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

43 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

The elevation is proportional to how many solute particles there are per kilogram of solvent. Kb is the solvent's constant (the elevation a

1

molal solution would show); everything else in the stem is either given or asked.

Definition

$m = \dfrac{\Delta T_b}{K_b}$ : $\dfrac{1.75}{3.5} = 0.5$ , $\dfrac{0.5}{2.40} = 0.21$ , $\dfrac{1.89}{3.15} = 0.6$ , $\dfrac{7.15}{2.75} = 2.6$ , $\dfrac{0.2}{0.52} = 0.385$ , $\dfrac{0.39}{0.52} = 0.75$ mol kg $^{-1}$ .
$K_b = \dfrac{\Delta T_b}{m}$ : boiling point $319.8$ K against the solvent's $319.5$ K gives $\Delta T_b = 0.3$ , and $0.12$ m gives $K_b = 2.5$ K kg mol $^{-1}$ . From masses: $1.5$ g of a $150$ g mol $^{-1}$ solute in $30$ g solvent is $m = 0.333$ , so $K_b = \dfrac{0.65}{0.333} = 1.95$ .
Moles $n = m \times W_{\text{solvent}}(\text{kg})$ : $\Delta T_b = 0.8$ , $K_b = 2$ , $0.5$ kg gives $n = 0.2$ . Solvent mass $= \dfrac{n}{m}$ : $0.01$ mol with $m = 0.3$ gives $0.033$ kg.
Elevation is proportional to molality: $0.1$ m glucose boils at $100.16^\circ$ C, so $0.5$ m boils at $100.8^\circ$ C. Equal boiling points mean equal molality: $18$ g dm $^{-3}$ glucose ( $0.1$ mol) and $6$ g dm $^{-3}$ of A give $M_A = 60$ .
Mass of solute from a boiling point: $85^\circ$ C against $76^\circ$ C with $K_b = 2.7$ : $m = \dfrac{9}{2.7}$ , moles $= \dfrac{10}{3} \times 0.16$ , mass $= 64$ g.

Boiling point elevation

\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

2

g of a non-volatile solute (

M = 80

g mol

^{-1}

) in

50

g of a solvent boils

0.9

K above the pure solvent. Find

K_b

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q51Easy

Example 1 · Solutions and Colligative Properties · Elevation of Boiling Point

A solution of non-volatile solute has boiling point elevation 1.75 K. Calculate molality of solution [

K_b=3.5\ \text{K kg mol}^{-1}

]

Grams where kilograms belong

Molality is per KILOGRAM of solvent.

30

g is

0.03

kg; leaving it as

30

drops the answer by a factor of

1000

, 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)

Substitute

m = \dfrac{W_2 \times 1000}{M_2 W_1}

into

\Delta T_b = K_b m

and solve for

M_2

W_2

is the solute mass,

W_1

the solvent mass in grams; the

1000

converts grams to kilograms.

Definition

$M_2 = \dfrac{1000\,K_b\,W_2}{\Delta T_b\,W_1}$ : $5$ g in $50$ g boiling $1.6$ K high with $K_b = 3.2$ : $\dfrac{1000 \times 3.2 \times 5}{1.6 \times 50} = 200$ ; $0.35$ g in $100$ g, $\Delta T_b = 0.01$ , $K_b = 0.5$ : $175$ ; $3.5$ g in $100$ g, $0.35$ K, $K_b = 2.5$ : $250$ .
Recognise the formula among rearrangements: $K_b$ and $W_2$ in the numerator, $\Delta T_b$ and $W_1$ in the denominator.
A boiling POINT ( $119.6^\circ$ C against $118^\circ$ C) must first become $\Delta T_b = 1.6$ K.

