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SOLUTIONS - NCERT Solutions | Chemistry Chapter 01 | Class 12th Boards

NCERT Wallah・184 minutes read

The text outlines the principles of solutions and colligative properties, focusing on calculations involving molarity, molality, vapor pressure, and freezing/boiling point depressions. It emphasizes the significance of understanding how solute concentration affects solution behavior, particularly through Raoult's Law and osmotic pressure, while providing examples and formulas for practical application in chemistry.

Insights

  • The text begins by discussing how pressure reduction impacts nitrogen solubility, highlighting a specific example involving the release of glucose, which has a molecular weight of 180 grams per mole.
  • Osmotic pressure is defined as the minimum pressure needed to halt osmosis, underlining its significance in understanding solutions and their behaviors.
  • The speaker introduces an NCERT series aimed at helping students prepare for exams by revisiting key physical chemistry concepts related to solutions.
  • The series intends to assist students in achieving full marks in board exams by focusing on essential concepts and numerical problems, emphasizing the consistency of values in exam questions.
  • Chapter 7 is dedicated to solutions, featuring various exam question formats, including multiple-choice and short answer questions, to prepare students effectively.
  • An analysis of the 2023-24 sample paper shows it includes multiple-choice, short answer, and long answer questions, with a particular focus on solutions and their classifications.
  • A solution is described as a homogeneous mixture of two or more components, with sugar and water serving as a practical example.
  • The speaker explains binary solutions, identifying the solute (the dissolved substance) and the solvent (the substance that dissolves).
  • Solutions are categorized into nine types based on the physical state of the solute and solvent, with practical examples such as air as a gas-gas solution and glucose in water as a solid-liquid solution.
  • The text mentions that hydrogen gas is used in reactions with alkenes to produce alkanes, with solid palladium, platinum, or nickel serving as catalysts for H2 gas adsorption.
  • Specific examples of solid solutions include alloys like 24-carat gold (gold and copper) and brass (copper and zinc), illustrating the diverse nature of solutions.
  • The mole fraction is defined as the ratio of the number of moles of a component to the total number of moles in the solution, with calculations provided for both solute and solvent.
  • Molarity is introduced as the number of moles of solute per liter of solution, with a formula to aid students in understanding how to calculate this measurement.
  • The text explains mass percentage as the mass of solute per 100 grams of solution, providing a formula for students to calculate this important property.
  • The discussion concludes with examples of calculations involving solutions, such as determining the molality and mole fraction of a glucose solution, emphasizing the importance of understanding these concepts for practical applications in chemistry.

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Recent questions

  • What is a solution in chemistry?

    A solution is a homogeneous mixture of two or more components, where the solute is dissolved in the solvent. For example, when sugar is mixed with water, the sugar dissolves, creating a uniform solution. Solutions can exist in various states, including gases, liquids, and solids, and are characterized by their consistent composition and properties throughout. Understanding solutions is fundamental in chemistry, as they play a crucial role in various chemical reactions and processes.

  • How do you calculate molarity?

    Molarity is calculated by dividing the number of moles of solute by the volume of the solution in liters. The formula is expressed as M = moles of solute / volume of solution (L). For instance, if you have 1 mole of sodium chloride dissolved in 1 liter of water, the molarity of the solution would be 1 M (molar). This measurement is essential for understanding the concentration of solutions in chemical reactions and laboratory settings.

  • What is osmotic pressure?

    Osmotic pressure is the minimum pressure required to stop the flow of solvent through a semi-permeable membrane during osmosis. It occurs when solvent molecules move from an area of lower solute concentration to an area of higher solute concentration, aiming to equalize concentrations on both sides of the membrane. The osmotic pressure can be mathematically defined as π = cRT, where c is the concentration of the solution, R is the gas constant, and T is the temperature in Kelvin. This concept is vital in biological systems and various industrial applications.

  • What is freezing point depression?

    Freezing point depression is a colligative property that describes the decrease in the freezing point of a solvent when a non-volatile solute is added. The extent of the depression is directly proportional to the number of solute particles in the solution. The formula for calculating freezing point depression is ΔTf = Kf * m, where Kf is the freezing point depression constant for the solvent, and m is the molality of the solution. This phenomenon is important in understanding how solutes affect the physical properties of solvents, such as in antifreeze solutions.

