Chapter 2 Solutions

Step 1: Calculate the moles of NaCl. Molar mass of NaCl = 58.5 g/mol. Moles of NaCl = $\frac{\text{mass of NaCl}}{\text{molar mass of NaCl}}$ :

Step 2: Convert the mass of water into kilograms.

Step 3: Calculate molality.

Therefore, the molality of the solution is 0.684 mol/kg.

7. What is the effect of temperature on the solubility of gases in liquids? Explain with examples.

Answer: The solubility of gases in liquids generally decreases with an increase in temperature. This can be explained based on the behavior of gas molecules and the principles of kinetic energy.

At higher temperatures, the kinetic energy of the gas molecules increases, which causes them to move faster. As a result, they are less likely to stay dissolved in the liquid because the increased molecular motion helps the gas molecules escape from the liquid phase into the gas phase.

For example, carbon dioxide (CO2) dissolved in carbonated drinks is more soluble at lower temperatures. When the drink is kept at room temperature or higher, the CO2 escapes, leading to the loss of carbonation.

This phenomenon is also explained by Henry's Law, which states that the solubility of a gas in a liquid is inversely proportional to the temperature:

where $S_g$ is the solubility, $k_H$ is Henry's law constant, and $T$ is the temperature.

8. What are ideal solutions? Explain with an example, and discuss the deviations from Raoult’s Law in non-ideal solutions.

Answer: An ideal solution is a solution that obeys Raoult’s Law over the entire concentration range. In an ideal solution, the intermolecular forces between the solute and solvent molecules are the same as the forces between the molecules of the solute and solvent in their pure states. This means there is no heat released or absorbed upon mixing, and the solution behaves ideally.

For example, a mixture of benzene and toluene is considered an ideal solution. Both benzene and toluene are non-polar, and the intermolecular forces between their molecules are very similar, so the solution follows Raoult’s Law.

However, in non-ideal solutions, deviations from Raoult’s Law occur. These deviations are classified into two types:

1. Positive deviation occurs when the solution has higher vapor pressure than predicted by Raoult’s Law. This happens when the intermolecular forces between solute and solvent molecules are weaker than the forces between the solvent molecules in the pure state. A typical example is the mixture of acetone and chloroform, where the intermolecular forces between acetone and chloroform are weaker than the forces in the pure components.

2. Negative deviation occurs when the solution has lower vapor pressure than predicted by Raoult’s Law. This happens when the intermolecular forces between solute and solvent are stronger than the forces between the molecules of the pure components. A common example is a solution of hydrochloric acid (HCl) in water, where strong hydrogen bonding between HCl and water molecules leads to a lower vapor pressure than expected.

9. Calculate the osmotic pressure of a solution containing 5.0 grams of urea (NH₂CONH₂) dissolved in 200 mL of water at 300 K.

Answer: We can calculate the osmotic pressure using the formula:

where:

• $\Pi$ is the osmotic pressure,
• $i$ is the Van't Hoff factor (for urea, $i = 1$ because it doesn’t dissociate),
• $M$ is the molarity of the solution,
• $R$ is the gas constant (0.0821 L·atm/mol·K),
• $T$ is the temperature in Kelvin.

Step 1: Calculate the molarity of the solution. Molar mass of urea (NH₂CONH₂) = 60 g/mol. Moles of urea = $\frac{5.0}{60} = 0.0833 \, \text{mol}$ .

The volume of the solution is 200 mL = 0.2 L.

Molarity $M = \frac{0.0833}{0.2} = 0.4165 \, \text{mol/L}$ .

Step 2: Use the osmotic pressure formula to calculate $\Pi$ :

Therefore, the osmotic pressure of the solution is 10.26 atm.

10. Define Raoult's Law and explain how it applies to non-volatile solutes.

Answer: Raoult's Law states that the partial vapor pressure of each volatile component in a solution is directly proportional to the mole fraction of that component in the solution. Mathematically:

where $P_A$ is the vapor pressure of component A in the solution, $P_A^0$ is the vapor pressure of the pure component A, and $X_A$ is the mole fraction of A in the solution.

For non-volatile solutes, Raoult's Law still holds but with an important modification. Since non-volatile solutes do not vaporize, their vapor pressure is zero. Therefore, the vapor pressure of the solution is determined solely by the solvent.

