MHT-CET Chemistry · Solutions and Colligative Properties

Osmotic Pressure

π = CRT = nRT/V, the van't Hoff equation: osmotic pressure is proportional to molar concentration and temperature; solved for n it gives the molar mass, and equal π at equal T means isotonic.

All Solutions and Colligative Properties notes MHT-CET Chemistry strategy

Why this matters

22 PYQs, none HARD. The van't Hoff equation solved for π, C, T or the molar mass (with R = 0.082 dm³ atm K⁻¹ mol⁻¹ and V in dm³), the isotonic pair of urea and sucrose solutions (three sittings), and rankings of electrolytes by i·m — the same ranking as the two previous pages. Units are the whole difficulty: litres for V, kelvin for T, and mass in grams over molar mass for n.

Concept 1 of 3

π = CRT: Solve for π, C, T or n

Intuition

Osmotic pressure behaves like the pressure of an ideal gas at the solution's molar concentration:

\pi V = nRT

. With

R = 0.082

^3

atm K

^{-1}

mol

^{-1}

V

in dm

^3

and

T

in kelvin,

\pi

comes out in atmospheres.

Definition

$\pi = CRT$ : $0.5$ M at $300$ K gives $0.5 \times 0.0821 \times 300 = 12.32$ atm; $0.025$ mol in $100$ mL gives $\dfrac{0.025 \times 0.082 \times 300}{0.1} = 6.15$ atm; $0.03$ mol in $0.1$ dm $^3$ : $7.4$ atm.
$C = \dfrac{\pi}{RT}$ : $12$ atm at $300$ K gives $\dfrac{12}{24.63} = 0.487$ M. $T = \dfrac{\pi}{CR}$ : $1.5$ atm for $0.05$ M gives $365.4$ K.
$\pi \propto C$ at fixed $T$ : $0.2$ M at $4.9$ atm means $1.5$ atm needs $0.06$ M; $1$ M urea has twice the osmotic pressure of $0.5$ M urea.
Osmotic pressure is the colligative property used for macromolecules because it is measurable at room temperature and large even for dilute solutions.

Van't Hoff equation

\pi = CRT = \frac{n}{V}RT \qquad (R = 0.082\ \text{dm}^3\,\text{atm K}^{-1}\text{mol}^{-1})

Worked example

Find the osmotic pressure of a solution of

0.04

mol of a non-electrolyte in

250

mL of water at

27^\circ

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Solutions and Colligative PropertiesEASY

Calculate the osmotic pressure of 0.5 M aqueous solution of nonvolatile solute at 300 K . [

R = 0.0821\text{ }atmdm^{3}{\text{ }K}^{- 1}{\text{ }mol}^{- 1}

]

[Q71 · 22 April Shift I · 2025]

Volume in millilitres

0.025

mol in

100

mL is

0.25

M, not

0.00025

M. Convert to dm

^3

first; the mL version gives

0.00615

atm and a wrong pick.

Drill 6 more on π = crt: solve for π, c, t or n

Concept 2 of 3

Molar Mass From Osmotic Pressure: M = W R T / (π V)

Intuition

Put

n = \dfrac{W}{M}

into

\pi V = nRT

M = \dfrac{WRT}{\pi V}

. The same equation solved for

W

tells how much solute produces a target osmotic pressure.

Definition

$4$ g in $1$ dm $^3$ , $2$ atm, $300$ K: $M = \dfrac{4 \times 0.082 \times 300}{2 \times 1} = 49.2$ . $0.8$ g in $0.3$ dm $^3$ , $0.2$ atm: $328$ . $0.4$ g in $300$ mL, $0.2$ atm: $164$ . $1$ g in $300$ mL, $0.2$ atm: $410$ . $8$ g in $2$ dm $^3$ , $0.6$ atm: $164$ .
$\pi$ from a known $M$ : $3$ g of a $60$ g mol $^{-1}$ solute in $2$ dm $^3$ at $300$ K: $\dfrac{0.05}{2} \times 24.63 = 0.62$ atm.
Mass for a target $\pi$ : $0.245$ atm in $2.5$ dm $^3$ at $300$ K needs $n = \dfrac{0.245 \times 2.5}{24.63} = 0.0249$ mol, i.e. $1.44$ g of a $58$ g mol $^{-1}$ solute.
Milligrams to grams ( $400$ mg $= 0.4$ g) and millilitres to dm $^3$ before substituting.

Molar mass

M = \frac{W\,R\,T}{\pi\,V}

Worked example

2

g of a protein in

500

mL of water has an osmotic pressure of

0.0246

atm at

300

K. Find its molar mass.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Solutions and Colligative PropertiesEASY

Calculate the molar mass of solute when 4 g of it dissolved in 1 dm³ solvent has osmotic pressure 2 atm at 300 K.