  • What is Raoult's Law?

    Raoult's Law states that the partial vapor pressure of each component in a solution is proportional to its mole fraction in the mixture. For an ideal solution, the total vapor pressure is the sum of the partial pressures of each component, expressed mathematically as P_total = P_A + P_B, where P_A and P_B are the partial pressures of components A and B, respectively. This law is crucial for understanding the behavior of liquid-liquid solutions and is applied in various fields, including distillation and the study of volatile substances.

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Summary

00:00

Understanding Solutions in Physical Chemistry

  • The discussion begins with the concept of pressure reduction and its effects on nitrogen solubility, specifically mentioning a release of 36 grams of glucose, which has a molecular weight of 180 grams per mole.
  • Osmotic pressure is introduced as a critical concept, defined as the minimum pressure required to stop osmosis, emphasizing its importance in understanding solutions.
  • The speaker welcomes students to an NCERT series focused on solutions, aiming to cover back questions from physical chemistry chapters to aid in exam preparation.
  • The series is designed to help students achieve full marks in board exams by revising key concepts and solving numerical problems, with an emphasis on the consistency of values in exam questions.
  • The chapter on solutions is identified as Chapter 7, which includes various types of questions in the exam format, such as multiple-choice questions (16 questions worth 1 mark each) and short answer questions (2 and 3 marks).
  • The analysis of the 2023-24 sample paper reveals that it includes multiple-choice questions, short answers, and long answer questions, with a focus on solutions and their types.
  • The definition of a solution is provided, describing it as a homogeneous mixture of two or more components, with sugar and water as a practical example of a solution.
  • The speaker explains binary solutions, which consist of two components, identifying the solute (the substance being dissolved) and the solvent (the substance doing the dissolving).
  • Nine types of solutions are categorized based on the physical state of the solute and solvent, including gaseous, liquid, and solid solutions, with examples provided for each type.
  • Specific examples of solutions are given, such as air as a gas-gas solution, chloroform and nitrogen as a gas-liquid solution, and glucose dissolved in water as a solid-liquid solution, illustrating the practical applications of these concepts.

17:17

Understanding Solutions and Their Calculations

  • Hydrogen gas (H2) is used in reactions involving alkenes to produce alkanes, utilizing solid palladium (Pd), platinum (Pt), or nickel (Ni) as catalysts for the adsorption of H2 gas.
  • An example of a liquid in solid solution is amalgam, which consists of liquid mercury combined with solid metals like sodium amalgam or zinc amalgam.
  • Solid in solid solutions are exemplified by alloys, such as 24-carat gold, which is a mixture of gold and copper, and other alloys like brass (copper and zinc) and bronze (copper and tin).
  • A solution is defined as a homogeneous mixture of two or more pure substances, characterized by uniform composition and properties throughout.
  • There are nine types of solutions categorized into three types each for gases, liquids, and solids, which can be represented in a table format for clarity in examinations.
  • An example of a solid solution where the solute is a gas is H2 adsorbed over platinum, palladium, or nickel, demonstrating gas in solid solution.
  • The mole fraction is defined as the ratio of the number of moles of a component (solute or solvent) to the total number of moles of all components in the solution, with specific calculations provided for both solute and solvent.
  • Molarity is defined as the number of moles of solute present in one liter of solution, with the formula expressed as M = moles of solute/volume of solution in liters.
  • Mass percentage is calculated as the mass of solute present in 100 grams of solution, with the formula being (mass of solute/total mass of solution) × 100, applicable for both solute and solvent.
  • To solve numerical problems involving solutions, it is essential to interpret given data correctly, such as the mass of solute and solution, and apply the appropriate formulas for calculations, including molarity and mass percentage.

