Concentration Terms

Real-Life Hook: From Ocean Salt to Medicine Dosage

Why is the ocean salty but your swimming pool isn’t? It’s all about concentration! Ocean water has about 35,000 ppm (parts per million) of dissolved salts, while pool water has less than 3,000 ppm. When doctors prescribe IV fluids, they carefully control molarity to match blood concentration (0.9% NaCl = 0.154 M). Understanding concentration isn’t just academic - it’s literally vital for survival. Too concentrated, and your cells shrink; too dilute, and they burst!

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Introduction

Concentration expresses how much solute is present in a given amount of solution or solvent. Different concentration units are useful in different contexts:

• Molarity (M): Chemical reactions, titrations (depends on temperature)
• Molality (m): Colligative properties, doesn’t change with temperature
• Mole fraction (χ): Ideal solutions, Raoult’s law, theoretical calculations
• Mass percentage: Commercial solutions, everyday use
• ppm/ppb: Trace amounts, environmental chemistry, pollutants

For JEE, interconversion between these units is heavily tested!

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Molarity (M)

Definition

Molarity is the number of moles of solute per liter of solution.

$$\boxed{M = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}}}$$ $$\boxed{M = \frac{n}{V(L)} = \frac{m}{M \times V(L)}}$$

Where:

• n = moles of solute
• m = mass of solute (g)
• M = molar mass (g/mol)
• V = volume of solution (L)

Important Points

1. Temperature dependent: Volume changes with temperature
2. Measured in mol/L or M
3. Used in: Volumetric analysis, reaction calculations
4. Solution volume, not solvent volume!

Solved Example 1: Basic Molarity

40 g of NaOH is dissolved in water to make 500 mL solution. Calculate molarity.

Solution:

Moles of NaOH = 40/40 = 1 mol Volume = 500 mL = 0.5 L

$$M = \frac{1}{0.5} = 2 \text{ M}$$ $$\boxed{M = 2 \text{ M}}$$

Solved Example 2: From Density

A solution has 20% H₂SO₄ by mass and density 1.14 g/mL. Calculate molarity.

Solution:

Assume 100 g solution:

• Mass of H₂SO₄ = 20 g
• Moles of H₂SO₄ = 20/98 = 0.204 mol

Volume of 100 g solution:

$$V = \frac{100}{1.14} = 87.72 \text{ mL} = 0.08772 \text{ L}$$ $$M = \frac{0.204}{0.08772} = 2.33 \text{ M}$$ $$\boxed{M = 2.33 \text{ M}}$$

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Molality (m)

Definition

Molality is the number of moles of solute per kilogram of solvent.

$$\boxed{m = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}}$$ $$\boxed{m = \frac{n}{m_{\text{solvent}}(kg)} = \frac{m_{\text{solute}}}{M \times m_{\text{solvent}}(kg)}}$$

Important Points

1. Temperature independent: Mass doesn’t change with temperature
2. Measured in mol/kg or m
3. Used in: Colligative properties (ΔTf, ΔTb, osmotic pressure)
4. Solvent mass, not solution mass!

Solved Example 3: Basic Molality

18 g glucose (M = 180) is dissolved in 500 g water. Calculate molality.

Solution:

Moles of glucose = 18/180 = 0.1 mol Mass of water = 500 g = 0.5 kg

$$m = \frac{0.1}{0.5} = 0.2 \text{ m}$$ $$\boxed{m = 0.2 \text{ m}}$$

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Mole Fraction (χ)

Definition

Mole fraction is the ratio of moles of one component to total moles of all components.

$$\boxed{\chi_A = \frac{n_A}{n_A + n_B + n_C + \ldots} = \frac{n_A}{n_{\text{total}}}}$$

Important Properties

1. Dimensionless (no units)
2. Range: 0 ≤ χ ≤ 1
3. Sum: χₐ + χᵦ + χᵧ + … = 1
4. Temperature independent

For Binary Solution

$$\boxed{\chi_{\text{solute}} + \chi_{\text{solvent}} = 1}$$ $$\boxed{\chi_{\text{solvent}} = 1 - \chi_{\text{solute}}}$$

Solved Example 4: Mole Fraction

A solution contains 36 g glucose (M=180) in 90 g water. Calculate mole fraction of each component.