Molar mass

M_2 = \frac{1000\,K_b\,W_2}{\Delta T_b\,W_1}

Worked example

2.4

g of a solute in

60

g of benzene (

K_b = 2.53

K kg mol

^{-1}

) raises the boiling point by

0.506

K. Find the molar mass.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q100Moderate

Example 2 · Solutions and Colligative Properties · Elevation of Boiling Point

Calculate the molar mass of a non-volatile solute when 5 g of it is dissolved in 50 g solvent, boils at

119.6^\circ\text{C}

. (

K_b = 3.2\,\text{K kg mol}^{-1}

, boiling point of pure solvent

= 118^\circ\text{C}

)

Swapping W₁ and W₂

W_2

(solute) is on top,

W_1

(solvent) below. The swap changes

200

2

— 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

For an electrolyte

\Delta T_b = i K_b m

, and with complete dissociation

i

is the number of ions per formula unit. At the same

K_b

, the solution with the largest

i \times m

has the largest elevation; compute the product for each option and rank.

Definition

$i$ : KCl, NaCl, KNO $_3$ , MgSO $_4$ , AlPO $_4$ → $2$ ; BaCl $_2$ , CaCl $_2$ , MgCl $_2$ , Na $_2$ SO $_4$ → $3$ ; AlCl $_3$ → $4$ ; Al $_2$ (SO $_4$ ) $_3$ → $5$ .
Maximum elevation among $0.1$ m KCl ( $0.2$ ), $0.05$ m NaCl ( $0.1$ ), $0.1$ m BaCl $_2$ ( $0.3$ ), $0.1$ m MgSO $_4$ ( $0.2$ ): BaCl $_2$ . Equimolal: AlCl $_3$ ( $i = 4$ ) beats BaCl $_2$ , KCl, NaCl.
Minimum: $0.01$ m MgCl $_2$ ( $0.03$ ) below $0.1$ m AlCl $_3$ , $1$ m KCl, $0.5$ m NaCl; $0.05$ m CaCl $_2$ ( $0.15$ ) below $0.1$ m NaCl, $0.2$ m KNO $_3$ , $0.1$ m Na $_2$ SO $_4$ ; $0.05$ m MgCl $_2$ ( $0.15$ ) below $0.2$ m KCl, $0.1$ m NaCl, $1$ m AlCl $_3$ .
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

\Delta T_b = i\,K_b\,m \qquad \text{rank by } i \times m

Worked example

Rank by boiling point elevation:

0.1

m Na

_2

_4

0.15

m KCl,

0.05

m Al

_2

(SO

_4

)

_3

0.2

m glucose.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q88Moderate

Example 3 · Solutions and Colligative Properties · Elevation of Boiling Point

Which of the following solutions on complete dissociation exhibits maximum elevation in boiling point?

Counting AlPO₄ as five ions

AlPO

_4

gives Al

^{3+}

and PO

_4^{3-}

: TWO ions,

i = 2

, like MgSO

_4

. Al

_2

(SO

_4

)

_3

is the one with five.

Drill 5 more on ranking electrolyte solutions: compare i × m

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

ΔTb = Kb · m: Solve for Any One of Molality, Kb, Moles or Solvent Mass
Boiling point elevation
$\Delta T_b = K_b\,m = K_b\,\frac{n_2}{W_1(\text{kg})} \qquad T_b = T_b^\circ + \Delta T_b$
Molar Mass From ΔTb: M₂ = 1000 Kb W₂ / (ΔTb W₁)
Molar mass
$M_2 = \frac{1000\,K_b\,W_2}{\Delta T_b\,W_1}$
Ranking Electrolyte Solutions: Compare i × m
Electrolyte elevation
$\Delta T_b = i\,K_b\,m \qquad \text{rank by } i \times m$

Watch out for (3)

Grams where kilograms belong
→ ΔTb = Kb · m: Solve for Any One of Molality, Kb, Moles or Solvent Mass
Swapping W₁ and W₂
→ Molar Mass From ΔTb: M₂ = 1000 Kb W₂ / (ΔTb W₁)
Counting AlPO₄ as five ions
→ Ranking Electrolyte Solutions: Compare i × m

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Related notes

Van't Hoff Factor — where i comes from when dissociation is incomplete →