33:34

Calculating Molarity and Mass Percentages in Solutions

  • The text discusses a calculation involving a solution of nitric acid (HNO3), specifically a solution containing 68 grams of HNO3 in 100 grams of solution, indicating that the mass of the solute is 68 grams and the total mass of the solution is 100 grams.
  • To find the molar mass of HNO3, the components are added: 1 hydrogen (1 g), 1 nitrogen (14 g), and 3 oxygens (16 g each), resulting in a total molar mass of 63 grams per mole.
  • The number of moles of HNO3 is calculated using the formula: number of moles = given mass (68 g) / molar mass (63 g), yielding approximately 1.07 moles.
  • The density of the solution is given as 1.54 grams per milliliter, and to find the volume of the solution, the mass of the solution (100 g) is divided by the density (1.5 g/mL), resulting in a volume of approximately 66.67 mL.
  • The molality of the solution is calculated using the formula: molality = number of moles of solute / volume of solvent in liters. The volume of solvent is derived from the total mass of the solution minus the mass of the solute.
  • The next calculation involves determining the mass percentage of aspirin in acetonitrile, where 6.5 grams of aspirin is dissolved in 450 grams of acetonitrile.
  • The mass percentage formula is applied: mass percentage = (mass of solute / total mass of solution) × 100, leading to the calculation of 6.5 g of aspirin divided by the total mass (6.5 g + 450 g) multiplied by 100, resulting in a mass percentage of approximately 1.4%.
  • A further question involves calculating the molality and mole fraction of a glucose solution, where 10 grams of glucose is present in 100 grams of solution, with a density of 1.2 g/mL.
  • The molar mass of glucose (C6H12O6) is established as 180 grams per mole, and the number of moles of glucose is calculated as 10 g / 180 g/mol, yielding approximately 0.0555 moles.
  • The mole fraction is determined by dividing the moles of glucose by the total moles of solute and solvent, with the solvent being water, calculated from the remaining mass of the solution (90 g) divided by its molar mass (18 g/mol).

49:17

Understanding Moles Mass and Concentration Calculations

  • The text begins with a mathematical problem involving the division of 555 by 5.0, suggesting that the answer can be approximated as 0.01, emphasizing the simplicity of the calculation.
  • It explains that mole fraction is a unitless quantity because both the solute and solvent are measured in moles, and it provides a method to calculate the moles of solvent using the formula: moles of solvent = 1 - mole fraction of solute.
  • The calculation of molality is introduced, defined as the number of moles of solute divided by the mass of solvent in kilograms, with a reminder that to convert grams to kilograms, one must divide by 1000.
  • The example provided states that if there are 0.555 moles of solute and 90 grams of solvent, the mass in kilograms is calculated as 0.09 kg (90 grams / 1000).
  • The text instructs to calculate molality using the formula: molality = moles of solute / mass of solvent (in kg), leading to a calculation of approximately 0.66 moles/kg.
  • It transitions to discussing molarity, defined as the number of moles of solute divided by the volume of solution in liters, with a note that if the volume is not given, the mass of the solution can be used if the density is known.
  • The mass of the solution is given as 100 grams with a density of 1.2 grams/mL, allowing the calculation of volume using the formula: volume = mass / density, resulting in a volume of approximately 83.33 mL (100 g / 1.2 g/mL).
  • The text emphasizes the importance of understanding the relationship between moles, mass, and molarity, leading to a calculation of molality using the previously determined values.
  • A problem involving a mixture of sodium carbonate and sodium bicarbonate is presented, where the total mass is 1 gram, and the molar masses of sodium carbonate (106 g/mol) and sodium bicarbonate (84 g/mol) are calculated.
  • The final calculations yield the masses of sodium carbonate and sodium bicarbonate in the mixture, approximately 0.558 grams and 0.442 grams, respectively, allowing for further calculations of moles for each compound.

