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SubjectsChemistry
असामान्य आणविक द्रव्यमान
In Class 12 Chemistry, the topic Abnormal Molecular Mass explains why the molar mass calculated from colligative properties may differ from the actual molecular mass. Students learn how solute particles associate, such as through dimerisation, or dissociate into ions in solution, changing the number of particles present. They use the van’t Hoff factor to relate these changes to relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure, and calculate the degree of association or dissociation in solutions.
TOPIC PRACTICE
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Association decreases the number of independent solute particles, so the van’t Hoff factor becomes less than one. Since the molar mass calculated from a colligative property is M_observed = M_true/i, a value of i < 1 makes the observed molar mass greater than the true molar mass. Dissociation and ionisation produce the opposite effect, giving i > 1.
For abnormal molar mass, the van’t Hoff factor is related to the normal and observed molar masses by i = normal molar mass / observed molar mass. If the observed molar mass is twice the normal value, i = M/(2M) = 0.5. A value smaller than one means that solute particles have combined, such as by dimerisation or another association process, reducing the number of particles in solution.
For colligative-property measurements, the observed molar mass is related to the true molar mass by M_observed = M_true/i. When solute particles associate, two or more particles combine into one species, so the van’t Hoff factor becomes less than 1. Dividing by a number below 1 makes the observed molar mass larger than the normal value. Dissociation has the opposite effect because it gives i greater than 1.
For abnormal molar mass, the van’t Hoff factor is i = normal molar mass ÷ observed molar mass = 120 ÷ 60 = 2. For dissociation of one particle into two ions, i = 1 + α, where α is the fraction dissociated. Thus, 2 = 1 + α, so α = 1, or 100%. Therefore, option D is correct.
Benzoic acid associates to form dimers in benzene. Association decreases the number of solute particles, so the van’t Hoff factor becomes less than one. Since ΔTf = iKf m, the observed freezing-point depression is smaller than expected. When the ordinary formula is used, this small depression gives an observed molar mass greater than the actual value. Thus, option A is correct.
For abnormal molar mass, the van’t Hoff factor is i = true molar mass/observed molar mass. Since the observed value is 4/7 of the true value, i = 7/4 = 1.75. On dissociation, AB₃ produces four particles, so i = 1 + (4 − 1)α = 1 + 3α. Therefore, 1.75 = 1 + 3α, giving α = 0.25 or 25%. Hence, option A is correct.
If 40% of the molecules dimerise, 60% remain monomeric. The effective particle factor is i = (0.60) + (0.40/2) = 0.80, because each dimer represents two original molecules but only one solution particle. The observed molar mass is Mobs = Mnormal/i = 100/0.80 = 125 g mol⁻¹. Thus option C is correct.
On dissociation, one formula unit of AB₂ forms three ions: A + 2B. For a dissociation degree α = 0.75, the van’t Hoff factor is i = 1 + α(n − 1), where n = 3. Therefore, i = 1 + 0.75(3 − 1) = 2.5. The observed molar mass is normal molar mass divided by i: 120 ÷ 2.5 = 48 g mol⁻¹. Thus, option B is correct.
First calculate the van’t Hoff factor: i = normal molar mass ÷ observed molar mass = 120 ÷ 72 = 1.6667. An AB solute dissociates into two particles, so for degree of dissociation α, i = 1 + α(2 − 1) = 1 + α. Therefore, α = 1.6667 − 1 = 0.6667, or approximately 66.7%. Hence, option C is correct.
The van’t Hoff factor is i = normal molar mass ÷ observed molar mass = 60 ÷ 96 = 0.625. For dimerisation, two solute molecules combine to form one dimer, so i = 1 − α/2, where α is the fraction associated. Substituting gives 0.625 = 1 − α/2; hence α/2 = 0.375 and α = 0.75, or 75%. Therefore, option C is correct.
The solute’s normal number of moles is n = 4.0/100 = 0.04 mol. Since the solvent mass is 200 g = 0.200 kg, the normal molality is m = 0.04/0.200 = 0.20 mol kg−1. The expected depression without abnormality is Kf m = 1.86 × 0.20 = 0.372 K, exactly equal to the observed value. Hence i = observed depression/normal depression = 1.0.
The relation between normal and observed molar mass is i = Mnormal/Mobserved. Hence Mobserved = 80/1.5 = 53.33 g mol−1, approximately 53.3 g mol−1. For an AB electrolyte dissociating as AB → A+ + B−, one formula unit produces two particles, so i = 1 + α. Therefore α = i − 1 = 0.5, or 50%.
Assume one mole of original molecules. Forty percent remain as monomers, contributing 0.40 effective particles, while 60% associate into trimers and contribute 0.60/3 = 0.20 particles. Thus the effective particle factor is i = 0.40 + 0.20 = 0.60. Association makes the observed molar mass larger, and Mobserved = Mnormal/i = 90/0.60 = 150 g mol−1.