Solution:

Moles of glucose = 36/180 = 0.2 mol Moles of water = 90/18 = 5 mol Total moles = 0.2 + 5 = 5.2 mol

$$\chi_{\text{glucose}} = \frac{0.2}{5.2} = 0.0385$$ $$\chi_{\text{water}} = \frac{5}{5.2} = 0.9615$$

Check: 0.0385 + 0.9615 = 1.000 ✓

$$\boxed{\chi_{\text{glucose}} = 0.0385, \quad \chi_{\text{water}} = 0.9615}$$

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Mass Percentage (% w/w)

Definition

Mass percentage is the mass of solute per 100 g of solution.

$$\boxed{\% \text{ (w/w)} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100}$$ $$\boxed{\% \text{ (w/w)} = \frac{m_{\text{solute}}}{m_{\text{solute}} + m_{\text{solvent}}} \times 100}$$

Solved Example 5: Mass Percentage

10 g salt is dissolved in 90 g water. Calculate mass percentage.

Solution:

Total mass = 10 + 90 = 100 g

$$\% \text{ (w/w)} = \frac{10}{100} \times 100 = 10\%$$ $$\boxed{\% \text{ (w/w)} = 10\%}$$

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Volume Percentage (% v/v)

Definition

Volume percentage is volume of liquid solute per 100 mL of solution.

$$\boxed{\% \text{ (v/v)} = \frac{\text{Volume of solute}}{\text{Volume of solution}} \times 100}$$

Solved Example 6: Volume Percentage

25 mL ethanol is mixed with water to make 100 mL solution. Calculate % v/v.

Solution:

$$\% \text{ (v/v)} = \frac{25}{100} \times 100 = 25\%$$ $$\boxed{\% \text{ (v/v)} = 25\%}$$

Note: “25% alcohol” on labels usually means 25% v/v.

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Mass by Volume (% w/v)

Definition

Mass by volume is grams of solute per 100 mL of solution.

$$\boxed{\% \text{ (w/v)} = \frac{\text{Mass of solute (g)}}{\text{Volume of solution (mL)}} \times 100}$$

Solved Example 7: Mass by Volume

5 g NaCl in 100 mL solution.

Solution:

$$\% \text{ (w/v)} = \frac{5}{100} \times 100 = 5\%$$ $$\boxed{\% \text{ (w/v)} = 5\%}$$

Medical use: “0.9% saline” means 0.9 g NaCl per 100 mL (physiological saline).

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Parts Per Million (ppm) and Parts Per Billion (ppb)

Definitions

$$\boxed{\text{ppm} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 10^6}$$ $$\boxed{\text{ppb} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 10^9}$$

Relationships

$$\boxed{1 \text{ ppm} = 1000 \text{ ppb}}$$ $$\boxed{1 \text{ ppm} = 0.0001\% = 10^{-4}\%}$$

For Dilute Aqueous Solutions

Density ≈ 1 g/mL, so:

$$\boxed{\text{ppm} = \frac{\text{mg solute}}{\text{L solution}}}$$

Solved Example 8: ppm

5 mg of fluoride is present in 10 L of water. Calculate ppm.

Solution:

Mass of solution ≈ 10 L × 1000 g/L = 10,000 g Mass of fluoride = 5 mg = 0.005 g

$$\text{ppm} = \frac{0.005}{10000} \times 10^6 = 0.5 \text{ ppm}$$

Alternative (for dilute aqueous):

$$\text{ppm} = \frac{5 \text{ mg}}{10 \text{ L}} = 0.5 \text{ mg/L} = 0.5 \text{ ppm}$$ $$\boxed{\text{ppm} = 0.5}$$

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Interconversions Between Concentration Terms

Molarity ↔ Molality

Given: Molarity (M), density (d), molar mass of solute (M)

$$\boxed{m = \frac{1000 \times M}{1000d - M \times M_{\text{solute}}}}$$ $$\boxed{M = \frac{1000d \times m}{1000 + m \times M_{\text{solute}}}}$$

Molarity ↔ Mole Fraction

For solvent (usually water, M = 18):

$$\boxed{\chi_{\text{solute}} = \frac{m \times M_{\text{solvent}}}{1000 + m \times M_{\text{solvent}}}}$$

Mass Percentage ↔ Molarity

$$\boxed{M = \frac{10 \times d \times \% \text{(w/w)}}{M_{\text{solute}}}}$$

Molality ↔ Mole Fraction

$$\boxed{\chi_{\text{solute}} = \frac{m \times M_{\text{solvent}}}{1000 + m \times M_{\text{solvent}}}}$$ $$\boxed{m = \frac{1000 \times \chi_{\text{solute}}}{M_{\text{solvent}} \times (1 - \chi_{\text{solute}})}}$$

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Solved Examples: Interconversions

Example 9: Molarity to Molality

A solution is 2 M in NaOH (M = 40) and has density 1.08 g/mL. Calculate molality.