01:05:54

Chemical Reactions and Solution Calculations

  • Sodium hydrogen carbonate (NaHCO3) reacts with hydrochlorous acid (HClO2) in a stoichiometric ratio, where 2 moles of sodium carbonate (Na2CO3) are equivalent to 1 mole of HClO2, indicating that the amount of Na2CO3 is double that of HClO2 in the reaction.
  • The calculation of moles shows that if 0.0025 moles of Na2CO3 are present, then 0.005 moles of HClO2 will be required for the reaction, as they react in a 1:1 ratio.
  • The total moles of reactants can be calculated by adding the moles of Na2CO3 and HClO2, resulting in approximately 0.01052 moles when Na2CO3 is 0.00526 moles.
  • To find the molarity of an acetic acid (AC) solution, the formula is Molarity = Moles of solute / Volume of solution in liters. For a 0.1 M solution, if 0.01578 moles of AC are present, the volume can be calculated as Volume = Moles / Molarity, leading to a volume of 157.8 mL.
  • When mixing two solutions, the first solution has a mass of 300 grams with a solute mass of 75 grams (25% concentration), and the second solution has a mass of 400 grams with a solute mass of 160 grams (40% concentration).
  • The total mass of solute after mixing both solutions is 235 grams, and the total mass of the resulting solution is 700 grams (300g + 400g).
  • The mass percentage of solute in the resulting solution is calculated as (mass of solute / total mass of solution) x 100, resulting in a mass percentage of 33.5%.
  • To find the mass percentage of solvent, subtract the mass percentage of solute from 100%, yielding a solvent mass percentage of 66.5%.
  • The molality of a solution is determined by the formula: Molality = Moles of solute / Mass of solvent in kg. For ethylene glycol (222.6 grams) mixed with 200 grams of water, the mass of water in kg is 0.2 kg, resulting in a molality of 17.95.
  • To calculate the volume of a solution when given its mass and density, use the formula: Volume = Mass of solution / Density. For a solution with a total mass of 422.6 grams and a density of 1.07 g/mL, the volume can be calculated accordingly.

01:23:27

Calculating Chloroform Concentration in Water

  • The molarity of the solution is expressed as MI Clear, indicating a concentration that can be converted into liters by dividing by 1000, which is a standard conversion for volume in chemistry.
  • The sample in question is drinking water, which contains chloroform, a carcinogenic substance, at a concentration of 15 parts per million (ppm) by mass.
  • To calculate the mass percentage of chloroform in the solution, the formula used is (mass of solute / mass of solution) x 100, where 15 grams of chloroform is present in 1,000,000 grams of solution.
  • The mass of the solvent is assumed to be nearly equal to the mass of the solution due to the negligible amount of solute, allowing for simplification in calculations.
  • The first calculation involves determining the mass percentage, which results in a value of 15 x 10^-6, or 0.0015% by mass of chloroform in the solution.
  • The second part of the calculation requires finding the molality, which is defined as the number of moles of solute per kilogram of solvent; this requires knowing the moles of chloroform and the mass of the solvent in kilograms.
  • The molar mass of chloroform (CHCl3) is calculated as 12 (C) + 1 (H) + 3 x 35.5 (Cl) = 119.5 grams per mole, which is essential for determining the number of moles from the mass of chloroform.
  • To find the moles of chloroform, the mass (15 grams) is divided by its molar mass (119.5 grams/mole), resulting in approximately 0.125 moles of chloroform.
  • The final molality is calculated by dividing the moles of solute (0.125) by the mass of the solvent (1,000 grams converted to kilograms), yielding a molality of 0.125 x 10^-3 or 1.25 x 10^-4 mol/kg.
  • A separate calculation involves the drug naloxone, which has a dosage of 1.5 mg and a molality of 1.5 x 10^-3 mol/kg; the molar mass of naloxone is 311 grams/mole, leading to a required mass of aqueous solution of approximately 4665 mg to achieve the desired dosage.

01:39:11

Calculating Concentrations in Chemical Solutions

  • The aqua solution mass is calculated to be 4665 grams, and if 1.5 mg is involved, the calculations must be precise to avoid errors in determining the required amounts.
  • For the preparation of a 250 mL solution of 0.15 molar benzoic acid, the molar mass of benzoic acid (C6H5COOH) is calculated to be 122 grams, which is essential for determining the mass needed for the solution.
  • To find the number of moles of benzoic acid, divide the mass of benzoic acid by its molar mass, and use the formula: moles = mass (g) / molar mass (g/mol).
  • The calculation for the required mass of benzoic acid involves multiplying the molarity (0.15 mol/L) by the volume (0.250 L) and the molar mass (122 g/mol), resulting in approximately 4.57 grams needed for the solution.
  • The density of lake water is given as 1.25 grams per mL, and the task is to find the sodium ion concentration in a 1 kg sample of this water.
  • To calculate the molality of sodium ions (Na+), use the formula: molality = moles of solute / mass of solvent (in kg), where the mass of Na+ is given as 92 grams.
  • The number of moles of sodium ions is calculated by dividing the mass of sodium (92 g) by its molar mass (approximately 23 g/mol), resulting in about 4 moles of Na+.
  • The final molality of the sodium ions in the solution is determined to be 4 moles per kg of solvent, indicating a concentrated solution.
  • The solubility of gases in liquids is influenced by temperature, where an increase in temperature shifts the equilibrium towards the endothermic direction, affecting gas solubility.
  • The solubility of a solute in a saturated solution is defined as the maximum amount of solute that can dissolve at a given temperature, and this equilibrium is crucial for understanding solution behavior.

