Mass % of a component=(Total mass of the solutionMass of the component in the solution)×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 = 10010×100=10
Common Mistake: Confusing the total mass of the solution with just the mass of the solvent.
Volume % of a component=(Total volume of solutionVolume of the component)×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.
M=Volume of solution in litersMoles of solute
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 = 0.45 L0.125 mol=0.278 M
Common Mistake: Forgetting to convert mL to L in the denominator.
m=Mass of solvent in kgMoles of solute
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 = 0.075 kg0.0417 mol=0.556 mol/kg
Common Mistake: Using the mass of the solution instead of the mass of the solvent.
pi=xi⋅pi0
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 pCHCl3=0.312×200=62.4 mm Hg
Common Mistake: Not considering the correct mole fraction when multiple components are present.
p=KH⋅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 KH value of 76.48 kbar, xN2=764800.987=1.29×10−5
Common Mistake: Confusing the constant KH for different gases or temperatures.
ΔTb=Kb⋅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, ΔTb=0.52×0.1=0.052 K, so the boiling point will be 373.202 K.
Common Mistake: Forgetting to use molality (mol/kg) instead of molarity.
ΔTf=Kf⋅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, ΔTf=1.86×1.2=2.2 K
Common Mistake: Incorrectly calculating the molality, especially in mixed solvent systems.
Kc=[Reactants][Products]
Explanation: The equilibrium constant expresses the ratio of product concentrations to reactant concentrations at equilibrium.
Example: For the reaction aA+bB⇌cC+dD,Kc=[A]a×[B]b[C]c×[D]d
Common Mistake: Ignoring the stoichiometric coefficients in the exponent when calculating the equilibrium constant.
Rate=k⋅[A]m⋅[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→C with rate law 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.
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.