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MHT-CET Chemistry · Solutions and Colligative Properties

Elevation of Boiling Point

A non-volatile solute raises the boiling point by ΔTb = Kb · m (times i for an electrolyte); rearranged, M₂ = 1000 Kb W₂/(ΔTb W₁) gives the solute's molar mass.

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Why this matters

28 PYQs, two HARD — the chapter's second-largest page and the purest formula drill on it. Molality from ΔTb and Kb (eleven sittings), Kb from a boiling point, moles or solvent mass from the same equation, the molar-mass rearrangement, and the ranking of electrolyte solutions by i·m (five sittings). Two things cost marks: the solvent mass must be in kilograms, and a stem that gives the boiling POINT needs ΔTb subtracted first.

Concept 1 of 3

ΔTb = Kb · m: Solve for Any One of Molality, Kb, Moles or Solvent Mass

Intuition

The elevation is proportional to how many solute particles there are per kilogram of solvent. Kb is the solvent's constant (the elevation a 11 molal solution would show); everything else in the stem is either given or asked.

Definition

  • m=ΔTbKbm = \dfrac{\Delta T_b}{K_b}: 1.753.5=0.5\dfrac{1.75}{3.5} = 0.5, 0.52.40=0.21\dfrac{0.5}{2.40} = 0.21, 1.893.15=0.6\dfrac{1.89}{3.15} = 0.6, 7.152.75=2.6\dfrac{7.15}{2.75} = 2.6, 0.20.52=0.385\dfrac{0.2}{0.52} = 0.385, 0.390.52=0.75\dfrac{0.39}{0.52} = 0.75 mol kg−1^{-1}.
  • Kb=ΔTbmK_b = \dfrac{\Delta T_b}{m}: boiling point 319.8319.8 K against the solvent's 319.5319.5 K gives ΔTb=0.3\Delta T_b = 0.3, and 0.120.12 m gives Kb=2.5K_b = 2.5 K kg mol−1^{-1}. From masses: 1.51.5 g of a 150150 g mol−1^{-1} solute in 3030 g solvent is m=0.333m = 0.333, so Kb=0.650.333=1.95K_b = \dfrac{0.65}{0.333} = 1.95.
  • Moles n=m×Wsolvent(kg)n = m \times W_{\text{solvent}}(\text{kg}): ΔTb=0.8\Delta T_b = 0.8, Kb=2K_b = 2, 0.50.5 kg gives n=0.2n = 0.2. Solvent mass =nm= \dfrac{n}{m}: 0.010.01 mol with m=0.3m = 0.3 gives 0.0330.033 kg.
  • Elevation is proportional to molality: 0.10.1 m glucose boils at 100.16∘100.16^\circC, so 0.50.5 m boils at 100.8∘100.8^\circC. Equal boiling points mean equal molality: 1818 g dm−3^{-3} glucose (0.10.1 mol) and 66 g dm−3^{-3} of A give MA=60M_A = 60.
  • Mass of solute from a boiling point: 85∘85^\circC against 76∘76^\circC with Kb=2.7K_b = 2.7: m=92.7m = \dfrac{9}{2.7}, moles =103×0.16= \dfrac{10}{3} \times 0.16, mass =64= 64 g.

Boiling point elevation

ΔTb=Kb m=Kb n2W1(kg)Tb=Tb∘+ΔTb\Delta T_b = K_b\,m = K_b\,\frac{n_2}{W_1(\text{kg})} \qquad T_b = T_b^\circ + \Delta T_b

Worked example

A solution of 22 g of a non-volatile solute (M=80M = 80 g mol−1^{-1}) in 5050 g of a solvent boils 0.90.9 K above the pure solvent. Find KbK_b.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Solutions and Colligative PropertiesEASY
A solution of non-volatile solute has boiling point elevation 1.75 K. Calculate molality of solution [Kb=3.5 K kg mol−1K_b=3.5\ \text{K kg mol}^{-1}]

[Q51 · 3rd May 2nd Shift · 2023]

Grams where kilograms belong

Molality is per KILOGRAM of solvent. 3030 g is 0.030.03 kg; leaving it as 3030 drops the answer by a factor of 10001000, and the option list is built around that.
Drill 17 more on δtb = kb · m: solve for any one of molality, kb, moles or solvent mass

Concept 2 of 3

Molar Mass From ΔTb: M₂ = 1000 Kb W₂ / (ΔTb W₁)

Intuition

Substitute m=W2×1000M2W1m = \dfrac{W_2 \times 1000}{M_2 W_1} into ΔTb=Kbm\Delta T_b = K_b m and solve for M2M_2. W2W_2 is the solute mass, W1W_1 the solvent mass in grams; the 10001000 converts grams to kilograms.

