Chemistry Chapter 1: Solutions - Comprehensive Formulae Guide

1. Stoichiometry and Concentration Units

1.1 Mass Percentage (w/w)

$\text{Mass \% of a component}=\left(\frac{\text{Mass of the component in the solution}}{\text{Total mass of the solution}}\right)\times 100$

Explanation: This formula calculates the percentage of a component by mass in a solution. It is commonly used in industrial chemical applications.

Example: A solution with 10 g of glucose dissolved in 90 g of water has a mass percentage of glucose = $\frac{10}{100}\times 100=10$

Common Mistake: Confusing the total mass of the solution with just the mass of the solvent.

1.2 Volume Percentage (v/v)

$\text{Volume \% of a component}=\left(\frac{\text{Volume of the component}}{\text{Total volume of solution}}\right)\times 100$

Explanation: Used to express the concentration of a liquid in a solution, particularly useful in mixtures of liquids.

Example: A 10% ethanol solution means 10 mL of ethanol is mixed with enough water to make 100 mL of solution.

Common Mistake: Miscalculating the total volume, especially in non-ideal solutions where volumes are not additive.

1.3 Molarity (M)

$M=\frac{\text{Moles of solute}}{\text{Volume of solution in liters}}$

Explanation: Molarity is the number of moles of solute per liter of solution. It is temperature-dependent because volume can change with temperature.

Example: For a solution with 5 g of NaOH in 450 mL of solution, Molarity = $\frac{0.125\text{ mol}}{0.45\text{ L}}=0.278\text{ M}$

Common Mistake: Forgetting to convert mL to L in the denominator.

1.4 Molality (m)

$m=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

Explanation: Molality is used when the concentration needs to be independent of temperature since it uses the mass of the solvent instead of the volume of the solution.

Example: For a solution with 2.5 g of ethanoic acid in 75 g of benzene, Molality = $\frac{0.0417\text{ mol}}{0.075\text{ kg}}=0.556\text{ mol/kg}$

Common Mistake: Using the mass of the solution instead of the mass of the solvent.

2. Thermodynamics and Colligative Properties

2.1 Raoult’s Law

$p_i=x_i\cdot p_i^0$

Explanation: Raoult's law states that the partial vapor pressure of each volatile component in a solution is directly proportional to its mole fraction.

Example: In a solution where the mole fraction of chloroform is 0.312, and the vapor pressure of pure chloroform is 200 mm Hg, the partial vapor pressure is $p_{CHCl3}=0.312\times 200=62.4\text{ mm Hg}$

Common Mistake: Not considering the correct mole fraction when multiple components are present.

2.2 Henry’s Law

$p=K_H\cdot x$

Explanation: Henry’s law relates the solubility of a gas in a liquid to the partial pressure of the gas above the liquid.

Example: For nitrogen at 293 K with a partial pressure of 0.987 bar and $K_H$ value of 76.48 kbar, $x_{N2}=\frac{0.987}{76480}=1.29\times 10^{-5}$

Common Mistake: Confusing the constant $K_H$ for different gases or temperatures.

2.3 Boiling Point Elevation

$\Delta T_b=K_b\cdot m$

Explanation: The boiling point of a solution is elevated compared to the pure solvent, proportional to the molality of the solution.

Example: If 18 g of glucose is dissolved in 1 kg of water, $\Delta T_b=0.52\times 0.1=0.052\text{ K}$, so the boiling point will be 373.202 K.

Common Mistake: Forgetting to use molality (mol/kg) instead of molarity.

2.4 Freezing Point Depression

$\Delta T_f=K_f\cdot m$

Explanation: The freezing point of a solution is lower than that of the pure solvent, and the depression is directly proportional to the molality of the solute.

Example: For a solution with 45 g of ethylene glycol in 600 g of water, $\Delta T_f=1.86\times 1.2=2.2\text{ K}$

Common Mistake: Incorrectly calculating the molality, especially in mixed solvent systems.

3. Chemical Kinetics and Equilibrium

3.1 Equilibrium Constant (K)

$K_c=\frac{[\text{Products}]}{[\text{Reactants}]}$

Explanation: The equilibrium constant expresses the ratio of product concentrations to reactant concentrations at equilibrium.

Example: For the reaction $aA+bB\rightleftharpoons cC+dD,K_c=\frac{[C{]}^c\times [D{]}^d}{[A{]}^a\times [B{]}^b}$

Common Mistake: Ignoring the stoichiometric coefficients in the exponent when calculating the equilibrium constant.

3.2 Rate Law

$\text{Rate}=k\cdot [A{]}^m\cdot [B{]}^n$

Explanation: The rate of a reaction is proportional to the product of the concentrations of the reactants, each raised to a power corresponding to their order in the reaction.

Example: For a reaction $A+B\to C$ with rate law $\text{Rate}=k[A{]}^2[B{]}^1$, the rate doubles if $[A]$ is doubled.

Common Mistake: Misidentifying the order of the reaction with respect to each reactant.

Common Misconceptions:

• Units Confusion: Always ensure that you are consistent with units—especially between mass, volume, and mole units.
• Temperature Dependence: Remember that molarity is temperature-dependent while molality is not.

Practice Problems:

1. Calculate the molarity of a solution with 10 g of NaOH in 250 mL of solution.
2. Determine the boiling point elevation for a solution containing 20 g of urea in 500 g of water.
3. Using Raoult’s law, find the total vapor pressure of a solution containing 30% ethanol and 70% water by mole fraction.

This summary provides a comprehensive overview of key formulae and concepts from the first chapter on Solutions, which is critical for NEET UG Chemistry preparation.