01:57:08

Temperature and Pressure Effects on Gas Solubility

  • Increasing temperature in a reaction favors the endothermic direction, causing gas and liquid to separate, which affects solubility.
  • As temperature rises, the solubility of gases decreases; conversely, lower temperatures increase gas solubility, making cold drinks more enjoyable due to higher CO2 solubility.
  • Aquatic animals thrive in colder water because lower temperatures enhance oxygen solubility, providing them with more oxygen compared to warmer water.
  • The solubility of a gas is influenced by its nature; polar gases have higher solubility due to higher critical temperatures, making them easier to liquefy than non-polar gases.
  • Henry's Law states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the liquid, which can be expressed mathematically as P = k * mole fraction.
  • The unit of Henry's constant (k) can vary, typically expressed in atm or bar, and is crucial for calculating solubility in terms of mass or molarity.
  • Dalton's approach simplifies expressing gas solubility in terms of mole fraction, leading to the formulation of Henry's constant (Ketch) for practical applications.
  • Applications of Henry's Law include the carbonation of beverages, where CO2 is dissolved under high pressure, enhancing its solubility in soda cans.
  • In deep-sea diving, increased pressure raises nitrogen solubility in divers' oxygen tanks, which can lead to decompression sickness if divers ascend too quickly, causing nitrogen bubbles to form in the bloodstream.
  • At high altitudes, lower atmospheric pressure reduces oxygen solubility in blood, leading to altitude sickness in climbers, characterized by weakness and impaired cognitive function.

02:12:49

Calculating Gas Partial Pressure and Solubility

  • The text discusses the calculation of partial pressure of a gas in a solution, starting with a scenario where the number of particles in the solution is 5 x 10² grams, leading to a partial pressure of 6.56 x 10⁻² grams for the gas involved, applying Henry's Law to determine the relationship between solubility and pressure.
  • In the first case, the mass of the gas is given as 5 x 10⁻² grams, and the resulting calculation shows that the partial pressure is 0.76 Pa, derived from the ratio of the two masses involved in the solution.
  • The second case reiterates that if the gas and temperature remain constant, the mass of the gas will affect the pressure, with the same mass of 5 x 10⁻² grams leading to a calculated pressure of 0.76 Pa after simplifying the ratio of the two masses.
  • The text introduces Henry's constant for methane in benzene, specified as 4.27 x 10⁵ Pa, and instructs to calculate the solubility of methane under a pressure of 760 mm Hg, using the formula p = k * x, where x represents the mole fraction of methane.
  • The calculation for the solubility of methane results in a value of 1.78 x 10⁻³, derived from dividing the given pressure (760 mm Hg) by Henry's constant (4.27 x 10⁵ Pa).
  • The text explains the composition of air, noting that nitrogen constitutes 79% and oxygen 20% of the air mixture, with a total pressure of 10 atm exerted by the air above water, leading to calculations of partial pressures for both gases.
  • The partial pressure of oxygen is calculated as 2 atm (20% of 10 atm), while the partial pressure of nitrogen is calculated as 7.9 atm (79% of 10 atm), which are then used to find their respective mole fractions in water.
  • The mole fraction calculations for both gases are performed using their partial pressures and Henry's constants, with the text emphasizing the importance of these values in determining solubility in water.
  • The text transitions to Raoult's Law, explaining that it applies to liquid-liquid solutions where both solute and solvent are volatile, and describes the equilibrium between the liquid and vapor phases in a closed container.
  • Raoult's Law states that the partial vapor pressure of each component in a mixture is proportional to its mole fraction, leading to the total vapor pressure being the sum of the partial pressures of each component, thus establishing a relationship for ideal solutions.