Definition

  • M2=1000 Kb W2ΔTb W1M_2 = \dfrac{1000\,K_b\,W_2}{\Delta T_b\,W_1}: 55 g in 5050 g boiling 1.61.6 K high with Kb=3.2K_b = 3.2: 1000×3.2×51.6×50=200\dfrac{1000 \times 3.2 \times 5}{1.6 \times 50} = 200; 0.350.35 g in 100100 g, ΔTb=0.01\Delta T_b = 0.01, Kb=0.5K_b = 0.5: 175175; 3.53.5 g in 100100 g, 0.350.35 K, Kb=2.5K_b = 2.5: 250250.
  • Recognise the formula among rearrangements: KbK_b and W2W_2 in the numerator, ΔTb\Delta T_b and W1W_1 in the denominator.
  • A boiling POINT (119.6∘119.6^\circC against 118∘118^\circC) must first become ΔTb=1.6\Delta T_b = 1.6 K.

Molar mass

M2=1000 Kb W2ΔTb W1M_2 = \frac{1000\,K_b\,W_2}{\Delta T_b\,W_1}

Worked example

2.42.4 g of a solute in 6060 g of benzene (Kb=2.53K_b = 2.53 K kg mol−1^{-1}) raises the boiling point by 0.5060.506 K. Find the molar mass.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Solutions and Colligative PropertiesMODERATE
Calculate the molar mass of a non-volatile solute when 5 g of it is dissolved in 50 g solvent, boils at 119.6∘C119.6^\circ\text{C}. (Kb=3.2 K kg mol−1K_b = 3.2\,\text{K kg mol}^{-1}, boiling point of pure solvent =118∘C= 118^\circ\text{C})

[Q90 · 2nd May Shift 2 · 2023]

Swapping W₁ and W₂

W2W_2 (solute) is on top, W1W_1 (solvent) below. The swap changes 200200 to 22 — not on the list — but on the formula-recognition stem the swapped option is (B).
Drill 3 more on molar mass from δtb: m₂ = 1000 kb w₂ / (δtb w₁)

Concept 3 of 3

Ranking Electrolyte Solutions: Compare i × m

Intuition

For an electrolyte ΔTb=iKbm\Delta T_b = i K_b m, and with complete dissociation ii is the number of ions per formula unit. At the same KbK_b, the solution with the largest i×mi \times m has the largest elevation; compute the product for each option and rank.

Definition

  • ii: KCl, NaCl, KNO3_3, MgSO4_4, AlPO4_4 → 22; BaCl2_2, CaCl2_2, MgCl2_2, Na2_2SO4_4 → 33; AlCl3_3 → 44; Al2_2(SO4_4)3_3 → 55.
  • Maximum elevation among 0.10.1 m KCl (0.20.2), 0.050.05 m NaCl (0.10.1), 0.10.1 m BaCl2_2 (0.30.3), 0.10.1 m MgSO4_4 (0.20.2): BaCl2_2. Equimolal: AlCl3_3 (i=4i = 4) beats BaCl2_2, KCl, NaCl.
  • Minimum: 0.010.01 m MgCl2_2 (0.030.03) below 0.10.1 m AlCl3_3, 11 m KCl, 0.50.5 m NaCl; 0.050.05 m CaCl2_2 (0.150.15) below 0.10.1 m NaCl, 0.20.2 m KNO3_3, 0.10.1 m Na2_2SO4_4; 0.050.05 m MgCl2_2 (0.150.15) below 0.20.2 m KCl, 0.10.1 m NaCl, 11 m AlCl3_3.
  • True/false stems: the boiling point of a solution of a non-volatile solute is ALWAYS higher than the pure solvent's; the relative lowering equals the mole fraction of the SOLUTE (not the solvent).

Electrolyte elevation

ΔTb=i Kb mrank by i×m\Delta T_b = i\,K_b\,m \qquad \text{rank by } i \times m

Worked example

Rank by boiling point elevation: 0.10.1 m Na2_2SO4_4, 0.150.15 m KCl, 0.050.05 m Al2_2(SO4_4)3_3, 0.20.2 m glucose.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Solutions and Colligative PropertiesMODERATE
Which of the following solutions on complete dissociation exhibits maximum elevation in boiling point?

[Q88 · 4th May Shift 1 · 2023]

Counting AlPO₄ as five ions

AlPO4_4 gives Al3+^{3+} and PO43−_4^{3-}: TWO ions, i=2i = 2, like MgSO4_4. Al2_2(SO4_4)3_3 is the one with five.
Drill 5 more on ranking electrolyte solutions: compare i × m

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