The total vapor pressure of a solution containing a non-volatile solute is the sum of the vapor pressures of the solvent components:

Here, $X_{\text{solvent}}$ is the mole fraction of the solvent, and $P_{\text{solvent}}^0$ is the vapor pressure of the pure solvent.

For example, when NaCl is dissolved in water, the vapor pressure of water decreases, as NaCl is a non-volatile solute. This reduction in vapor pressure is one of the colligative properties of the solution.

11. What is meant by a saturated solution? How does the concentration of a solute change with temperature in a saturated solution?

Answer: A saturated solution is one in which the maximum amount of solute has been dissolved at a given temperature and pressure. In this state, the solution is in dynamic equilibrium; the rate at which the solute dissolves in the solvent is equal to the rate at which the solute crystallizes out of the solution. Any additional solute added will not dissolve.

The concentration of solute in a saturated solution depends on the temperature. For most solid solutes:

• As temperature increases, the solubility of the solute generally increases. This is because at higher temperatures, the solvent molecules have more energy, which allows them to break apart the solute molecules more effectively.
• For example, the solubility of NaCl in water increases slightly with temperature, while the solubility of KNO₃ increases significantly as temperature rises.

However, there are exceptions like gas solubility in liquids, where solubility decreases with increasing temperature.

12. How is the freezing point depression related to the molality of the solution? Explain using a numerical example.

Answer: The freezing point depression is directly proportional to the molality of the solution and is given by the equation:

where:

• $\Delta T_f$ is the depression in freezing point,
• $K_f$ is the cryoscopic constant (freezing point depression constant),
• $m$ is the molality of the solution.

Example: Calculate the freezing point depression when 10 grams of urea (molar mass = 60 g/mol) is dissolved in 100 g of water. Assume the cryoscopic constant $K_f$ for water is 1.86 K·kg/mol.

Step 1: Calculate the molality of the solution. Moles of urea = $\frac{10}{60} = 0.1667 \, \text{mol}$ .

Mass of solvent (water) = 100 g = 0.1 kg.

Molality $m = \frac{0.1667}{0.1} = 1.667 \, \text{mol/kg}$ .

Step 2: Calculate the freezing point depression.

Thus, the freezing point of the solution will be lowered by 3.1 K.

13. What is the relation between the vapor pressure of a solution and the mole fraction of the solvent? Explain with the help of Raoult’s Law.

Answer: Raoult’s Law states that the partial vapor pressure of each volatile component in a solution is directly proportional to its mole fraction in the solution. Mathematically, for a solution containing volatile solute and solvent, the vapor pressure of the solvent is given by:

where:

• $P_A$ is the vapor pressure of the solvent in the solution,
• $X_A$ is the mole fraction of the solvent in the solution,
• $P_A^0$ is the vapor pressure of the pure solvent.

This relationship shows that when the mole fraction of the solvent $X_A$ increases, the vapor pressure of the solvent also increases. In other words, the vapor pressure of the solution is lower than that of the pure solvent, depending on the amount of solute present in the solution. As more solute is added, the mole fraction of the solvent decreases, resulting in a further reduction in the vapor pressure.

For example, when NaCl is dissolved in water, the vapor pressure of water decreases due to the presence of NaCl, and this can be calculated using Raoult's Law.

14. What are the colligative properties of solutions? Discuss the effects of the presence of a solute on these properties.

Answer: Colligative properties are properties of solutions that depend on the number of solute particles, rather than the identity or chemical nature of the solute. These properties are affected by the concentration of solute particles in a solution. The four primary colligative properties are:

1. Lowering of Vapor Pressure: When a non-volatile solute is added to a solvent, the vapor pressure of the solvent decreases. This happens because the solute particles occupy the surface area of the solvent, thereby reducing the number of solvent molecules able to escape into the vapor phase.

2. Boiling Point Elevation: The presence of a non-volatile solute causes the boiling point of the solution to be higher than that of the pure solvent. This occurs because the vapor pressure of the solution is lower, requiring more heat to raise the temperature to the boiling point.

   $$\Delta T_b = K_b \cdot m$$

   where $\Delta T_b$ is the elevation in boiling point, $K_b$ is the ebullioscopic constant, and $m$ is the molality of the solution.