[R = 0.082\text{ dm}^3\text{ atm K}^{-1}\text{ mol}^{-1}]

[Q56 · 4th May Shift 2 · 2023]

Leaving V in millilitres

300

mL is

0.3

^3

; with

300

in the denominator the molar mass comes out

1000

times too small. The options are spaced to catch a factor-of-ten slip, not this one — so the mistake shows as 'none of the options', a sign to recheck units.

Drill 6 more on molar mass from osmotic pressure: m = w r t / (π v)

Concept 3 of 3

Isotonic and Hypertonic Solutions, and π = iCRT for Electrolytes

Intuition

Two solutions with the same osmotic pressure at the same temperature are isotonic and no solvent flows between them across a membrane; the one with higher

\pi

is hypertonic. For a non-electrolyte that means equal molarity; for electrolytes, equal

i \times C

Definition

$6$ g dm $^{-3}$ urea ( $0.1$ M) and $34.2$ g dm $^{-3}$ sucrose ( $0.1$ M) are isotonic; $3$ g dm $^{-3}$ urea and $17.1$ g dm $^{-3}$ sucrose ( $0.05$ M each) likewise. $6$ g dm $^{-3}$ urea against $17.12$ g dm $^{-3}$ sucrose ( $0.05$ M): urea is hypertonic.
$\pi = iCRT$ : $0.2$ M KCl with $i = 1.83$ at $273$ K gives $8.2$ atm; $1.7$ g CaCl $_2$ ( $111$ g mol $^{-1}$ ) in $1.25$ dm $^3$ with $i = 2.47$ at $300$ K gives $0.744$ atm.
Rankings by $i \times m$ (complete dissociation): $0.1$ m BaCl $_2$ ( $0.3$ ) $<$ $0.5$ m KCl ( $1$ ) $<$ $0.5$ m Li $_2$ SO $_4$ ( $1.5$ ) $<$ $0.5$ m Al $_2$ (SO $_4$ ) $_3$ ( $2.5$ ); equimolar KCl $<$ BaCl $_2$ $<$ AlCl $_3$ $<$ Al $_2$ (SO $_4$ ) $_3$ ( $i = 2, 3, 4, 5$ ); decreasing: $0.5$ m Al $_2$ (SO $_4$ ) $_3$ ( $2.5$ ) $>$ $0.3$ m MgSO $_4$ ( $0.6$ ) $>$ $0.2$ m KCl ( $0.4$ ) $>$ $0.1$ m BaCl $_2$ ( $0.3$ ).
Isotonic in the exam's sense compares molar concentrations, not masses per litre; convert every 'g dm $^{-3}$ ' by the molar mass first.

Electrolytes and isotonicity

\pi = i\,C\,R\,T \qquad \text{isotonic} \iff \pi_1 = \pi_2 \iff (iC)_1 = (iC)_2

Worked example

Are

9

g dm

^{-3}

glucose (

180

) and

3

g dm

^{-3}

urea (

60

) isotonic? If not, which is hypertonic?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Solutions and Colligative PropertiesMODERATE

Which of the following solutions will not show flow of solvent in either direction when separated by semipermeable membrane?

[Q58 · 11th May Shift 2 · 2023]

Comparing grams per litre directly

6

g of urea and

34.2

g of sucrose are the SAME number of moles. Isotonicity is about molarity; the mass-matched pair (

6

and

6

) is never the answer.

Drill 7 more on isotonic and hypertonic solutions, and π = icrt for electrolytes

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)

π = CRT: Solve for π, C, T or n
Van't Hoff equation
$\pi = CRT = \frac{n}{V}RT \qquad (R = 0.082\ \text{dm}^3\,\text{atm K}^{-1}\text{mol}^{-1})$
Molar Mass From Osmotic Pressure: M = W R T / (π V)
Molar mass
$M = \frac{W\,R\,T}{\pi\,V}$
Isotonic and Hypertonic Solutions, and π = iCRT for Electrolytes
Electrolytes and isotonicity
$\pi = i\,C\,R\,T \qquad \text{isotonic} \iff \pi_1 = \pi_2 \iff (iC)_1 = (iC)_2$

Watch out for (3)

Volume in millilitres
→ π = CRT: Solve for π, C, T or n
Leaving V in millilitres
→ Molar Mass From Osmotic Pressure: M = W R T / (π V)
Comparing grams per litre directly
→ Isotonic and Hypertonic Solutions, and π = iCRT for Electrolytes

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