Chemistry Lesson Plan Objective in Hindi for Students
A non-ideal solution is formed by mixing 30 mL of chloroform with 50 mL of acetone. What will be the total volume of the resulting mixture?
a) Greater than 80 mL
b) Less than 80 mL
c) Exactly 80 mL
d) Greater than or equal to 80 mL
Explanation: In solutions involving Molecular interactions, especially between different liquids, the final volume after mixing does not always equal the simple sum of individual volumes. This happens due to changes in intermolecular forces such as hydrogen Bonding, dipole–dipole interactions, and van der Waals forces. When two liquids like chloroform and acetone are mixed, they may interact strongly, causing molecules to pack more efficiently than in their pure states. This can lead to contraction or expansion in volume depending on whether attractive or repulsive forces dominate. In non-ideal solutions, deviations from ideal behavior occur because the interactions between unlike molecules differ significantly from those between like molecules. If attraction between different molecules is stronger, the mixture becomes more tightly packed, reducing the expected total volume. If repulsion dominates, expansion may occur. Therefore, the actual volume must be understood in terms of Molecular interaction behavior rather than simple arithmetic addition. This concept is commonly used in physical Chemistry to explain real solution behavior and deviations from Raoult’s law.
Option b - Less than 80 mL
The vapour pressure will be the lowest for which of the following?
a) Pure water
b) 0.1 m aqueous urea solution
c) 0.2 m aqueous urea solution
d) 0.3 m aqueous urea solution
Explanation: Vapour pressure of a solution depends on the number of solute particles present and their effect on the escaping tendency of solvent molecules. When a non-volatile solute is added to a solvent like water, it reduces the surface availability of solvent molecules, thereby lowering the vapour pressure. This phenomenon is a colligative property, meaning it depends only on the number of solute particles rather than their chemical nature. As solute concentration increases, more solvent molecules are effectively “blocked” from escaping into the vapour phase. This leads to a progressively greater reduction in vapour pressure. The relationship is described by Raoult’s law, where vapour pressure decreases in proportion to the mole fraction of the solvent. Thus, among solutions with the same solvent but different concentrations, the one with the highest solute concentration will exhibit the strongest reduction in vapour pressure. The concept is widely used in predicting solution behavior in Chemistry and understanding how solutes influence physical properties of solvents.
Option d - 0.3 m aqueous urea solution
When 18 g of glucose is dissolved in 178.28 g of water, what is the vapour pressure of water at 100 °C for this solution?
a) 759 torr
b) 7.60 torr
c) 76 torr
d) 752.40 torr
Explanation: Vapour pressure in a solution depends on the relative proportion of solvent molecules present at the surface compared to the pure solvent. When a non-volatile solute like glucose is dissolved in water, it reduces the mole fraction of water in the mixture. Since only solvent molecules contribute to vapour formation, fewer available solvent molecules lead to a reduced escaping tendency into the vapour phase. This effect is governed by Raoult’s law, which states that vapour pressure is directly proportional to the mole fraction of the solvent. In such problems, glucose does not contribute to vapour pressure because it does not volatilize. Instead, it only dilutes the solvent’s effective concentration. The overall vapour pressure is therefore lower than that of pure water at the same temperature. The calculation typically involves converting given masses into moles, determining mole fraction of water, and then applying the proportional relationship with the vapour pressure of pure water at 100 °C. This concept is important in understanding boiling point behavior and how dissolved substances influence phase equilibrium in liquid systems.
Option d - 752.40 torr
The relative lowering of vapour pressure caused by dissolving 71.5 g of a substance in 100 g of water is 0.00713. What is the molar Mass of the solute?
a) 18
b) 342
c) 60
d) 180
Explanation: Relative lowering of vapour pressure is a colligative property that depends on the ratio of solute particles to solvent molecules. It is defined as the decrease in vapour pressure of a solvent when a non-volatile solute is added, relative to the vapour pressure of the pure solvent. This property is directly proportional to the mole fraction of the solute in dilute solutions. For a non-electrolyte, the number of solute particles is directly related to its moles, which in turn depends on its molar Mass. By converting the given Mass of solute and solvent into moles, one can establish a relationship between mole fraction and relative lowering. Since the solvent is water, its moles are calculated from its molar Mass, while the solute’s moles are expressed in terms of its unknown molar Mass. The proportionality allows determination of molar Mass from measurable vapour pressure changes. This approach is widely used in physical Chemistry for determining Molecular weights of unknown substances, especially when they do not easily form volatile compounds or suitable crystals for other methods.