3. Freezing Point Depression: The addition of a solute to a solvent lowers the freezing point of the solution. This is because the solute particles interfere with the formation of the crystalline solid structure of the solvent.

   $$\Delta T_f = K_f \cdot m$$

   where $\Delta T_f$ is the depression in freezing point, $K_f$ is the cryoscopic constant, and $m$ is the molality of the solution.

4. Osmotic Pressure: Osmotic pressure is the pressure exerted by a solution when it is separated from pure solvent by a semipermeable membrane. The osmotic pressure depends on the solute concentration in the solution and is given by the formula:

   $$\Pi = i \cdot M \cdot R \cdot T$$

   where $\Pi$ is the osmotic pressure, $i$ is the Van’t Hoff factor, $M$ is the molarity, $R$ is the gas constant, and $T$ is the temperature in Kelvin.

All of these properties depend only on the number of solute particles in the solution, not the identity of the solute, which is why they are classified as colligative properties.

15. Explain the concept of ideal and non-ideal solutions with examples.

Answer: An ideal solution is one in which the enthalpy of mixing is zero, and the solution obeys Raoult’s Law throughout the entire concentration range. In ideal solutions, the intermolecular forces between the solute and solvent molecules are similar to the forces present in the pure components. This means that there is no heat change upon mixing, and the solution behaves in a perfectly predictable manner.

Example of an ideal solution: A mixture of benzene and toluene is an ideal solution. Both are non-polar solvents, and the intermolecular forces between benzene molecules and toluene molecules are similar to those in pure benzene or pure toluene, making the solution obey Raoult’s Law.

A non-ideal solution, on the other hand, is one that does not obey Raoult’s Law and exhibits deviations in the vapor pressure behavior. Non-ideal solutions occur when the intermolecular interactions between solute and solvent differ significantly from the interactions in the pure components. These deviations can either be positive or negative:

• Positive deviation occurs when the vapor pressure of the solution is higher than predicted, typically when the solute-solvent interactions are weaker than the solvent-solvent or solute-solute interactions. An example is the mixture of acetone and chloroform.
• Negative deviation occurs when the vapor pressure of the solution is lower than predicted, typically when the solute-solvent interactions are stronger than the solvent-solvent or solute-solute interactions. An example is the solution of water and hydrochloric acid.

16. How does the addition of a non-volatile solute affect the vapor pressure of a solvent?

Answer: The addition of a non-volatile solute to a solvent results in a decrease in the vapor pressure of the solvent. This is because non-volatile solutes do not evaporate into the vapor phase, so they occupy the surface area of the solvent, thus reducing the number of solvent molecules that can escape into the gas phase.

According to Raoult’s Law, the vapor pressure of the solvent in the solution is proportional to the mole fraction of the solvent:

where $P_A$ is the vapor pressure of the solvent in the solution, $X_A$ is the mole fraction of the solvent, and $P_A^0$ is the vapor pressure of the pure solvent.

The presence of the non-volatile solute decreases the mole fraction of the solvent, which in turn lowers the vapor pressure of the solvent. This lowering of vapor pressure is a colligative property.

17. What is the effect of adding an electrolyte to a solution in terms of colligative properties?

Answer: When an electrolyte is added to a solution, it dissociates into ions, which increases the number of particles in the solution. Since colligative properties depend on the number of solute particles, the presence of an electrolyte enhances the effects of colligative properties compared to a non-electrolyte solution. Specifically:

1. Freezing Point Depression: The freezing point will be lowered more than it would be for a non-electrolyte solution because the electrolyte dissociates into multiple ions. For example, if NaCl dissociates into Na⁺ and Cl⁻, the number of particles in solution is effectively doubled.

2. Boiling Point Elevation: Similarly, the boiling point will be raised more for an electrolyte solution. The greater the number of particles (ions), the greater the elevation in boiling point.

3. Osmotic Pressure: The osmotic pressure will also increase more because the number of solute particles is higher.

For example, if 1 mole of NaCl is dissolved in water, it dissociates into 2 moles of ions, thereby doubling the effect on the colligative properties compared to a non-electrolyte solution.

18. Explain the concept of the Van’t Hoff factor (i) and its importance in colligative properties.

Answer: The Van’t Hoff factor (denoted as $i$ ) is a measure of the number of particles into which a solute dissociates in solution. For non-electrolytes, which do not dissociate in solution, $i = 1$ . For electrolytes, $i$ is greater than 1 because the solute dissociates into multiple ions.

