6: Molar Mass Determination

For a dilute nonelectrolyte solution, osmotic pressure follows \(\pi=CRT\). Therefore, \(C=\pi/(RT)=0.615/(0.082\times300)=0.025\,\mathrm{mol\,L^{-1}}\). The volume is \(500\,\mathrm{mL}=0.500\,\mathrm{L}\), so moles of solute equal \(0.025\times0.500=0.0125\,\mathrm{mol}\). Hence, molar mass is \(1.5/0.0125=120\,\mathrm{g\,mol^{-1}}\), option C.

Use the freezing-point equation \(\Delta T_f=K_fm\). The molality is \(m=0.186/1.86=0.10\,\mathrm{mol\,kg^{-1}}\). Since the solvent mass is \(400\,\mathrm{g}=0.400\,\mathrm{kg}\), the moles of solute are \(0.10\times0.400=0.040\,\mathrm{mol}\). Consequently, the molar mass is \(3.2/0.040=80\,\mathrm{g\,mol^{-1}}\). Thus, option C is the only correct answer.

For a non-dissociating solute, \(\Delta T_b=K_bm\). Thus, the molality is \(m=0.208/0.52=0.40\,\mathrm{mol\,kg^{-1}}\). The solvent weighs \(100\,\mathrm{g}=0.100\,\mathrm{kg}\), so the amount of solute is \(0.40\times0.100=0.040\,\mathrm{mol}\). The molar mass is therefore \(2/0.040=50\,\mathrm{g\,mol^{-1}}\), making option B correct.

For a non-dissociating dilute solution, \(\pi=CRT\). Taking \(R=0.082\,\mathrm{L\,atm\,mol^{-1}\,K^{-1}}\), the molar concentration is \(C=1.23/(0.082\times300)=0.050\,\mathrm{mol\,L^{-1}}\). A \(1\,\mathrm{L}\) solution therefore contains \(0.050\,\mathrm{mol}\) of solute. Its molar mass is \(6/0.050=120\,\mathrm{g\,mol^{-1}}\), so option C is correct.

For abnormal molar mass, the van’t Hoff factor is calculated as i = true molar mass ÷ observed molar mass. Therefore, i = 58.5 ÷ 39 = 1.5. For NaCl, complete dissociation into Na⁺ and Cl⁻ would ideally give i close to 2, so the value 1.5 indicates partial dissociation or non-ideal behavior in the solution. Hence, option B is correct.

The van’t Hoff factor is i = Mtrue ÷ Mobserved. Since Mobserved = (5/8)Mtrue, i = 8/5 = 1.6. For dissociation of AB into A and B, the number of particles increases from one to two, so i = 1 + α. Thus, 1.6 = 1 + α, giving α = 0.6. The percentage dissociation is therefore 60%, so option C is correct.

First calculate the van’t Hoff factor: i = Mtrue ÷ Mobserved = 111 ÷ 55.5 = 2. CaCl₂ dissociates as CaCl₂ ⇌ Ca²⁺ + 2Cl⁻, producing three particles from one formula unit. Therefore, i = 1 + (3 − 1)α = 1 + 2α. Substitution gives 2 = 1 + 2α, so α = 0.5, or 50%. Hence option B is correct.

For trimerisation, three original solute particles combine to form one trimer. If α is the fraction associated, the van’t Hoff factor is i = 1 − 2α/3. Substituting i = 0.6 gives 0.6 = 1 − 2α/3. Hence 2α/3 = 0.4, so α = 0.4 × 3/2 = 0.6. Thus, the degree of association is 60%, and option C is correct.

Because the observed molar mass is half the true molar mass, i = Mtrue/Mobserved = 2. For AB₂ dissociation, AB₂ ⇌ A + 2B produces three particles from one formula unit. Therefore, i = 1 + (3 − 1)α = 1 + 2α. Solving 2 = 1 + 2α gives α = 0.5, or 50% dissociation. Hence, option B is correct.

The correct solvent mass is 200 g = 0.200 kg. If the student uses 200 kg, the solvent mass is taken as 1000 times too large. In formulas such as ΔTf = Kf m, the calculated number of solute moles becomes 1000 times too large for the same measured effect. Since molar mass equals solute mass divided by moles, the calculated molar mass becomes 1000 times smaller. Therefore, option A is correct.

First calculate the effective molality from ΔTf = iKf m: m = 0.372/(2 × 1.86) = 0.100 mol kg−1. The solvent mass is 250 g = 0.250 kg, so moles of solute = 0.100 × 0.250 = 0.0250 mol. Thus the true molar mass is 2.0/0.0250 = 80 g mol−1. Option B is correct.

For a solution showing abnormal behaviour, osmotic pressure is π = iCRT. Therefore, C = π/(iRT) = 0.615/(1.25 × 0.082 × 300) = 0.020 mol L−1. In 0.100 L, the moles are 0.020 × 0.100 = 0.0020 mol. Hence molar mass = 1.0/0.0020 = 500 g mol−1, so C is correct.

