Chemistry Bridge Course - Lecture 6 | CHEMCA JEE & NEET

CHEMCA

JEE & NEET Academy by Abhishek Sengar

Lecture 6 Solution Chemistry Target: Class 10 to 11 Transition (JEE/NEET)

Concentration Terms of Solutions

Welcome to Lecture 6 of the CHEMCA Bridge Course! Solution stoichiometry and concentration methods are vital parts of physical chemistry. In this session, Abhishek Sengar Sir introduces how solution compositions are quantitatively expressed, including Molarity, Molality, Mole Fraction, and Mass Percentage, focusing on temperature dependencies and conversion keys.

Video Lecture Broadcast

Instructor: Abhishek Sengar Sir Published: April 18, 2026 Subject: Concentration Terms

Interactive Lecture Timestamps

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In-Depth Lecture Notes & Summary

Concept of Solutions

A Solution is a homogeneous mixture of two or more chemically non-reacting substances. For our study, we focus on Binary Solutions (consisting of exactly two components):

Solute

The component present in a smaller quantity, which is dissolved in the solvent. E.g., salt, sugar, urea.

Solvent

The component present in the largest quantity, which dissolves the solute. E.g., water, alcohol, benzene.

Mass % and Volume %

These terms express the relative amount of solute in terms of percentages:

Mass Percentage ($w/w\%$): Mass of solute present in $100\text{ g}$ of solution: $$\text{Mass } \% = \frac{\text{Mass of Solute (g)}}{\text{Total Mass of Solution (g)}} \times 100$$
Volume Percentage ($v/v\%$): Volume of solute present in $100\text{ mL}$ of solution: $$\text{Volume } \% = \frac{\text{Volume of Solute (mL)}}{\text{Total Volume of Solution (mL)}} \times 100$$

Molarity ($M$)

The most common term in chemical laboratories. It is defined as the number of moles of solute dissolved in $1\text{ Liter}$ ($1000\text{ mL}$) of solution.

$$\text{Molarity (M)} = \frac{\text{Moles of Solute (n)}}{\text{Volume of Solution (L)}} = \frac{w_B \times 1000}{M_B \times V_{\text{soln}}\text{ (mL)}}$$

Where $w_B$ is given solute weight, $M_B$ is molar mass of solute, and $V_{\text{soln}}$ is the solution volume.

Temperature Dependency: Molarity depends on volume ($V$), and since volume expands or contracts with temperature, Molarity changes with temperature!

Molality ($m$)

Molality is defined as the number of moles of solute dissolved in exactly $1\text{ kg}$ ($1000\text{ g}$) of the solvent (not solution!).

$$\text{Molality (m)} = \frac{\text{Moles of Solute (n)}}{\text{Mass of Solvent (kg)}} = \frac{w_B \times 1000}{M_B \times w_A\text{ (g)}}$$

Where $w_A$ is the mass of the solvent in grams.

Why Molality is Preferred: Since mass does not change with temperature, Molality is independent of temperature! This makes it ideal for thermodynamics experiments.

Mole Fraction ($\chi$)

The ratio of the number of moles of a particular component to the total number of moles of all components in the solution. For a binary solution with solute ($B$) and solvent ($A$):

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

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

Key Rule: The sum of mole fractions of all components in a solution is always equal to 1: $$\chi_A + \chi_B = 1$$

Dilution and Mixing Formulas

When solvent is added to dilute a solution, or when multiple solutions of different strengths are combined:

1. Dilution Formula

$$M_1V_1 = M_2V_2$$

2. Mixing Formula (Same Solutes)

$$M_f = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}$$

Solution Analyzer

Input solute parameters and solvent mass to instantly calculate the corresponding Concentration Terms.

Select Solute

Solute Mass ($w_B$ in grams)

Water Mass ($w_A$ in grams)

Solution Density ($d$ in g/mL)

Analysis Results:

Solute Moles ($n_B$): 0.500

Mass % ($w/w$): 20.00 %

Molarity ($M$): 6.00 M

Molality ($m$): 6.25 m

Mole Fraction ($\chi_B$): 0.101

Dilution Matrix Solver

Calculate final concentrations following dilution ($M_1V_1 = M_2V_2$).

Molarity 1 ($M_1$)

Volume 1 ($V_1$ mL)

Volume 2 ($V_2$ mL)

Final Molarity ($M_2$) 0.67 M

Lecture 6 Quiz Challenge

Validate your conversion and dilution math skills instantly.

Question 1 of 5

Score: 0/0

Have solutions questions?

If you are struggling to understand why Molarity depends on temperature while Molality remains constant, drop a message to Abhishek Sengar Sir!

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Chemistry Bridge Course Lecture 6