02:30:03

Understanding Raoult's Law and Solution Behavior

  • The text discusses the principles of Raoult's Law, which relates to the vapor pressure of solutions, specifically focusing on ideal and non-ideal solutions. It states that for a binary solution with two components, the partial vapor pressure of each component can be calculated using the formula: \( P_A = P^0_A \cdot x_A \) and \( P_B = P^0_B \cdot x_B \), where \( P^0 \) represents the vapor pressure of the pure components and \( x \) represents their mole fractions.
  • In an ideal solution, Raoult's Law holds true, meaning the total vapor pressure \( P_{total} \) is equal to the sum of the partial pressures of the components: \( P_{total} = P_A + P_B \). If the solution deviates from this behavior, it is classified as non-ideal.
  • Non-ideal solutions exhibit either positive or negative deviations from Raoult's Law. Positive deviation occurs when the experimental vapor pressure is greater than predicted by Raoult's Law, indicating weaker interactions between the molecules of different components.
  • Conversely, negative deviation occurs when the experimental vapor pressure is lower than that predicted by Raoult's Law, suggesting stronger interactions between the molecules of different components.
  • The text emphasizes that the mixing of components can lead to changes in volume and enthalpy. In positive deviation, the volume increases due to weaker interactions, while in negative deviation, the volume decreases due to stronger interactions.
  • It is noted that the entropy change (\( \Delta S \)) during mixing is always positive, regardless of whether the solution is ideal, positively deviating, or negatively deviating, as mixing increases disorder.
  • The text provides examples of ideal solutions, such as mixtures of similar molecules like ethyl bromide and ethyl chloride, which exhibit ideal behavior due to similar molecular interactions.
  • It also discusses specific cases of positive and negative deviations, such as the mixing of water with alcohol, where the hydrogen bonding interactions are weaker after mixing, leading to positive deviation.
  • The importance of molecular attractions in solutions is highlighted, particularly how the strength of interactions between solute and solvent affects the overall behavior of the solution.
  • Finally, the text encourages reviewing questions related to Raoult's Law and deviations, suggesting that understanding these concepts is crucial for grasping the behavior of solutions in chemistry.

02:46:34

Understanding Raoult's Law in Vapor Pressure

  • The discussion begins with a focus on understanding the importance of complete numerical data in problem-solving, emphasizing that students should keep their answers thorough and clear, particularly in relation to the subject matter being taught.
  • The instructor introduces the concept of Raoult's Law, explaining that it can be used to derive numerical results, specifically in calculating the total vapor pressure of a mixture of heptanol and octane, with given vapor pressures of 105.2 kPa for heptanol and 46.8 kPa for octane.
  • To find the total vapor pressure, students are instructed to calculate the mole fractions of heptanol and octane using the formula: number of moles = mass (grams) / molar mass (grams/mol), with heptanol's mass given as 26 grams and octane's mass as 35 grams.
  • The molar mass of heptanol is not explicitly stated but is implied to be necessary for calculations, while the molar mass of octane is given as 114 grams. The instructor guides students to divide the mass by the molar mass to find the number of moles for both substances.
  • The mole fraction of heptanol is calculated as 0.31, and the mole fraction of octane is derived as approximately 0.57, leading to the conclusion that the mole fractions should sum to 1, confirming the calculations are correct.
  • Students are then instructed to calculate the partial pressures of heptanol and octane using their respective mole fractions and the given vapor pressures, with the formula: partial pressure = mole fraction × vapor pressure.
  • The partial pressure of heptanol is calculated to be approximately 45.6 kPa, while the partial pressure of octane is calculated to be around 25.4 kPa, leading to a total vapor pressure of 73.4 kPa when both partial pressures are added together.
  • The instructor transitions to a new problem involving benzene and toluene, where the vapor pressures of pure benzene (507.1 mmHg) and pure toluene (32.0 mmHg) are provided, and students are tasked with calculating the mole fraction of benzene in the vapor phase.
  • The number of moles for benzene and toluene is calculated using their respective masses and molar masses (benzene: 78 g/mol, toluene: 92 g/mol), leading to mole fractions of benzene and toluene being derived from the total moles.
  • Finally, the instructor emphasizes the importance of understanding the relationship between partial pressures and mole fractions in vapor phases, reiterating that the mole fraction of a component in vapor can be calculated by dividing its partial pressure by the total pressure, ensuring students grasp the practical application of Raoult's Law in real-world scenarios.