Option d - 180
Colligative properties are commonly used to determine the
Explanation: Colligative properties depend solely on the number of solute particles present in a solution and not on their chemical identity. These properties include relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. Since these effects are directly related to the concentration of particles in solution, they provide a powerful experimental method for determining Molecular characteristics of solutes. In particular, they are widely used to determine molar Mass because the magnitude of the change in a colligative property can be mathematically related to the number of moles of solute present. By measuring any one of these properties and knowing the Mass of solute used, the molar mass can be calculated. This method is especially useful for substances that are difficult to analyze using direct chemical or spectroscopic methods. It is also applicable for non-volatile and non-electrolyte solutes where ideal behavior assumptions hold reasonably well.
Option a - Molar mass
A solution contains 2 moles of a solute dissolved in 2 moles of water. If the vapour pressure of pure water is 24 mm Hg, what is the vapour pressure of the solution?
a) 24 mm Hg
b) 32 mm Hg
c) 48 mm Hg
d) 12 mm Hg
Explanation: Vapour pressure lowering occurs when a non-volatile solute is added to a solvent, reducing the number of solvent molecules available at the liquid surface. According to Raoult’s law, the vapour pressure of the solution is proportional to the mole fraction of the solvent. In this case, both solute and solvent are present in equal moles, meaning the solvent’s mole fraction is significantly reduced compared to pure water. Since only water contributes to vapour formation, its effective concentration determines the vapour pressure. The solute does not contribute to vapour pressure but dilutes the solvent. The ratio of moles directly determines how much the vapour pressure is reduced relative to the pure solvent value. This principle is fundamental in colligative property problems and helps predict how strongly solutes influence phase equilibrium in liquid systems.
Option d - 12 mm Hg
An aqueous glucose solution is prepared by dissolving 18 g of glucose in 90 g of water. What is the relative lowering in vapour pressure?
a) 0.02
b) 1.1
c) 20
d) 1.96
Explanation: Relative lowering of vapour pressure is a colligative property that depends on the ratio of solute particles to solvent molecules in a solution. Glucose is a non-volatile solute, so it does not contribute to vapour pressure. Instead, it reduces the fraction of solvent molecules at the surface, lowering the escaping tendency of water molecules. The extent of lowering depends on the mole fraction of solute relative to solvent. To analyze such a system, masses are converted into moles using molar masses, and then mole fractions are used to determine the proportional decrease in vapour pressure. Since glucose does not dissociate, the number of solute particles equals the number of moles. The relationship is linear for dilute solutions, making it a useful method for understanding solution behavior and intermolecular effects in liquid mixtures.
Option a - 0.02
Which of the following is a colligative property of a solution?
a) Vapour pressure
b) Boiling point
c) Osmotic pressure
d) Freezing point
Explanation: Colligative properties are physical properties of solutions that depend only on the number of dissolved particles and not on their chemical nature. These include vapour pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. Each of these arises due to the reduction in effective solvent molecules caused by the presence of solute particles. For example, in vapour pressure lowering, fewer solvent molecules escape into the vapour phase; in boiling point elevation, more energy is required to reach boiling; in freezing point depression, the formation of Solid structure is hindered; and in osmotic pressure, solvent movement across a semipermeable membrane is affected. Among the given choices, only one represents such a colligative effect, while others may represent general physical properties of substances. Understanding this distinction is important in Thermodynamics and solution Chemistry.
Option c - Osmotic pressure
A non-volatile solute weighing 18 g is dissolved in 900 g of water, producing a relative lowering in vapour pressure of 0.002. What is the molar mass of the solute?
a) 60
b) 342
c) 18
d) 180
Explanation: Relative lowering of vapour pressure depends on the mole fraction of solute in a dilute solution. Since the solute is non-volatile, it does not contribute to vapour pressure; it only reduces the number of solvent molecules at the surface. This reduction is directly proportional to the number of solute particles present. By expressing moles of solute in terms of its molar mass and converting solvent mass into moles, a proportional relationship can be formed. The given relative lowering value allows determination of how many moles of solute are present. This is then used to calculate the molar mass. Such calculations are a standard application of colligative properties and are widely used in physical Chemistry to determine Molecular weights of unknown substances, especially Organic compounds that do not ionize or volatilize.