Solution:

Consider 1 L solution:

• Moles of NaOH = 2 mol
• Mass of NaOH = 2 × 40 = 80 g
• Mass of solution = 1000 × 1.08 = 1080 g
• Mass of water = 1080 - 80 = 1000 g = 1 kg

$$m = \frac{2}{1} = 2 \text{ m}$$ $$\boxed{m = 2 \text{ m}}$$

Alternative using formula:

$$m = \frac{1000 \times 2}{1000 \times 1.08 - 2 \times 40} = \frac{2000}{1080 - 80} = \frac{2000}{1000} = 2 \text{ m}$$

Example 10: Mole Fraction to Molality

Mole fraction of ethanol (M = 46) in water is 0.2. Calculate molality.

Solution:

χ(ethanol) = 0.2, so χ(water) = 0.8

Let moles of ethanol = 0.2, water = 0.8

Mass of water = 0.8 × 18 = 14.4 g = 0.0144 kg

$$m = \frac{0.2}{0.0144} = 13.89 \text{ m}$$ $$\boxed{m = 13.89 \text{ m}}$$

Alternative using formula:

$$m = \frac{1000 \times 0.2}{18 \times 0.8} = \frac{200}{14.4} = 13.89 \text{ m}$$

Example 11: % w/w to Molarity

10% H₂SO₄ (w/w) solution has density 1.07 g/mL. Calculate molarity.

Solution:

Assume 100 g solution:

• Mass of H₂SO₄ = 10 g
• Moles = 10/98 = 0.102 mol
• Volume = 100/1.07 = 93.46 mL = 0.09346 L

$$M = \frac{0.102}{0.09346} = 1.09 \text{ M}$$ $$\boxed{M = 1.09 \text{ M}}$$

Alternative using formula:

$$M = \frac{10 \times 1.07 \times 10}{98} = \frac{107}{98} = 1.09 \text{ M}$$

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Dilution Formula

When solution is diluted by adding more solvent:

$$\boxed{M_1V_1 = M_2V_2}$$

Where:

• M₁ = initial molarity
• V₁ = initial volume
• M₂ = final molarity
• V₂ = final volume

Solved Example 12: Dilution

How much water should be added to 100 mL of 2 M HCl to make it 0.5 M?

Solution:

$$M_1V_1 = M_2V_2$$ $$2 \times 100 = 0.5 \times V_2$$ $$V_2 = 400 \text{ mL}$$

Water to be added = 400 - 100 = 300 mL

$$\boxed{\text{Add 300 mL water}}$$

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Mixing Solutions

Same Solute, Different Concentrations

$$\boxed{M_{\text{final}} = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}}$$

Assuming volumes are additive (dilute solutions).

Solved Example 13: Mixing

50 mL of 0.2 M HCl + 100 mL of 0.5 M HCl. Find final molarity.

Solution:

$$M_{\text{final}} = \frac{0.2 \times 50 + 0.5 \times 100}{50 + 100}$$ $$= \frac{10 + 50}{150} = \frac{60}{150} = 0.4 \text{ M}$$ $$\boxed{M = 0.4 \text{ M}}$$

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Common Calculation Mistakes

Mistake 1: Solution vs Solvent Mass/Volume

❌ Wrong: Molarity = moles/volume of solvent ✓ Right: Molarity = moles/volume of solution

❌ Wrong: Molality = moles/mass of solution ✓ Right: Molality = moles/mass of solvent

Mistake 2: Wrong Density Units

❌ Wrong: Using density in kg/L instead of g/mL ✓ Right: Check units! d in g/mL or g/cm³ (numerically equal)

Mistake 3: Forgetting 1000 in Formulas

❌ Wrong: m = n/mass (giving wrong units) ✓ Right: m = n/mass(kg) or = 1000n/mass(g)

Mistake 4: Adding Volumes Incorrectly

❌ Wrong: 50 mL alcohol + 50 mL water = 100 mL (not always true!) ✓ Right: Volumes may not be additive for concentrated solutions

Mistake 5: Mole Fraction > 1

❌ Wrong: Calculating χ > 1 and accepting it ✓ Right: Check calculation - χ must be ≤ 1, and sum of all χ = 1

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Memory Tricks

“MoMol” Rule

Molarity uses volume of solution (Mo-solution) Molality uses mass of solvent (Mol-vent)

“MPPM” Trick

Molarity: mol/L (M-1000) Percentage: × 100 Ppm: × 10⁶ (million) Ppb: × 10⁹ (billion)

“TINT” for Temperature

Temperature Independent: molality, mole fraction, mass % Temperature dependent: molarity, normality (volume changes)

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Practice Problems

Level 1: Basic (JEE Main)

Q1. Calculate molarity of 4 g NaOH in 250 mL solution.

