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

44 PYQs, none HARD, and now the chapter's largest page. 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

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift I · Q71Easy

Example 1 · Solutions and Colligative Properties · Osmotic Pressure

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}

]

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)

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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 2 · Q56Easy

Example 2 · Solutions and Colligative Properties · Osmotic Pressure

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}]

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

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?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q58Moderate

Example 3 · Solutions and Colligative Properties · Osmotic Pressure

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

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

Test yourself on Solutions and Colligative Properties

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Solutions and Colligative Properties chapter test Drill all 44 questions