Solutions — Class 12 Chemistry Notes

Solutions · Class 12 Chemistry · 7 topics.

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Topics covered in Solutions

  1. 1.The Formation of Different Types of Solutions

    Short Answer

    In simple terms, the formation of solutions involves mixing two substances together until they are evenly distributed. Solutions can be of different types based on the state of matter of the solute and the solvent, like solid in liquid (salt in water), gas in liquid (carbon dioxide in water), or liquid in liquid (alcohol in water).

    Long Answer

    1. Introduction:
    A solution is a homogeneous mixture of two or more substances. The substance in the smallest amount and the one that dissolves is called the solute. The substance in the larger amount is called the solvent.

    2. Types of Solutions:

    • Solid in Liquid: This is the most common type of solution, where a solid (solute) dissolves in a liquid (solvent). Example: Salt or sugar dissolving in water. The process involves the solute particles being surrounded by solvent molecules, breaking the solute into smaller pieces until it is evenly distributed.


    • Gas in Liquid: Here, a gas dissolves in a liquid. Example: Carbonated beverages where carbon dioxide is dissolved in water under pressure. The solubility of gases in liquids increases with an increase in pressure and decreases with an increase in temperature.

    • Liquid in Liquid: In this type, two liquids mix to form a solution. Example: Alcohol in water. The molecules of the two liquids interact and spread out evenly.

    3. How Solutions Form:

    • Step 1: Breaking of solute into individual components.
    • Step 2: Overcoming the intermolecular forces in the solvent to make room for the solute.
    • Step 3: Interaction between solute and solvent molecules, leading to a uniform distribution.

    4. Real-life Examples and Applications:

    • In Everyday Life: Preparing tea or coffee involves dissolving sugar and coffee or tea in water, a liquid in liquid and solid in liquid solution, respectively.
    • In Industries: The pharmaceutical industry uses solutions for creating various medications in the form of syrups (a solid in liquid solution).
    • In Healthcare: Saline solutions (salt in water) are used for IV fluids.

    5. Conclusion:
    Understanding the formation of solutions is fundamental in chemistry and is widely applicable in daily life, from cooking to sophisticated industrial processes.

  2. 2.Express Concentration of Solution In Different Units

    Short Answer

    Concentration of a solution can be expressed in several units, such as molarity (M), molality (m), mass percent, volume percent, and parts per million (ppm). Molarity is moles of solute per liter of solution, molality is moles of solute per kilogram of solvent, mass percent is the mass of solute divided by the total mass of the solution multiplied by 100, volume percent is the volume of solute divided by the total volume of the solution multiplied by 100, and ppm is the mass of solute divided by the total mass of the solution multiplied by 106106.

    Long Answer

    Let's explore each unit with examples and real-life applications:

    1. Molarity (M): This is defined as the number of moles of solute per liter of solution. It's calculated as:

      Molarity (M)=Moles of soluteLiters of solutionMolarity (M)=Liters of solutionMoles of solute​
      • Example: If you dissolve 1 mole of sodium chloride (NaCl) in enough water to make 1 liter of solution, the molarity is 1 M.
      • Real-life application: Molarity is widely used in laboratories for preparing solutions for chemical reactions or titrations.

    2. Molality (m): This is the number of moles of solute per kilogram of solvent. It's calculated as:

      Molality (m)=Moles of soluteKilograms of solventMolality (m)=Kilograms of solventMoles of solute​
      • Example: If you dissolve 1 mole of glucose in 1 kilogram of water, the molality is 1 m.
      • Real-life application: Molality is used in scenarios where temperature variations affect the volume of the solution, like in boiling point elevation or freezing point depression studies.

    3. Mass Percent: This is the mass of the solute divided by the total mass of the solution, multiplied by 100. It's calculated as:

      Mass Percent=(Mass of soluteTotal mass of solution)×100Mass Percent=(Total mass of solutionMass of solute​)×100
      • Example: If you mix 25 grams of salt with 75 grams of water, the mass percent of salt is (25/(25+75))×100=25%(25/(25+75))×100=25%.
      • Real-life application: Mass percent is used in cooking recipes and pharmaceutical formulations.

    4. Volume Percent: This is the volume of the solute divided by the total volume of the solution, multiplied by 100. It's calculated as:

      Volume Percent=(Volume of soluteTotal volume of solution)×100Volume Percent=(Total volume of solutionVolume of solute​)×100
      • Example: In an alcohol beverage containing 50 mL of ethanol mixed with water to make a total of 200 mL of solution, the volume percent of ethanol is (50/200)×100=25%(50/200)×100=25%.
      • Real-life application: Volume percent is often used in the alcohol industry to describe the strength of alcoholic beverages.