Solution

Moles = 4/40 = 0.1 mol Volume = 0.25 L

M = 0.1/0.25 = 0.4 M

Answer: 0.4 M

Q2. Calculate molality of 18 g glucose (M=180) in 1 kg water.

Solution

Moles = 18/180 = 0.1 mol

m = 0.1/1 = 0.1 m

Answer: 0.1 m

Q3. A solution contains 0.1 mol ethanol and 0.9 mol water. Find mole fraction of ethanol.

Solution

Total moles = 0.1 + 0.9 = 1

χ(ethanol) = 0.1/1 = 0.1

Answer: 0.1

Q4. 5 g NaCl in 95 g water. Calculate mass percentage.

Solution

% w/w = 5/(5+95) × 100 = 5%

Answer: 5%

Level 2: Intermediate (JEE Main/Advanced)

Q5. A solution is 1 M and 1.2 m in NaCl. Calculate density of solution. M(NaCl) = 58.5

Solution

Consider 1 L solution (from molarity): Moles of NaCl = 1 mol Mass of NaCl = 58.5 g

From molality: 1.2 m means 1.2 mol per kg solvent For 1 mol solute: Mass of solvent = 1/1.2 = 0.833 kg = 833 g

Mass of solution = 58.5 + 833 = 891.5 g Volume = 1 L = 1000 mL

Density = 891.5/1000 = 0.892 g/mL

Answer: 0.892 g/mL

Q6. Mole fraction of glucose in solution is 0.1. Calculate molality. M(H₂O) = 18

Solution

Using formula: $$m = \frac{1000 \times 0.1}{18 \times 0.9} = \frac{100}{16.2} = 6.17 \text{ m}$$

Answer: 6.17 m

Q7. 100 mL of 1 M HCl is diluted to 500 mL. Find final molarity.

Solution

M₁V₁ = M₂V₂ 1 × 100 = M₂ × 500 M₂ = 0.2 M

Answer: 0.2 M

Q8. 20% H₂SO₄ (w/w) solution has density 1.14 g/mL. Calculate molality. M(H₂SO₄) = 98

Solution

Assume 100 g solution: Mass of H₂SO₄ = 20 g → 20/98 = 0.204 mol Mass of water = 80 g = 0.08 kg

m = 0.204/0.08 = 2.55 m

Answer: 2.55 m

Level 3: Advanced (JEE Advanced - Tricky)

Q9. (JEE Advanced Pattern) Two solutions of same solute have molarities M₁ and M₂, and densities d₁ and d₂. They are mixed in volume ratio V₁:V₂. Find molarity of mixture. (Assume additive volumes)

Solution

Let volumes be V₁ and V₂.

Total moles = M₁V₁ + M₂V₂ Total volume = V₁ + V₂

$$M_{\text{mix}} = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}$$

Answer: $M = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}$

(Note: Density given but not needed if volumes are additive)

Q10. (Tricky) A solution is labeled “1 M H₂SO₄”. For neutralization calculations, should you treat it as: a) 1 M or 2 N? b) Why?

Answer

**a) 2 N**

H₂SO₄ is dibasic (can donate 2 H⁺ ions) n-factor = 2

N = M × n-factor = 1 × 2 = 2 N

b) Reason: Normality accounts for the actual “reacting capacity.” Each molecule of 1 M H₂SO₄ provides 2 equivalents of H⁺.

For neutralization: N₁V₁(acid) = N₂V₂(base) Use normality, not molarity!

Answer: a) 2 N, b) n-factor = 2 for H₂SO₄

Q11. (JEE Advanced 2019 Pattern) The molality of a solution formed by dissolving 1.2 g of compound (M = 60) in 100 g benzene is x. If the same compound when dissolved in water forms a dimer, what would be the molality when 1.2 g dissolves in 100 g water?

Solution

**In benzene (no association):** Moles = 1.2/60 = 0.02 mol m = 0.02/0.1 = 0.2 m

In water (dimerization: 2A → A₂): Moles of monomer = 0.02 After dimerization, moles of particles = 0.02/2 = 0.01 mol m = 0.01/0.1 = 0.1 m

Answer: 0.1 m (half of 0.2 m)

Q12. (Conceptual) Why is molality preferred over molarity for colligative properties?

Answer

**Reasons:**

1. Temperature independent: Molality uses mass of solvent (doesn’t change with T). Molarity uses volume (expands/contracts with T).