Option d - 180
The vapour pressure of pure water at 30 °C is 31.8 mm Hg. What will be the vapour pressure of a solution containing 90 g of glucose dissolved in 900 g of water at the same temperature?
a) 30.24 mm Hg
b) 31.482 mm Hg
c) 32.46 mm Hg
d) 30.74 mm Hg
Explanation: Vapour pressure of a solution is always lower than that of the pure solvent when a non-volatile solute is present. This occurs because solute particles occupy space at the liquid surface and reduce the number of solvent molecules that can escape into the vapour phase. Glucose, being non-volatile, only affects vapour pressure through dilution of water molecules. According to Raoult’s law, vapour pressure depends on the mole fraction of the solvent. Therefore, as the amount of solute increases, the mole fraction of water decreases, leading to a proportional reduction in vapour pressure. The calculation involves converting masses into moles and applying the proportional relationship between mole fraction and vapour pressure. This principle is essential in understanding solution Thermodynamics and phase equilibrium behavior.
Option b - 31.482 mm Hg
Benzene and toluene form an almost ideal solution. At 25 °C, the vapour pressures of pure benzene and toluene are 75 torr and 22 torr respectively. What is the partial vapour pressure of benzene in a solution containing 78 g of benzene and 46 g of toluene?
a) 50 torr
b) 25 torr
c) 37.5 torr
d) 53.5 torr
Explanation: In ideal solutions, each component obeys Raoult’s law independently. The partial vapour pressure of a component depends on its mole fraction in the liquid phase multiplied by its pure vapour pressure. Benzene and toluene form nearly ideal mixtures because their intermolecular interactions are similar to those in the pure liquids. Therefore, no significant deviation occurs from Raoult’s law. To determine partial vapour pressure, the mole fraction of benzene is first found using its amount relative to total moles of both components. This mole fraction is then multiplied by the vapour pressure of pure benzene. This approach reflects how each component contributes independently to the total vapour pressure of the mixture. Understanding this behavior is essential for studying binary liquid mixtures and phase equilibrium in physical Chemistry.
Option a - 50 torr
The lowering of vapour pressure caused by adding 100 g of sucrose to 1000 g of water is calculated when the vapour pressure of pure water at 25 °C is 28.8 mm Hg. What is the vapour pressure lowering?
a) 1.25 mm Hg
b) 0.125 mm Hg
c) 1.15 mm Hg
d) 0.12 mm Hg
Explanation: Vapour pressure lowering occurs when a non-volatile solute such as sucrose is dissolved in a solvent like water. This reduces the mole fraction of water molecules at the surface, thereby decreasing their tendency to escape into the vapour phase. The magnitude of this decrease depends on the ratio of solute particles to solvent molecules. According to Raoult’s law, vapour pressure lowering is proportional to the solute mole fraction in dilute solutions. Since sucrose is non-volatile and does not dissociate, each Molecule contributes as a single particle. The calculation involves converting given masses into moles, determining mole fraction of solute, and then applying the proportional relationship with vapour pressure of pure water. This principle is widely used in colligative property analysis and helps explain how dissolved substances influence phase behavior in solutions.
Option b - 0.125 mm Hg
When a liquid is heated, its vapour pressure
a) Increases
b) Decreases
c) Increases and becomes constant at its boiling point
d) Cannot be predicted
Explanation: Vapour pressure is the pressure exerted by vapour molecules when a liquid is in dynamic equilibrium with its vapour phase. This equilibrium depends strongly on temperature because temperature directly affects the kinetic energy of molecules in the liquid. As temperature increases, more molecules gain sufficient energy to overcome intermolecular attractions and escape from the liquid surface into the vapour phase. This increases the number of vapour molecules in the container, thereby increasing vapour pressure. The relationship between temperature and vapour pressure is exponential in nature and is often described by the Clausius–Clapeyron equation in Thermodynamics. At higher temperatures, the rate of evaporation increases significantly compared to condensation, leading to a higher equilibrium vapour pressure. At the boiling point, vapour pressure becomes equal to external atmospheric pressure, allowing bubbles of vapour to form throughout the liquid. This concept is fundamental in phase change behavior and explains why liquids evaporate faster when heated.
Option c - Increases and becomes constant at its boiling point
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