03:05:40

Vapor Pressure and Non-Volatile Solutes Explained

  • The discussion focuses on solutions containing a non-volatile solid solute, which does not exert vapor pressure, such as sugar or urea, and how this affects the vapor pressure of the solution compared to the pure solvent.
  • Raoult's Law applies to solutions with non-volatile solutes, indicating that the vapor pressure of the solution (PS) is lower than that of the pure solvent (P0), leading to a relative lowering of vapor pressure.
  • The formula for relative lowering of vapor pressure is given as RLVP = (P0 - PS) / P0, where P0 is the vapor pressure of the pure solvent and PS is the vapor pressure of the solution.
  • In a solution with a non-volatile solute, the vapor pressure decreases because the solute particles occupy space at the surface, preventing some solvent molecules from escaping into the vapor phase.
  • The mole fraction of the solute (Xb) can be calculated using the formula Xb = nb / (na + nb), where nb is the number of moles of solute and na is the number of moles of solvent.
  • An example is provided where the vapor pressure of a solution is given as 1.004 times that of the pure solvent, which has a normal boiling point of 100 degrees Celsius at 1 ATM pressure.
  • To find the molar mass of the solute, the vapor pressure of the solution and the mass percentage of the solute are used, with calculations involving the number of moles of solute and solvent.
  • The mass of the solvent is determined by subtracting the mass of the solute from the total mass of the solution, with a specific example stating that if 2 grams of solute are present in 100 grams of solution, the solvent mass is 98 grams.
  • The calculations involve converting pressure units from bar to ATM, where 1 ATM equals 1.013 bar, and using this to find the number of moles of solute based on the vapor pressure data.
  • The final calculations yield a molar mass of approximately 41.5 g/mol for the solute, demonstrating the process of determining the properties of a solution containing a non-volatile solute.

03:23:21

Calculating Vapor Pressure and Colligative Properties

  • The calculation begins with a distance of 56.5 miles and a pressure difference of 12.3, leading to a division of 12.3 by 56.5 to find a value of approximately 0.217. This value is essential for further calculations regarding pressure (P).
  • The next step involves subtracting 0.27 from 12.3, resulting in an approximate value of 12.03, which is used for subsequent calculations. The importance of maintaining consistent units, specifically kilopascals for pressure, is emphasized.
  • A new question is introduced, focusing on finding the mass of a non-volatile solute. The given mass of the solute is 40 grams, and the vapor pressure of pure octane is stated to be 100 atm, which is reduced to 80% due to the presence of the solute.
  • The formula for relative lower vapor pressure is introduced: \( P_0 - P_s \), where \( P_0 \) is the vapor pressure of the pure solvent (100 atm) and \( P_s \) is the vapor pressure of the solution (80 atm). This leads to the calculation of the mass of the solute.
  • The mass of the solvent is given as 114 grams, and the molar mass of the solute is calculated based on the molecular formula, resulting in a total of 114 grams for the solvent.
  • The calculation continues with the formula \( \text{Given Mass of Solute} / \text{Molar Mass} \), where the given mass is 40 grams. The calculations yield a result of 10 grams for the mass of the solute.
  • A similar question is presented regarding the vapor pressure of water, where 25 grams of glucose is dissolved in 450 grams of water. The vapor pressure of pure water is given as 17.535 mm of Hg.
  • The molar mass of glucose is noted as 180 grams, and the molar mass of water is 18 grams. The number of moles for both solute and solvent is calculated to determine the vapor pressure of the solution.
  • The concept of colligative properties is introduced, explaining that these properties depend on the number of solute particles in a solution rather than their type. This is crucial for understanding how solute concentration affects vapor pressure.
  • The elevation in boiling point is discussed, stating that the boiling point of a pure solvent increases when a non-volatile solute is added. The relationship is defined as directly proportional to the number of moles of solute present in 1 kg of solvent, establishing a foundational principle in colligative properties.