    5. Parts Per Million (PPM): This is the mass of the solute divided by the total mass of the solution, multiplied by 106106. It's used for very dilute solutions.

      PPM=(Mass of soluteTotal mass of solution)×106PPM=(Total mass of solutionMass of solute​)×106
      • Example: If 1 gram of a pollutant is dissolved in 1,000,000 grams of water, the concentration is 1 ppm.
      • Real-life application: PPM is used in environmental chemistry to measure pollution levels in air and water.

    These concentration units help scientists and engineers to precisely describe and control the composition of solutions in various fields, including chemistry, biology, environmental science, and the food and beverage industry.

  3. 3.Solubility

    Short Answer

    Solubility of a Solid in a Liquid: Solubility is the maximum amount of a solute (solid) that can dissolve in a given amount of solvent (liquid) at a specific temperature. Mathematically, it is expressed in terms of concentration, such as grams of solute per 100 grams of solvent.

    Solubility of a Gas in a Liquid: The solubility of a gas in a liquid is defined by Henry’s Law, which states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. The mathematical expression for Henry's Law is S = kP, where S is the solubility of the gas, P is the pressure, and k is Henry’s law constant.

    Examples:

    • Salt in water is an example of the solubility of a solid in a liquid.
    • The fizz in soda is due to carbon dioxide gas dissolved in the liquid, showcasing the solubility of a gas in a liquid.

    Long Answer

    Solubility of a Solid in a Liquid

    Explanation: Solubility of a solid in a liquid is an important concept in chemistry that describes how much of a solid substance (the solute) can be dissolved in a liquid (the solvent) at a certain temperature. The solubility varies with temperature; generally, it increases as the temperature increases.

    Mathematical Expression: If we denote the solubility of a solid in a liquid as S (in grams of solute per 100 grams of solvent), then it can be represented as:

    =Mass of SoluteMass of Solvent×100S=Mass of SolventMass of Solute​×100

    Real-Life Example: When you add sugar to tea, the sugar is the solute, and the tea is the solvent. The amount of sugar that can be dissolved increases as the tea's temperature increases.

    Applications: This concept is widely used in cooking, pharmaceuticals, and chemical industries for preparing solutions of desired concentrations.

    Activity: Try dissolving different amounts of salt in a cup of water at room temperature. Then, heat the water and try dissolving more salt to observe how solubility changes with temperature.

    Solubility of a Gas in a Liquid

    Explanation: The solubility of gases in liquids is an interesting phenomenon that depends largely on the pressure of the gas above the liquid and the temperature. According to Henry’s Law, the amount of gas that can dissolve in a liquid at a given temperature increases with the pressure of the gas.

    Mathematical Expression: Henry’s Law can be mathematically expressed as:

    =S=kP

    where S is the solubility of the gas in the liquid (usually in moles per liter), P is the partial pressure of the gas above the liquid, and k is Henry’s Law constant, which varies for different solute-solvent pairs and temperatures.

    Real-Life Example: Carbonated beverages like soda are a classic example of gas solubility. Carbon dioxide is dissolved in these drinks under high pressure, and when the pressure is released upon opening the bottle, the gas escapes, creating fizz.

    Applications: This principle is crucial in the beverage industry, in scuba diving to prevent decompression sickness, and in industrial processes involving gases.

    Activity: Open a carbonated drink and observe the bubbles forming. This is the carbon dioxide gas coming out of solution as the pressure is released.

  4. 4.Vapour Pressure of Liquid Solutions

    Short Answer

    Vapour Pressure of Liquid-Liquid Solutions: The vapor pressure of a solution made from two liquids is the sum of the partial vapor pressures of each component, each multiplied by its mole fraction in the solution.

    Raoult’s Law as a Special Case of Henry’s Law: Raoult's Law states that the vapor pressure of an ideal solution is directly proportional to the mole fraction of the solvent. It can be seen as a special case of Henry's Law for solutions where the solute concentration is very low.

    Vapour Pressure of Solutions of Solids in Liquids: The vapor pressure of a solution with a non-volatile solute is lower than the vapor pressure of the pure solvent. This decrease is proportional to the mole fraction of the solute in the solution.

