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Solutions - Types of solutions

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits.

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A binary solution consists of two components: the Solvent (the component present in the largest quantity) and the Solute (the component present in a smaller quantity).

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Solutions are classified into nine types based on the physical state of the solute and the solvent (Solid, Liquid, or Gas).

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Gaseous Solutions: The solvent is always a gas. Examples include gas in gas (Air), liquid in gas (Chloroform mixed with nitrogen gas), and solid in gas (Camphor in nitrogen gas).

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Liquid Solutions: The solvent is always a liquid. Examples include gas in liquid (O2O_2 dissolved in water), liquid in liquid (Ethanol dissolved in water), and solid in liquid (Glucose dissolved in water).

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Solid Solutions: The solvent is always a solid. Examples include gas in solid (Solution of hydrogen in palladium), liquid in solid (Amalgam of mercury with sodium), and solid in solid (Copper dissolved in gold).

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The concentration of a solution is the amount of solute dissolved in a known amount of solvent or solution, expressed through various units like Molarity (MM), Molality (mm), and Mole Fraction (χ\chi).

📐Formulae

Mass Percentage (w/w)=Mass of component in solutionTotal mass of solution×100\text{Mass Percentage (w/w)} = \frac{\text{Mass of component in solution}}{\text{Total mass of solution}} \times 100

Mole Fraction of component A (χA)=nAnA+nB\text{Mole Fraction of component A } (\chi_A) = \frac{n_A}{n_A + n_B}

Molarity (M)=Moles of soluteVolume of solution in Litres\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in Litres}}

Molality (m)=Moles of soluteMass of solvent in kilograms\text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kilograms}}

Parts per million (ppm)=Number of parts of the componentTotal number of parts of all components of the solution×106\text{Parts per million (ppm)} = \frac{\text{Number of parts of the component}}{\text{Total number of parts of all components of the solution}} \times 10^6

💡Examples

Problem 1:

Identify the type of solution and the physical state of the solute and solvent in 'Amalgam of mercury with sodium'.

Solution:

Type: Liquid in Solid solution. Solute: Mercury (HgHg, liquid); Solvent: Sodium (NaNa, solid).

Explanation:

In an amalgam, mercury is the liquid component dispersed within the solid metal lattice of sodium.

Problem 2:

Calculate the mole fraction of ethylene glycol (C2H6O2C_2H_6O_2) in a solution containing 20%20\% of C2H6O2C_2H_6O_2 by mass.

Solution:

Mass of C2H6O2=20 gC_2H_6O_2 = 20\text{ g}, Mass of water = 80 g80\text{ g}. Moles of C2H6O2=2062.07≈0.322 molC_2H_6O_2 = \frac{20}{62.07} \approx 0.322\text{ mol}. Moles of water = \frac{80}{18} \approx 4.444\text{ mol}.. \chi_{glycol} = \frac{0.322}{0.322 + 4.444} \approx 0.068$.

Explanation:

Mole fraction is the ratio of the number of moles of a particular component to the total number of moles of all components in the solution.

Problem 3:

Give an example of a solid in gas solution.

Solution:

Camphor in nitrogen gas.

Explanation:

In this case, the solid particles of camphor (solute) are sublimated and distributed homogeneously within the nitrogen gas (solvent).

Types of solutions - Revision Notes & Key Formulas | ICSE Class 12 Chemistry