6: Molar Mass Determination

The molar mass formula is M = m/n, where m is the mass in grams and n is the amount in moles. Using the supplied data, M = 4.4 g ÷ 0.025 mol = 176 g mol⁻¹. Thus, the correct choice is option C. A common error is to multiply mass and moles or to misplace the decimal point; checking that 0.025 × 176 equals 4.4 confirms the result.

Molarity is defined as the number of moles of solute present in one litre of solution. First convert the volume: 500 mL = 0.500 L. The number of moles of solute is n = M × V = 0.01 mol L⁻¹ × 0.500 L = 0.005 mol. Molar mass is mass divided by moles, so it is 0.9 g ÷ 0.005 mol = 180 g mol⁻¹. Therefore, option C is correct.

The van’t Hoff factor i represents the ratio of the number of particles actually present in solution to the number expected from the formula units initially dissolved. A completely non-dissociated solute remains as intact molecules, so one formula unit gives one particle. Consequently, i = 1, and the colligative property shows normal behaviour. Therefore, option C is correct.

For a dilute solution, osmotic pressure is π = nRT/V, and n = w/M. Combining these gives M = wRT/(πV). Thus, when osmotic pressure, temperature, volume, and R are held constant, the calculated molar mass is directly proportional to the solute mass w. Increasing w therefore increases the calculated value of M.

Large biomolecules such as proteins and polymers can be decomposed or structurally altered by heating or cooling. The osmotic-pressure method is performed at essentially constant temperature and does not require a substantial temperature change. It also works well with dilute solutions, making it especially suitable for sensitive, high-molar-mass biomolecules. Therefore, option A is correct.

For a dilute non-electrolyte solution, osmotic pressure is given by πV = nRT, or π = nRT/V. Substituting n = 0.01 mol, R = 0.082 L atm K⁻¹ mol⁻¹, T = 300 K, and V = 1 L gives π = (0.01 × 0.082 × 300)/1 = 0.246 atm. Therefore option A is correct. The litre-atmosphere value of R is consistent with the volume unit.

Molality is defined as the number of moles of solute present per kilogram of solvent. The solute amount is n = mass/molar mass = 1/50 = 0.02 mol. The solvent mass is 200 g = 0.200 kg. Therefore, molality m = 0.02/0.200 = 0.10 mol kg⁻¹. Hence, option B is the correct answer.

For a non-dissociating solute, the van’t Hoff factor is i = 1. The boiling-point elevation is therefore ΔTb = iKb m = Kb m. Substituting Kb = 0.52 K kg mol⁻¹ and m = 0.5 mol kg⁻¹ gives ΔTb = 0.52 × 0.5 K = 0.26 K. The units of kg and mol cancel, so option B is correct.

The ebullioscopic constant K_b is a characteristic property of the solvent, not a universal constant. The relation ΔT_b = K_b m uses the K_b value belonging to the actual solvent. If a value for another solvent is substituted, the calculated molality is wrong; consequently, the calculated number of solute moles and molar mass are also wrong. Thus the correct solvent-specific K_b must always be used.

Absolute temperature must be used in the osmotic-pressure equation \(\pi=CRT\), because the gas constant is defined for the Kelvin scale. Convert the given Celsius temperature by adding 273 (more precisely, 273.15): \(T=27+273=300\,\mathrm{K}\). Using 27 directly would make the numerical calculation dimensionally and physically incorrect. Therefore, option C is correct.

For a non-dissociating solute, the freezing-point-depression relation is \(\Delta T_f=iK_fm\), and \(i=1\). Thus, \(m=\Delta T_f/K_f=0.93/1.86=0.50\,\mathrm{mol\,kg^{-1}}\). The units also confirm the result because kelvin cancels, leaving moles per kilogram of solvent. Therefore, option B is correct.

The amount of solute in moles is obtained from \(n=m/M\), where \(m\) is the mass and \(M\) is the molar mass. Thus, \(n=2\,\mathrm{g}/(100\,\mathrm{g\,mol^{-1}})=0.02\,\mathrm{mol}\). The volume is not needed for this particular calculation; it would be needed if molarity or osmotic pressure itself were being calculated. Therefore, option B is correct.