Because the solute does not dissociate, i = 1. Thus ΔTf = Kf m, so the molality is m = 0.372/1.86 = 0.200 mol kg−1. The solvent mass is 300 g = 0.300 kg, giving moles of solute = 0.200 × 0.300 = 0.0600 mol. Therefore, molar mass = 4.8/0.0600 = 80 g mol−1, option C.

For a non-electrolytic solute, i = 1, so ΔTb = Kb m. The molality is m = 0.156/0.52 = 0.300 mol kg−1. The solvent mass is 150 g = 0.150 kg, hence moles of solute = 0.300 × 0.150 = 0.0450 mol. The molar mass is therefore 2.7/0.0450 = 60 g mol−1, making B correct.

Molarity is moles of solute per litre of solution. First convert the volume: 250 mL = 0.250 L. Moles of solute = M × V = 0.01 mol L⁻¹ × 0.250 L = 0.0025 mol. Molar mass = mass ÷ moles = 0.75 g ÷ 0.0025 mol = 300 g mol⁻¹. Therefore, option C is correct.

Use the relation moles = molarity × volume in litres. Convert 750 mL to 0.750 L. Thus, moles of solute = 0.02 mol L⁻¹ × 0.750 L = 0.015 mol. The molar mass is mass divided by amount: 2.7 g ÷ 0.015 mol = 180 g mol⁻¹. Hence, option D is the correct answer.

For a non-electrolyte, the freezing-point relation is ΔT_f = K_f m, where m is molality. Thus, m = 0.186/1.86 = 0.100 mol kg⁻¹. The solvent mass is 100 g = 0.100 kg, so moles of solute = 0.100 × 0.100 = 0.0100 mol. Therefore, molar mass = mass/moles = 1.86/0.0100 = 186 g mol⁻¹. Hence, option C is correct.

Since the solute does not dissociate, ΔT_f = K_f m. Therefore, the molality is m = 0.558/1.86 = 0.300 mol kg⁻¹. The solvent mass is 250 g = 0.250 kg, so the amount of solute is n = 0.300 × 0.250 = 0.0750 mol. Its molar mass is M = 6/0.0750 = 80 g mol⁻¹. Thus, option B is the only correct answer.

For a non-dissociated solute, boiling-point elevation follows ΔT_b = K_b m. Hence, m = 0.312/0.52 = 0.600 mol kg⁻¹. The solvent mass is 100 g = 0.100 kg, so the solute amount is n = 0.600 × 0.100 = 0.0600 mol. The molar mass is therefore M = 3/0.0600 = 50 g mol⁻¹. Thus, option B is correct.

Molarity is moles of solute per litre of solution. First convert the volume: 125 mL = 0.125 L. The number of moles in the sample is n = Molarity × volume = 0.04 × 0.125 = 0.005 mol. Therefore, molar mass = mass/moles = 0.9/0.005 = 180 g mol⁻¹. Hence, option C is correct.

Molality is defined as moles of solute per kilogram of solvent. The solvent mass is 400 g = 0.400 kg. Therefore, moles of solute = molality × solvent mass = 0.05 × 0.400 = 0.020 mol. Using the given solute mass, molar mass = 2.4/0.020 = 120 g mol⁻¹. Thus, option C is correct.

For abnormal molar masses, the van’t Hoff factor relates the observed and true values by \(M_{\text{observed}}=M_{\text{true}}/i\). Therefore, \(M_{\text{true}}=iM_{\text{observed}}\). Substituting the data gives \(M_{\text{true}}=1.25\times80=100\,g\,mol^{-1}\). Since \(i>1\), dissociation has increased the number of solute particles and makes the observed molar mass smaller than the true value. Thus, option C is correct.

The relation between observed and true molar mass is \(M_{\text{observed}}=M_{\text{true}}/i\), so \(M_{\text{true}}=iM_{\text{observed}}\). Using the given values, \(M_{\text{true}}=0.75\times160=120\,g\,mol^{-1}\). Because \(i<1\), association has reduced the number of solute particles; consequently, the observed molar mass is greater than the true molar mass. Therefore, option C is correct.

An \(AB\) molecule dissociates into two particles. Therefore, for degree of dissociation \(\alpha\), \(i=1+\alpha(2-1)=1+\alpha\). With \(\alpha=0.25\), \(i=1.25\). Since \(M_{\text{observed}}=M_{\text{true}}/i\), the true molar mass is \(M_{\text{true}}=1.25\times80=100\,g\,mol^{-1}\). Hence, option C is correct.

The van’t Hoff relation for abnormal molar mass is \(M_{\text{observed}}=M_{\text{true}}/i\). Rearranging gives \(M_{\text{true}}=iM_{\text{observed}}\). Substitution gives \(M_{\text{true}}=1.5\times72=108\,g\,mol^{-1}\). Since the factor is greater than one, dissociation increases the particle count and causes the observed molar mass to be lower than the true value. Thus, C is correct.