2. Accurate at different T: Colligative properties are often measured at different temperatures (boiling point elevation, freezing point depression).

3. Theoretical basis: Raoult’s law and colligative property equations are derived using mole fraction, which is directly related to molality, not molarity.

4. No volume measurement needed: Weighing is more accurate than volume measurement.

Answer: Molality is temperature-independent and more accurate for colligative property measurements.

Q13. (Ultra-tricky) A solution has density d g/mL, contains x% solute (w/w), molar mass M. Derive expression for molarity.

Solution

Assume 100 g solution: Mass of solute = x g Moles of solute = x/M

Volume of solution = 100/d mL = 100/(1000d) L = 0.1/d L

$$M = \frac{x/M}{0.1/d} = \frac{xd}{0.1M} = \frac{10xd}{M}$$

Answer: $M = \frac{10xd}{M}$ where x is % w/w, d is density in g/mL, M is molar mass

Q14. (Multi-concept) 1 L of 1 M NaCl (d = 1.04 g/mL) is mixed with 1 L of 2 M NaCl (d = 1.08 g/mL). Assuming volumes are not additive but masses are, find: a) Mass of mixture b) Moles of NaCl c) Volume of mixture (d = 1.06 g/mL) d) Final molarity

Solution

**a) Mass of mixture:** Mass₁ = 1000 mL × 1.04 = 1040 g Mass₂ = 1000 mL × 1.08 = 1080 g Total mass = 2120 g

b) Moles of NaCl: Moles₁ = 1 L × 1 M = 1 mol Moles₂ = 1 L × 2 M = 2 mol Total moles = 3 mol

c) Volume of mixture: V = mass/density = 2120/1.06 = 2000 mL = 2 L

d) Final molarity: M = 3 mol / 2 L = 1.5 M

Answer: a) 2120 g, b) 3 mol, c) 2 L, d) 1.5 M

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Cross-Links and Further Study

Related Topics

• Mole Concept - Foundation for all concentration terms
• Solutions - Detailed solution chemistry
• Equivalent Concept - Normality and equivalents
• Stoichiometry - Using molarity in reactions
• Colligative Properties - Applications of molality
• Raoult’s Law - Using mole fraction

For JEE Advanced Students

• Van’t Hoff factor in colligative properties
• Osmotic pressure calculations
• Non-ideal solutions
• Activity vs concentration
• Buffer solutions and pH calculations

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Quick Revision Table

All Concentration Terms

Term	Formula	Units	T-dependent?	Use
Molarity	n/V(L)	mol/L, M	Yes	Reactions, titrations
Molality	n/m(kg)	mol/kg, m	No	Colligative properties
Mole fraction	n₁/n_total	None	No	Raoult’s law, ideal solutions
% w/w	(m_solute/m_soln)×100	%	No	Commercial labels
% v/v	(V_solute/V_soln)×100	%	Yes	Liquid mixtures
% w/v	(m_solute/V_soln)×100	%	Yes	Medical solutions
ppm	(m_solute/m_soln)×10⁶	ppm	No	Trace amounts
ppb	(m_solute/m_soln)×10⁹	ppb	No	Ultra-trace

Important Formulas Summary

$$\boxed{M = \frac{n}{V(L)} \quad m = \frac{n}{m_{\text{solvent}}(kg)} \quad \chi = \frac{n_i}{n_{\text{total}}}}$$ $$\boxed{M_1V_1 = M_2V_2 \text{ (dilution)}}$$ $$\boxed{M = \frac{10 \times d \times \%}{M_{\text{solute}}} \text{ (% to M)}}$$ $$\boxed{m = \frac{1000 \times \chi}{M_{\text{solvent}}(1-\chi)} \text{ (χ to m)}}$$

Interactive Demo: Visualize Solution Preparation

See how different concentration terms relate to each other in solution preparation.

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Tips for JEE

1. Identify what’s given: Solution or solvent? Mass or volume?
2. Choose right concentration: Colligative → molality; Reaction → molarity
3. Always assume 100 g/mL: Makes % calculations easy
4. Check temperature dependence: If T changes, avoid molarity
5. Interconversion path: % → M requires density; M → m requires density and M
6. Mole fraction check: Sum must equal 1
7. ppm for dilute aqueous: 1 ppm ≈ 1 mg/L
8. Memorize common densities: Water ≈ 1 g/mL, dilute solutions ≈ 1.0-1.2 g/mL
9. Mixing formulas: Same solute only! Different solutes → use mole balance
10. Units, units, units: L vs mL, g vs kg, mol vs mmol

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Last updated: May 2025 Part of JEE Chemistry - Basic Concepts series