    Mathematical Expressions:

    • Liquid-Liquid Solutions: =+Ptotal​=PA​xA​+PB​xB​
    • Raoult’s Law: =0Psolution​=Psolvent0​xsolvent​
    • Solutions of Solids in Liquids: Δ=0−ΔP=Psolvent0​−Psolution​

    Numerical Example: For a solution of ethanol and water:

    • Let the vapor pressure of pure ethanol (ℎ0Pethanol0​) be 100 mmHg and its mole fraction (ℎxethanol​) in the solution be 0.5.
    • The vapor pressure of the solution would be =ℎ0ℎ=100×0.5=50Psolution​=Pethanol0​xethanol​=100×0.5=50 mmHg.

    Long Answer

    Vapour Pressure of Liquid-Liquid Solutions

    Explanation: In a solution composed of two liquids, each component contributes to the total vapor pressure based on its volatility and concentration. The vapor pressure of each component (A and B) is directly proportional to its mole fraction in the solution.

    Mathematical Expression: The total vapor pressure (Ptotal​) of the solution is calculated as: =+Ptotal​=PA​xA​+PB​xB​ where PA​ and PB​ are the vapor pressures of the pure components, and xA​ and xB​ are their mole fractions in the solution.

    Real-Life Example: Mixing alcohol with water for a beverage. The total vapor pressure of the mixture is a combination of the vapor pressures of alcohol and water, each influenced by their respective amounts in the solution.

    Raoult’s Law as a Special Case of Henry’s Law

    Explanation: Raoult's Law applies to ideal solutions and states that the vapor pressure of the solvent over the solution is proportional to its mole fraction. It can be viewed as a special case of Henry's Law, emphasizing the behavior of the solvent in dilute solutions.

    Mathematical Expression: For a solvent in a solution, Raoult’s Law is expressed as: =0Psolution​=Psolvent0​xsolvent​ where 0Psolvent0​ is the vapor pressure of the pure solvent, and xsolvent​ is the mole fraction of the solvent in the solution.

    Real-Life Example: Diluting perfume with ethanol. The vapor pressure of the diluted perfume depends on the concentration of ethanol according to Raoult’s Law.

    Vapour Pressure of Solutions of Solids in Liquids

    Explanation: When a non-volatile solid is dissolved in a liquid, the vapor pressure of the resulting solution is lower than that of the pure solvent. This is because the solute particles occupy space at the surface, reducing the number of solvent molecules that can escape to the vapor phase.

    Mathematical Expression: The change in vapor pressure (ΔΔP) due to the solute is: Δ=0−ΔP=Psolvent0​−Psolution​ where 0Psolvent0​ is the vapor pressure of the pure solvent, and Psolution​ is the vapor pressure of the solution.

    Real-Life Example: Adding salt to water. The vapor pressure of the saltwater solution is lower than that of pure water because the salt ions reduce the number of water molecules that can evaporate.

    Numerical Example:

    • Consider a solution where the vapor pressure of pure water (0Pwater0​) is 23.8 mmHg and the mole fraction of water (xwater​) in the solution is 0.9.
    • Using Raoult’s Law, the vapor pressure of the solution is =0=23.8×0.9=21.42Psolution​=Pwater0​xwater​=23.8×0.9=21.42 mmHg.

    These principles and expressions provide a foundational understanding of how solutes affect the vapor pressure of solutions, important for various scientific and industrial applications, such as formulating products in the pharmaceutical and food industries, and understanding environmental phenomena.

  5. 5.An Ideal Solution & Non-Ideal Solution

    Short Answer: An ideal solution is one that follows Raoult's Law at all concentrations for both solvent and solute. It has a vapor pressure that is directly proportional to the mole fraction of the component in the solution. Non-ideal solutions do not follow Raoult's Law strictly. They have interactions between molecules that are different from those in the pure substances, leading to deviations from the expected vapor pressure.

    Let’s take a closer look at these diagrams:

    • Diagram (a): This is a graphical representation of Raoult's Law for an ideal solution. The straight lines represent the vapor pressure of pure components (P₁ and P₂). According to Raoult's Law, the vapor pressure of each component in a solution is directly proportional to its mole fraction. The upper curved line shows the total vapor pressure of the solution, which is the sum of the partial pressures of the components. This line is a straight line from the vapor pressure of pure component 1 (P₁) to that of component 2 (P₂), indicating that the solution behaves ideally.

    • Diagram (b): This shows a non-ideal solution. The curved lines again represent the vapor pressures of pure components (P₁ and P₂), but the total vapor pressure of the solution now shows a curve that deviates from the straight line connecting P₁ and P₂. This deviation occurs because the interactions between the molecules of different components are not the same as the interactions in the pure components. The curve above the straight line represents a positive deviation (the total vapor pressure is higher than expected), and the curve below represents a negative deviation (the total vapor pressure is lower than expected).