The importance of the Van’t Hoff factor in colligative properties is that it allows the effects of the solute to be taken into account when calculating properties like boiling point elevation, freezing point depression, and osmotic pressure.

For example:

• In a solution of NaCl in water, $i = 2$ because NaCl dissociates into two ions: Na⁺ and Cl⁻.
• In a solution of Na₂SO₄ in water, $i = 3$ because it dissociates into three ions: Na⁺, Na⁺, and SO₄²⁻.

The Van’t Hoff factor is used in equations for colligative properties:

where $\Delta T_f$ is the depression in freezing point, $K_f$ is the cryoscopic constant, and $m$ is the molality of the solution.

19. Calculate the molar mass of a solute from the depression in freezing point.

Answer: To calculate the molar mass of a solute using the depression in freezing point, we use the formula:

where $\Delta T_f$ is the depression in freezing point, $K_f$ is the cryoscopic constant, and $m$ is the molality of the solution. The molality $m$ is defined as:

Step 1: Calculate the molality $m$ .

For example, if the depression in freezing point is 2.0 K, the cryoscopic constant $K_f$ is 1.86 K·kg/mol, and the solvent mass is 100 g (0.1 kg), the molality is:

Step 2: Calculate the moles of solute.

For 0.1 kg of solvent, the moles of solute are:

Step 3: Calculate the molar mass. If the mass of the solute is 5.4 g, the molar mass $M$ is:

Thus, the molar mass of the solute is 50 g/mol.

20. Define osmotic pressure. Derive the relation for osmotic pressure.

Answer: Osmotic pressure is the pressure required to stop the flow of solvent molecules through a semipermeable membrane that separates a solution from pure solvent. It is a colligative property, meaning it depends on the number of solute particles in the solution, not on the type of solute.

The formula for osmotic pressure is derived from the ideal gas law:

where:

• $\Pi$ is the osmotic pressure,
• $i$ is the Van’t Hoff factor (number of particles the solute dissociates into),
• $M$ is the molarity of the solution (moles of solute per liter of solution),
• $R$ is the gas constant (0.0821 L·atm/mol·K),
• $T$ is the temperature in Kelvin.

This equation shows that osmotic pressure increases with increasing solute concentration (molarity), temperature, and the number of dissociated particles (represented by $i$ ).

21. What is the significance of the cryoscopic constant and ebullioscopic constant in colligative properties?

Answer: The cryoscopic constant (denoted as $K_f$ ) and the ebullioscopic constant (denoted as $K_b$ ) are two constants that play a crucial role in the calculation of freezing point depression and boiling point elevation, respectively, for solutions.

• Cryoscopic Constant (K_f): This constant is specific to a given solvent and determines the freezing point depression for a solution. It has units of K·kg/mol, and the depression in freezing point ($\Delta T_f$ ) is directly proportional to the molality of the solute and the cryoscopic constant:

  $$\Delta T_f = K_f \cdot m$$

  where $m$ is the molality of the solution. A higher $K_f$ means that the solvent will experience a greater depression in freezing point for a given solute concentration.

• Ebullioscopic Constant (K_b): This constant is also specific to the solvent and determines the extent to which the boiling point of a solution is elevated. The boiling point elevation ($\Delta T_b$ ) is given by:

  $$\Delta T_b = K_b \cdot m$$

  where $m$ is the molality of the solution. A larger $K_b$ value means a greater increase in boiling point for the same concentration of solute.

These constants are important in understanding and predicting how a solute affects the boiling and freezing points of solvents, which is critical in various industrial and scientific applications.

22. Explain the concept of reverse osmosis.

Answer: Reverse osmosis is the process by which solvent molecules move from an area of lower solute concentration to an area of higher solute concentration, against the natural osmotic pressure, by applying an external pressure greater than the osmotic pressure. This process is the reverse of natural osmosis, where solvent naturally moves from a region of lower solute concentration to a region of higher solute concentration.

In reverse osmosis, a semipermeable membrane is used to separate a solvent from a solution. When external pressure is applied to the solution side, solvent molecules are forced through the membrane, leaving the solute behind. The purified solvent on the other side is collected as "pure water," while the concentrated solution (with higher solute concentration) is discarded.

