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Solutions - Types of Solutions

Grade 12CBSEChemistry

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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The component that is present in the largest quantity is known as the solvent. The solvent determines the physical state in which the solution exists. One or more components present in the solution other than solvent are called solutes.

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In a binary solution, there are only two components: one solute and one solvent.

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Gaseous solutions involve a gas as the solvent. Examples include mixtures of N2N_2 and O2O_2, or chloroform vapor mixed with nitrogen gas (CHCl3CHCl_3 and N2N_2).

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Liquid solutions involve a liquid as the solvent. Examples include ethanol (C2H5OHC_2H_5OH) dissolved in water (H2OH_2O) or glucose (C6H12O6C_6H_{12}O_6) dissolved in water.

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Solid solutions involve a solid as the solvent. Examples include alloys like Brass (CuCu and ZnZn), or the solution of hydrogen gas (H2H_2) in palladium (PdPd).

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Concentration of a solution is the amount of solute present in a given quantity of the solvent or solution.

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Mass percentage (w/ww/w) is defined as the mass of the solute per 100100 units of mass of the solution.

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

📐Formulae

\text{Mass % of a component} = \frac{\text{Mass of the component in the solution}}{\text{Total mass of the solution}} \times 100

\text{Volume % of a component} = \frac{\text{Volume of the component}}{\text{Total volume of solution}} \times 100

Mole Fraction (xA)=nAnA+nB\text{Mole Fraction } (x_A) = \frac{n_A}{n_A + n_B}

∑xi=x1+x2+...+xn=1\sum x_i = x_1 + x_2 + ... + x_n = 1

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

💡Examples

Problem 1:

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:

Assume 100 g100 \text{ g} of solution. Mass of C2H6O2=20 gC_2H_6O_2 = 20 \text{ g}, Mass of H2O=80 gH_2O = 80 \text{ g}. Molar mass of C2H6O2=62 g mol−1C_2H_6O_2 = 62 \text{ g mol}^{-1}. Moles of C2H6O2(n1)=2062=0.322 molC_2H_6O_2 (n_1) = \frac{20}{62} = 0.322 \text{ mol}. Moles of H2O(n2)=8018=4.444 molH_2O (n_2) = \frac{80}{18} = 4.444 \text{ mol}. xglycol=0.3220.322+4.444=0.068x_{glycol} = \frac{0.322}{0.322 + 4.444} = 0.068.

Explanation:

To find the mole fraction, we first determine the mass of each component from the percentage, convert masses to moles using molar masses, and then apply the mole fraction formula xA=nAntotalx_A = \frac{n_A}{n_{total}}.

Problem 2:

Calculate the molarity of a solution containing 5 g5 \text{ g} of NaOHNaOH in 450 mL450 \text{ mL} solution.

Solution:

Moles of NaOH=5 g40 g mol−1=0.125 molNaOH = \frac{5 \text{ g}}{40 \text{ g mol}^{-1}} = 0.125 \text{ mol}. Volume of solution in litres =4501000=0.450 L= \frac{450}{1000} = 0.450 \text{ L}. M=0.125 mol0.450 L=0.278 mol L−1 or 0.278 MM = \frac{0.125 \text{ mol}}{0.450 \text{ L}} = 0.278 \text{ mol L}^{-1} \text{ or } 0.278 \text{ M}.

Explanation:

Molarity is defined as moles of solute divided by the volume of solution in Litres. The molar mass of NaOHNaOH is calculated as 23+16+1=40 g/mol23 + 16 + 1 = 40 \text{ g/mol}.

Types of Solutions - Revision Notes & Key Formulas | CBSE Class 12 Chemistry