    In both diagrams, the x-axis represents the mole fraction (x₁ and x₂ for component 1 and 2, respectively), which ranges from 0 to 1. The y-axis represents the vapor pressure. As the mole fraction changes, the vapor pressure over the solution also changes.

    This concept is important in industrial applications where the vapor pressures of mixtures must be controlled, such as in distillation processes or when designing separation processes in chemical engineering.

    Long Answer: In an ideal solution, the interactions between different particles are similar to those between like particles. This means that the enthalpy of mixing is zero, and the volume of mixing is also ideally zero. The vapor pressure of each component in the solution is directly proportional to its mole fraction. Mathematically, Raoult's Law can be expressed as:

    =⋆⋅P=P⋆⋅X

    Where P is the vapor pressure of the component in the solution, ⋆P⋆ is the vapor pressure of the pure component, and X is the mole fraction of the component in the solution.

    For a non-ideal solution, interactions between particles of different components are either stronger or weaker than those between like particles, which leads to a positive or negative deviation from Raoult's Law. The enthalpy of mixing is not zero and the volume change upon mixing is not ideally zero either.

    The vapor pressure can be higher or lower than what is predicted by Raoult's Law, which is represented by:

    =⋆⋅+ΔP=P⋆⋅X+ΔP

    Here, ΔΔP represents the deviation from the ideal behavior. This deviation can be quantified by an activity coefficient, γ, which adjusts the mole fraction:

    =⋅⋆⋅P=γ⋅P⋆⋅X

    Real-life example: When you mix water and ethanol to make a drink, they form a non-ideal solution. The vapor pressure of the solution is not what you'd expect if they were ideal, because the molecules interact in a special way due to hydrogen bonding.

    Activity: To understand this better, try mixing equal parts of alcohol and water, and compare the volume before and after mixing. You will observe that the volume after mixing is slightly less than the total volume of the two liquids before mixing, indicating a non-ideal solution behavior.

    Use in real life: Understanding vapor pressure is crucial for many industries, like the beverage industry for distillation processes, and for chemists who design medicines and need precise control over solvents and their behaviors.

    Now let's look at a numerical example:

    If we have a solution made of two components, A and B, with the mole fraction of A (XA​) being 0.5 and the vapor pressure of pure A (⋆PA⋆​) being 100 mmHg, then the expected vapor pressure of A in an ideal solution would be:

    =⋆⋅=100⋅0.5=50 mmHgPA​=PA⋆​⋅XA​=100⋅0.5=50 mmHg

    For non-ideal solutions, if A and B interact more strongly than A with itself, the vapor pressure might be lower, say 45 mmHg instead of 50 mmHg. This indicates a negative deviation from Raoult's Law.

  6. 6.Colligative Properties and Determination of Molar Mass

    Short Answer

    Colligative Properties are properties of solutions that depend on the number of particles in a given volume of solvent and not on the type of particles. These properties include:

    1. Relative Lowering of Vapour Pressure: When a solute is added to a solvent, the vapour pressure of the solution decreases compared to the pure solvent.
    2. Elevation of Boiling Point: The boiling point of a solution is higher than that of the pure solvent because the solute particles disrupt the escape of solvent particles from the liquid to the gas phase.
    3. Depression of Freezing Point: The freezing point of a solution is lower than that of the pure solvent because the solute particles interfere with the formation of the solid structure of the solvent.
    4. Osmosis and Osmotic Pressure: Osmosis is the movement of solvent particles through a semi-permeable membrane from a less concentrated solution to a more concentrated one. Osmotic pressure is the pressure required to stop this flow.
    5. Reverse Osmosis and Water Purification: Reverse osmosis involves applying pressure to overcome osmotic pressure, allowing water to pass through a semi-permeable membrane while solutes are retained, thus purifying the water.

    These properties are used to determine the molar mass of solutes in solutions.

    Long Answer

    1. Relative Lowering of Vapour Pressure

    Explanation: The vapour pressure of a solution is lower than that of the pure solvent because the solute particles occupy space at the surface, reducing the number of solvent molecules that can escape into the vapour phase. The relative lowering of vapour pressure is directly proportional to the mole fraction of the solute in the solution.

    Formula: Δ=0−=0×ΔP=P0​−Ps​=P0​×Xs​

    Where 0P0​ is the vapour pressure of the pure solvent, Ps​ is the vapour pressure of the solution, and Xs​ is the mole fraction of the solute.

    Example: If the vapour pressure of pure water is 23.8 mmHg and the vapour pressure of a sugar solution is 23 mmHg, calculate the mole fraction of sugar in the solution.

