremely small and difficult to measure accurately. Osmotic pressure can still be measured reliably at low concentration, and the relation \(\pi=CRT\) allows the molar concentration and hence the molar mass to be calculated. Therefore, option A is correct.

For a non-dissociating solute, \(\Delta T_f=K_fm\). Hence the molality is \(m=0.372/1.86=0.20\,\mathrm{mol\,kg^{-1}}\). The solvent mass is \(200\,\mathrm{g}=0.200\,\mathrm{kg}\), so moles of solute are \(0.20\times0.200=0.040\,\mathrm{mol}\). Therefore, molar mass is \(2.4/0.040=60\,\mathrm{g\,mol^{-1}}\), making option B correct.

For a non-dissociating solute, the boiling-point relation is \(\Delta T_b=K_bm\). Thus, \(m=0.104/0.52=0.20\,\mathrm{mol\,kg^{-1}}\). The solvent mass is \(250\,\mathrm{g}=0.250\,\mathrm{kg}\), so the amount of solute is \(0.20\times0.250=0.050\,\mathrm{mol}\). Its molar mass is therefore \(5/0.050=100\,\mathrm{g\,mol^{-1}}\), so option C is correct.