For an ideal solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as (p° − p)/p° = xsolute, where xsolute is the mole fraction of the solute. Since the given relative lowering is 0.05, the mole fraction of solute is also 0.05. The value 0.95 is the approximate solvent mole fraction, not the solute mole fraction.

Mole fraction is defined as xsolute = nsolute/ntotal. Rearranging gives nsolute = xsolute × ntotal. Therefore, nsolute = 0.1 × 2 = 0.2 mol. The value 1.8 mol would be the number of moles of the other component if there were only solute and solvent, because the solvent mole fraction would be 1 − 0.1 = 0.9 and its moles would be 1.8 mol.

For osmotic pressure, the equation is usually written as πV = nRT, or M = wRT/(πV), when R is expressed in L atm K⁻¹ mol⁻¹. Therefore, the solution volume must be expressed in litres. Since 1000 mL = 1 L, 250 mL = 250/1000 = 0.25 L. Using 250 directly would introduce a factor-of-1000 error.

Molar mass is the mass of one mole of a substance. It is calculated using M = mass/number of moles. Here, M = 3.6 g ÷ 0.02 mol = 180 g mol⁻¹. Therefore, the correct answer is option C. Since the amount in moles is directly provided, no colligative-property or van’t Hoff-factor formula is required.

For a nonelectrolyte, the depression in freezing point is ΔT_f = K_f m, where m is the molality. If the solvent mass and the solute’s molar mass remain unchanged, doubling the solute mass doubles the number of solute moles and therefore doubles the molality. Since K_f is constant for the solvent, ΔT_f also becomes double. Hence, option B is correct.

The elevation in boiling point is the difference between the boiling point of the solution and that of the pure solvent: ΔT_b = T_b(solution) − T_b°(solvent). Substituting the data gives ΔT_b = 373.26 K − 373.00 K = 0.26 K. Therefore, option B is correct. The value 373.26 K is the solution’s actual boiling point, not the elevation.

Freezing-point depression is defined as the freezing point of the pure solvent minus the freezing point of the solution: ΔT_f = T_f° − T_f. Thus, ΔT_f = 273.00 K − 272.44 K = 0.56 K. Option B is correct. The solution freezes at the lower temperature, and 272.44 K is the actual solution freezing point rather than the magnitude of the depression.

For freezing-point depression, ΔT_f = iK_fm. A nonelectrolyte does not dissociate, so i = 1. Therefore, m = ΔT_f/K_f = 0.56/1.86 = 0.301 mol kg⁻¹, approximately 0.30 mol kg⁻¹. Hence, option C is correct. Using 0.60 mol kg⁻¹ would predict a depression nearly twice as large, while the other values do not satisfy the given ratio.

The osmotic-pressure method requires a semipermeable membrane, which allows solvent molecules to pass through but prevents solute particles from crossing. This selective movement creates osmotic pressure, and the measured pressure is related to concentration by π = cRT. The freezing-point, boiling-point, and vapour-pressure methods do not require a semipermeable membrane. Therefore, the correct answer is the osmotic-pressure method.

Freezing-point depression is defined as \(\Delta T_f=T_f^0-T_f\), where \(T_f^0\) is the freezing point of the pure solvent and \(T_f\) is that of the solution. Therefore, the pure solvent’s freezing point is essential for calculating the temperature depression and then the molar mass. The solute’s melting point, solution boiling point, and solute density are not required in the basic method.

Osmotic pressure is preferred for proteins and other high-molar-mass substances because it can be measured in very dilute solutions at or near room temperature. For such substances, the changes in boiling point, freezing point, or vapour pressure are often extremely small and difficult to measure accurately. Osmotic pressure therefore gives a more reliable molar-mass determination.

Association combines several solute molecules into fewer effective particles. Because colligative properties depend on particle number, the observed effect becomes smaller than expected. If this smaller effect is interpreted using the ideal formula, the calculated molar mass becomes too large. In terms of the van’t Hoff factor, association gives i < 1, whereas dissociation generally gives i > 1.

The relation between the van’t Hoff factor and molar masses is i = Mnormal/Mobserved. Substituting the given values gives i = 100/125 = 0.8. Since i is less than one, the result is consistent with association of solute particles, which reduces the number of effective particles. A value above one would instead suggest dissociation or another particle-increasing effect.