    2. Elevation of Boiling Point

    Explanation: The boiling point of a solution is higher than the pure solvent's boiling point. This is because the addition of a solute lowers the solvent's vapour pressure, requiring more heat to reach the vapour pressure necessary for boiling.

    Formula: Δ=×ΔTb​=Kb​×m

    Where ΔΔTb​ is the elevation of boiling point, Kb​ is the ebullioscopic constant, and m is the molality of the solution.

    Example: If the boiling point elevation constant for water is 0.512 K kg/mol, and we have a solution with a molality of 2 mol/kg, the boiling point elevation is 0.512×2=1.0240.512×2=1.024 K.

    3. Depression of Freezing Point

    Explanation: The freezing point of a solution is lower than that of the pure solvent. Solute particles disrupt the orderly arrangement of solvent molecules into a solid, thus lowering the freezing point.

    Formula: Δ=×ΔTf​=Kf​×m

    Where ΔΔTf​ is the depression of freezing point, Kf​ is the cryoscopic constant, and m is the molality of the solution.

    Example: For water, =1.86Kf​=1.86 K kg/mol. If a solution has a molality of 1 mol/kg, the freezing point depression is 1.86×1=1.861.86×1=1.86 K.

    4. Osmosis and Osmotic Pressure

    Explanation: Osmosis is the movement of solvent molecules through a semi-permeable membrane from a region of lower solute concentration to a region of higher solute concentration. Osmotic pressure is the pressure required to prevent this movement, directly proportional to the solute concentration.

    Formula: Π=×××Π=i×M×R×T

    Where ΠΠ is the osmotic pressure, i is the van't Hoff factor, M is the molarity of the solution, R is the gas constant, and T is the temperature in Kelvin.

    Example: For a 1 M NaCl solution at 25°C (298298K), assuming =2i=2 (Na++ and Cl−−), =0.0821R=0.0821 L atm K−1−1 mol−1−1, the osmotic pressure is 2×1×0.0821×298=48.72×1×0.0821×298=48.7 atm.

    5. Reverse Osmosis and Water Purification

    Explanation: Reverse osmosis is a process where pressure is applied to a solution to force water molecules through a semi-permeable membrane, leaving the solute behind, effectively purifying the water.

    Application: It is widely used for desalination of seawater, making it drinkable, and in the purification of water for industrial and domestic use.

  7. 7.Abnormal Molar Masses

    Short Answer

    Abnormal molar masses occur when the observed molar mass of a solute in a solution differs from its expected value. This discrepancy is often due to the solute undergoing dissociation (breaking into smaller particles) or association (combining to form larger particles) in the solution. The van't Hoff factor, i, helps account for this by comparing the actual number of particles in solution to the number expected if the solute did not dissociate or associate.

    Long Answer

    Understanding Abnormal Molar Masses

    Explanation: In solutions, solutes can behave differently than expected. Instead of remaining as individual molecules, they may split into ions (dissociation) or combine to make bigger molecules or aggregates (association). This affects their colligative properties, leading to abnormal molar masses when calculated. The van't Hoff factor (i) quantifies this effect, showing how many times more or fewer particles are present compared to what was initially dissolved.

    Mathematical Expression for van't Hoff factor (i): =Actual number of particles in solutionNumber of formula units dissolvedi=Number of formula units dissolvedActual number of particles in solution​

    For Dissociation (e.g., NaCl in water): =1i=1n​ Where n is the number of particles the compound dissociates into. For NaCl, which dissociates into Na++ and Cl−−, =2i=2.

    For Association (e.g., dimerization of acetic acid): =1i=n1​ Where n is the number of molecules that combine to form one larger molecule. For acetic acid dimers, =1/2i=1/2 since two molecules of acetic acid combine to form one dimer.

    Numerical Example for Dissociation: If 1 mole of NaCl is dissolved in water, it dissociates into 1 mole of Na++ and 1 mole of Cl−−, resulting in a total of 2 moles of particles. If the expected molar mass was calculated assuming no dissociation, the observed colligative property would suggest the presence of 2 moles of particles, doubling the expected value. Thus, for NaCl, =2i=2.

    Numerical Example for Association: If 1 mole of acetic acid in a solution forms 0.5 moles of dimers, the total number of entities in solution is 0.5 moles of dimers. This would mean the observed molar mass is double the expected value for monomeric acetic acid because now, 1 mole of acetic acid appears to behave like 0.5 moles of solute. Thus, for acetic acid dimerization, =1/2i=1/2.

    Application: Understanding abnormal molar masses is crucial for accurately determining the molar mass of solutes in solutions, especially in biochemical and industrial processes where solute behavior significantly impacts the system's properties.

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