03:40:13

Colligative Properties and Their Practical Applications

  • The discussion begins with the concept of colligative properties, specifically focusing on the relationship between the change in boiling point (ΔT_b) and the number of moles of solute, indicating that ΔT_b is directly proportional to the molality (m) of the solution, expressed as ΔT_b = k_b * m, where k_b is the boiling point elevation constant for a specific solvent.
  • The boiling point elevation constant (k_b) for water is specified as 0.52 Kelvin kg/mol, emphasizing that this value is fixed for a particular solvent and is essential for calculating the change in boiling point when a solute is added.
  • The text transitions to the concept of freezing point depression, defining the freezing point as the temperature at which the vapor pressures of the solid and liquid forms of a substance are equal, using water as an example with a freezing point of 0 degrees Celsius.
  • When a non-volatile solute is added to water, it reduces the vapor pressure of the solution compared to pure water, leading to a decrease in the freezing point, which is mathematically represented as ΔT_f = T_f1 - T_f3, where T_f1 is the freezing point of pure water and T_f3 is the freezing point of the solution.
  • The relationship between freezing point depression (ΔT_f) and the molality of the solute is established, indicating that ΔT_f is directly proportional to the number of moles of solute per kilogram of solvent, with a proportionality constant (k_f) specific to the solvent, which for water is -1.86 °C kg/mol.
  • The text introduces osmotic pressure, explaining that it occurs when solvent molecules move through a semi-permeable membrane from a region of lower solute concentration to a region of higher solute concentration, aiming to equalize the concentrations on both sides.
  • Osmotic pressure (π) is defined mathematically as π = cRT, where c is the concentration (molarity) of the solution, R is the gas constant (0.0821 atm L/K mol), and T is the temperature in Kelvin, highlighting the relationship between osmotic pressure and solute concentration.
  • Isotonic solutions are described as solutions that have the same osmotic pressure, with an example given of a 0.91% saline solution that is isotonic with human blood cells, ensuring that cells neither swell nor shrink when placed in this solution.
  • A practical example is provided involving a sugar solution with a freezing point of 271 Kelvin, where the freezing point of pure water is noted as 273.15 Kelvin, indicating the need to calculate the mass percentage of sucrose in the solution based on the freezing point depression.
  • The summary concludes with a reminder that the freezing point depression and boiling point elevation are both colligative properties that depend on the number of solute particles in a solution, reinforcing the importance of understanding these concepts for practical applications in chemistry.

03:56:11

Freezing Point Depression Calculations Explained

  • The text discusses calculating the freezing point depression of solutions, specifically focusing on cane sugar (sucrose) and glucose in water, with the first solution containing 5 grams of cane sugar in 100 grams of water, resulting in a solvent mass of 95 grams.
  • The formula for freezing point depression (ΔTf) is introduced, which is ΔTf = Kf * molality, where molality (m) is defined as the number of moles of solute divided by the mass of solvent in kilograms.
  • For the cane sugar solution, the molar mass of sucrose is given as 342 grams per mole, and the calculation for molality involves converting the mass of water from grams to kilograms (95 grams = 0.095 kg) and determining the number of moles of sucrose present.
  • The freezing point of pure water is stated as 273.15 K, and the depression in freezing point is calculated using the formula ΔTf = 273.15 - Tf, where Tf is the freezing point of the solution.
  • The second solution involves 5 grams of glucose in water, with glucose having a molar mass of 180 grams per mole. The same method is applied to calculate the freezing point depression for this solution.
  • The calculations for both solutions yield a freezing point of approximately 269.065 K for the glucose solution, with the final answer being rounded to a range of 269.1 to 269.2 K.
  • The text transitions to a new problem involving two compounds formed by elements A and B, specifically AB2 and AB4, and discusses how the freezing point depression of benzene is affected when 1 gram of AB2 is dissolved, resulting in a depression of 2.3 K.
  • The molar mass of the solute is calculated using the freezing point depression formula, with the first case yielding a molar mass of approximately 11.8 grams per mole for AB2 and the second case yielding approximately 196.15 grams per mole for AB4, based on a depression of 1.3 K.
  • The final calculations involve determining the atomic masses of elements A and B based on the molar masses derived from the freezing point depression data, emphasizing the importance of accurate calculations and unit conversions throughout the process.
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