Using the freezing-point equation, the molar mass calculated without correcting for association is M_observed = K_f × w × 1000/(ΔT_f × W) = 1.86 × 0.5 × 1000/(0.093 × 100) = 100 g mol⁻¹. For 50% dimerisation, the van’t Hoff factor is i = 1 − α/2 = 1 − 0.50/2 = 0.75. Since M_observed = M_normal/i, M_normal = iM_observed = 0.75 × 100 = 75 g mol⁻¹. Therefore, option B is correct.
If the AB salt did not dissociate, its molality would be (1.0/100)/0.100 = 0.10 mol kg⁻¹. The corresponding freezing-point depression would be K_fm = 1.86 × 0.10 = 0.186 K. The observed value is 0.372 K, exactly twice this value, so i = 0.372/0.186 = 2. For AB → A⁺ + B⁻, i = 1 + α. Therefore α = 2 − 1 = 1, or 100%. Option D is correct.
One formula unit of A₂B produces three ions on complete dissociation: 2A⁺ and B²⁻. For a compound giving ν = 3 particles, the van’t Hoff factor is i = 1 + α(ν − 1). With α = 0.40, i = 1 + 0.40(3 − 1) = 1.80. The observed molar mass is related to the normal molar mass by M_observed = M_normal/i, so M_observed = 150/1.80 = 83.3 g mol⁻¹. Option B is correct.
For dimer association, two original molecules combine to form one particle. If the degree of association is α = 0.25, the number of particles becomes 1 − α/2 of the original number. Thus the van’t Hoff factor is i = 1 − 0.25/2 = 0.875. Since the observed molar mass is M_observed = M_normal/i, the normal value is M_normal = iM_observed = 0.875 × 200 = 175 g mol⁻¹. Therefore, option B is correct.
The number of moles of solute is 0.8/40 = 0.02 mol, while the solvent mass is 0.200 kg. Therefore, the molality is 0.02/0.200 = 0.10 mol kg⁻¹. For a normally behaving solute, ΔTf = Kf m = 1.86 × 0.10 = 0.186 K, exactly equal to the observed depression. Hence the van’t Hoff factor is i = 1, showing neither association nor dissociation; the solute exhibits normal behaviour.
For osmotic pressure, π = iCRT. Substituting the data gives i = 2.46/(0.05 × 0.082 × 300). The denominator is 1.23 atm, so i = 2. A van’t Hoff factor of two means that the solution produces twice as many effective particles as the undissociated solute formula units. Thus one formula unit behaves as two particles, as in complete dissociation of a 1:1 electrolyte such as AB into A⁺ and B⁻.
For abnormal molar mass, the van’t Hoff factor is related to molar masses by i = Mnormal/Mobserved = 120/80 = 1.5. Effective molality is i times the normal or analytical molality. Therefore, normal molality : effective molality = m : 1.5m = 1:1.5 = 2:3. The supplied masses are not needed because the requested ratio depends only on the two molar masses.
Using π = iCRT, i = 7.38/(0.2 × 0.082 × 300). The denominator is 4.92 atm, so i = 1.5. For an AB electrolyte, complete dissociation would produce two particles and therefore i = 2 under ideal conditions. Since the measured value lies between 1 and 2, the electrolyte undergoes partial, not complete, dissociation. Hence option A is correct.
One formula unit of AB₃ produces four ions on complete dissociation: A³⁺ and three B⁻ ions. For degree of dissociation α = 0.5, the van’t Hoff factor is i = 1 + α(ν − 1), where ν = 4. Thus i = 1 + 0.5(3) = 2.5. Since Mobserved = Mnormal/i, the observed molar mass is 200/2.5 = 80 g mol⁻¹. Therefore, option B is correct.
The relation between normal and observed molar masses is Mobserved = Mnormal/i. If the observed molar mass is half the normal value, then i = 2. For AB₃, complete dissociation gives four particles, so i = 1 + α(4 − 1) = 1 + 3α. Substituting i = 2 gives 2 = 1 + 3α, hence α = 1/3 = 0.333. Therefore, the degree of dissociation is 33.3%.
The van’t Hoff factor for an association process is i = Mtrue/Mobserved, because association reduces the number of solute particles. Therefore i = 90/120 = 0.75. For dimerisation, two original molecules combine to form one dimer, and the relation is i = 1 − α/2, where α is the fraction associated. Substitution gives 0.75 = 1 − α/2, so α/2 = 0.25 and α = 0.50. The degree of association is therefore 50%.
The van’t Hoff factor is i = Mtrue/Mobserved. Since the observed molar mass is (4/7)Mtrue, i = Mtrue/[(4/7)Mtrue] = 7/4 = 1.75. An AB3 solute gives four particles on complete dissociation: AB3 → A + 3B. Therefore i = 1 + 3α. Substituting 1.75 = 1 + 3α gives α = 0.75/3 = 0.25. Expressed as a percentage, the degree of dissociation is 25%.
QUIZ COMPLETE