Reverse osmosis is widely used for water purification, desalination (removing salts from seawater), and in various industrial applications where purification of liquids is required.

23. Discuss the significance of the Van’t Hoff factor in the calculation of colligative properties.

Answer: The Van’t Hoff factor (i) is a measure of the number of particles into which a solute dissociates in solution. It is an important factor in the calculation of colligative properties, as these properties depend on the number of solute particles present in the solution.

For non-electrolytes (substances that do not dissociate in solution), the Van’t Hoff factor $i = 1$ . However, for electrolytes (which dissociate into ions), $i$ will be greater than 1. For example, for sodium chloride (NaCl), which dissociates into two ions, $i = 2$ .

The Van’t Hoff factor affects the magnitude of colligative properties, such as:

• Freezing Point Depression: The depression in freezing point is proportional to $i$ , the number of particles in the solution.
• Boiling Point Elevation: The elevation in boiling point is similarly affected by $i$ .
• Osmotic Pressure: Osmotic pressure is directly proportional to the Van’t Hoff factor.

The general formula for the change in colligative properties considering the Van’t Hoff factor is:

where:

• $\Delta T$ is the change in the property (e.g., freezing point depression or boiling point elevation),
• $i$ is the Van’t Hoff factor,
• $K$ is the constant specific to the property,
• $m$ is the molality of the solution.

Thus, the Van’t Hoff factor allows us to account for dissociation or association of solute molecules in solution, making it essential for accurately determining colligative properties.

24. What is the relationship between molality and molarity? Derive the relationship.

Answer: The relationship between molality (m) and molarity (M) is given by:

Where:

• $m$ is the molality (mol/kg of solvent),
• $M$ is the molarity (mol/L of solution),
• $\text{density of solution}$ is the mass of the solution per unit volume (g/L),
• $\text{mol. mass of solute}$ is the molar mass of the solute,
• $\text{density of solvent}$ is the density of the solvent.

Derivation:

1. Molality is defined as the number of moles of solute per kilogram of solvent:

   $$m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$$

2. Molarity is defined as the number of moles of solute per liter of solution:

   $$M = \frac{\text{moles of solute}}{\text{volume of solution (L)}}$$

To express molality in terms of molarity, we need to consider the total volume of solution and its density. By using the density of the solution, we can relate mass and volume, and therefore molality and molarity.

The relationship shows that molality is a function of molarity, solvent density, and the solute’s molar mass.

25. What is the role of solvent-solute interactions in the formation of solutions? Explain with examples.

Answer: The formation of a solution depends on the interactions between solute and solvent molecules. These interactions determine whether a solute will dissolve in a solvent and how much solute can dissolve in a given amount of solvent. There are three main types of interactions that play a crucial role in solution formation:

1. Solvent-Solvent Interactions: These are the forces between solvent molecules. For a solution to form, the solute molecules must be able to break the solvent-solvent interactions. For example, water molecules are held together by hydrogen bonds. For a solute like NaCl to dissolve in water, the solvent-solvent hydrogen bonds must be overcome.

2. Solute-Solute Interactions: Solute molecules are also held together by intermolecular forces. When a solute dissolves, these interactions must be broken apart. For instance, when salt (NaCl) dissolves in water, the ionic bonds between the Na⁺ and Cl⁻ ions must be broken.

3. Solvent-Solute Interactions: The most crucial interactions in solution formation are the interactions between the solute and solvent molecules. For a solute to dissolve, the solvent must interact with the solute in such a way that it stabilizes the solute particles in the solution. For example:

   • Ionic compounds in polar solvents: When NaCl is dissolved in water, the Na⁺ ions are surrounded by water molecules with partial negative charges, and the Cl⁻ ions are surrounded by water molecules with partial positive charges. This process is known as hydration.
   • Non-polar solutes in non-polar solvents: Oil dissolves in hexane because both are non-polar, and the molecules interact through London dispersion forces.

The "like dissolves like" principle is a general rule: polar solvents generally dissolve polar solutes, and non-polar solvents dissolve non-polar solutes.

The strength of these solvent-solute interactions determines the solubility and other properties of the solution. If the solvent-solute interactions are strong, the solute will dissolve readily, and the solution will be stable.