Concentration of Solutions: Complete Guide for ISC & CBSE Class 11 Chemistry

Understanding Concentration Methods in Solutions

Grade 11 ISC & CBSE Chemistry Guide

When preparing a solution in chemistry, we often need to describe how much solute is present in a given amount of solvent. This quantitative expression is called the concentration of a solution.

Different methods are used depending on the type of problem and experimental conditions.

MOLARITY (M)

Definition:
“The number of moles of solute dissolved per litre of solution is known as the molarity of the solution.”
It is denoted by M.

$$M = \frac{\text{moles of solute}}{\text{volume of solution in litres}}$$

✔ Key Points

• Volume is taken in litres of solution, not solvent.
• Temperature affects molarity (because volume changes).

✔ Example

If 1 mole of NaCl is dissolved to make 1 L solution → 1 M NaCl solution

MOLALITY (m)

Definition:
“The number of moles of solute present in 1000 g of solvent is known as molality of the solution.”
It is denoted by m.

$$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$$

✔ Key Points

• Uses mass of solvent, not solution.
• Independent of temperature → preferred in colligative properties.

✔ Example

1 mole glucose in 1 kg water → 1 m solution

MOLE FRACTION (χ)

Definition:
“The ratio of the number of moles of a constituent to the total number of moles of all the constituents present in the solution is the mole fraction of that constituent.”
It is denoted by χ (chi).

For component A:

$$\chi_A = \frac{n_A}{n_A + n_B}$$

For component B:

$$\chi_B = \frac{n_B}{n_A + n_B}$$

✔ Important Relations

For two components:

$$\chi_A + \chi_B = 1$$

For three or more components:

$$\chi_A + \chi_B + \chi_C + \dots = 1$$

✔ Key Points

• Unitless quantity.
• Useful in vapour pressure & Raoult’s law calculations.

 PERCENTAGE METHODS

(i) Mass by Mass Percentage (W/W)

Amount of solute in grams dissolved in 100 g of solution.

$$\% (w/w) = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100$$

✔ Used in solid–solid mixtures and alloys.

(ii) Mass by Volume Percentage (W/V)

Amount of solute in grams dissolved in 100 mL of solution.

$$\% (w/v) = \frac{\text{mass of solute}}{\text{volume of solution}} \times 100$$

✔ Common in medical and biological solutions.

 STRENGTH OF SOLUTION

Definition:
“The amount of the solute (in grams) present in one litre of the solution is known as the strength of the solution.”

$$\text{Strength}=\frac{\text{mass of solute (g)}}{\text{volume of solution (L)}}$$

Unit: g L⁻¹

✔ Relation with Molarity

$$\text{Strength} = M \times \text{Molar mass}$$

PARTS PER MILLION (ppm)

Used when the solute concentration is extremely small (trace amounts).

Definition:
The quantity of solute per million (10⁶) parts of the system.

$$ppm = \frac{\text{amount of solute}}{\text{amount of solution}} \times 10^6$$

✔ In aqueous solutions:

$$ppm = \frac{\text{mass of solute (mg)}}{\text{volume of solution (L)}}$$

✔ Uses

• Water pollution measurement
• Mineral content in drinking water
• Environmental chemistry

 IMPORTANT NOTE

$$1 \text{ dm}^3 = 1 \text{ L} = 1000 \text{ mL} = 1000 \text{ cm}^3$$

 Quick Comparison

Method	Based On	Temperature Effect	Unit
Molarity	Volume of solution	Affected	mol L⁻¹
Molality	Mass of solvent	Not affected	mol kg⁻¹
Mole Fraction	Moles	Not affected	No unit
% w/w	Mass	Not affected	%
% w/v	Mass & volume	Slight	%
Strength	g per litre	Affected	g L⁻¹
ppm	Trace quantity	Not significant	ppm

Exam Tips & Common Mistakes

✔ Always convert volume to litres in molarity problems.
✔ Use mass of solvent (not solution) for molality.
✔ Mole fraction has no units.
✔ ppm is used for very dilute solutions.
✔ Strength is not the same as molarity.