Solutions

The chapter “Solutions” from 12th standard chemistry part 1 NCERT Board covers the following key concepts:

Definition and Types of Solutions:
- Solutions can be classified based on the physical state of solute and solvent (gaseous, liquid, or solid solutions).
- Binary solutions contain two components: solute (present in smaller quantity) and solvent (present in larger quantity).
Expressing Concentration of Solutions:
- Various methods to express the amount of solute in a solution:
  - Mass percentage (% w/w)
  - Volume percentage (% v/v)
  - Mass by volume percentage (% w/v)
  - Parts per million (ppm)
  - Mole fraction
  - Molarity (M)
  - Molality (m)
  - Normality (N)
Solubility:
- Factors affecting solubility:
  - Nature of solute and solvent (like dissolves like)
  - Temperature
  - Pressure (for gases in liquids)
Vapour Pressure of Solutions:
- Vapour pressure: Pressure exerted by the vapor of a liquid in equilibrium with its liquid phase.
- Raoult’s Law: For a solution of volatile liquids, the partial vapor pressure of each component is directly proportional to its mole fraction in the solution.
- Ideal and Non-ideal Solutions:
  - Ideal solutions obey Raoult’s law.
  - Non-ideal solutions show deviations from Raoult’s law (positive or negative).
Colligative Properties:
- Properties of solutions that depend only on the number of solute particles, not their nature.
- Four main colligative properties:
  - Relative lowering of vapor pressure
  - Elevation of boiling point
  - Depression of freezing point
  - Osmotic pressure¹
Abnormal Molar Mass:
- Sometimes, the molar mass of a solute determined using colligative properties is different from the expected value.
- This is due to association or dissociation of solute particles in the solution.
- Van’t Hoff factor (i) is used to account for this.

Exercise

1. Define the terra solution. How many types of solutions are formed? Write briefly about each type with an example.

Ans :

A terra solution isn’t a standard scientific term like “solution” in chemistry. It seems like a phrase that might be used in a specific context, perhaps related to environmental science or a particular project.

Gas in Gas:

Solute: Gas
Solvent: Gas
Example: Air (oxygen in nitrogen)

Gas in Liquid:

Solute: Gas
Solvent: Liquid
Example: Carbonated drinks (carbon dioxide in water)

Gas in Solid:

Solute: Gas
Solvent: Solid
Example: Hydrogen adsorbed in palladium

Liquid in Gas:

Solute: Liquid
Solvent: Gas
Example: Water vapor in air

Liquid in Liquid:

Solute: Liquid
Solvent: Liquid
Example: Alcohol in water

Liquid in Solid:

Solute: Liquid
Solvent: Solid
Example: Mercury in amalgamated zinc

Solid in Gas:

Solute: Solid
Solvent: Gas
Example: Iodine vapor in air

Solid in Liquid:

Solute: Solid
Solvent: Liquid
Example: Sugar in water

Solid in Solid:

Solute: Solid
Solvent: Solid
Example: Alloys like brass (zinc in copper

2. Give an example of a solid solution in which the solute is a gas?

Ans :

Hydrogen in Palladium: Palladium metal can absorb significant amounts of hydrogen gas. The hydrogen atoms fit into the spaces between the palladium atoms, forming a solid solution. In this case, hydrogen (the gas) is the solute, and palladium (the solid) is the solvent.

3. Define the following terms:

(i) Mole fraction

(ii) Molality

(iii) Molarity

(iv) Mass percentage

Ans :

(i) Mole Fraction (x)

The mole fraction of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all components in the solution.

(ii) Molality (m)

Molality is defined as the number of moles of solute dissolved in 1 kilogram (1000 grams) of solvent.

(iii) Molarity (M)

Molarity is defined as the number of moles of solute dissolved in 1 liter of solution.

(iv) Mass Percentage (% w/w)

Mass percentage is the mass of the solute expressed as a percentage of the total mass of the solution.

4. Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of acid if the density of the solution is 1·504 g mL-1 ?

Ans :

1. Understand the given information:

The solution is 68% nitric acid by mass. This means that 100 g of the solution contains 68 g of HNO₃.
The density of the solution is 1.504 g/mL.

2. Calculate the mass of the solution:

We’ll assume we have 100 g of the solution for easier calculation since the percentage is based on mass.

3. Calculate the volume of the solution:

Density = mass / volume
Volume = mass / density
Volume = 100 g / 1.504 g/mL
Volume ≈ 66.5 mL

4. Convert the volume to liters:

66.5 mL * (1 L / 1000 mL) ≈ 0.0665 L

5. Calculate the moles of HNO₃:

Molar mass of HNO₃ = 1 (H) + 14 (N) + 3 * 16 (O) = 63 g/mol
Moles of HNO₃ ≈ 1.08 moles

6. Calculate the molarity:

Molarity = moles of HNO₃ / volume of solution (in L)
Molarity = 1.08 moles / 0.0665 L
Molarity ≈ 16.2 M

5. A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1 .2 g m L-1, then what shall be the molarity of the solution?

Ans :

1. Understand the given information:

10% w/w glucose solution: This means 100 g of solution contains 10 g of glucose and 90 g of water.
Density of the solution: 1.2 g/mL

2. Calculate the moles of glucose and water:

Molar mass of glucose (C₆H₁₂O₆)
= 180 g/mol
Moles of glucose = (10 g) / (180 g/mol) ≈ 0.056 moles
Molar mass of water (H₂O)
= 18 g/mol
Moles of water = (90 g) / (18 g/mol)
= 5 moles

3. Calculate the molality:

Molality (m)
= moles of glucose / mass of water (in kg)
Molality (m) = 0.056 moles / (90 g * 1 kg / 1000 g)
Molality (m) ≈ 0.617 mol/kg

4. Calculate the mole fractions:

Total moles = moles of glucose + moles of water = 0.056 + 5 = 5.056 moles
Mole fraction of glucose (x_glucose)
= moles of glucose / total moles
x_glucose = 0.056 / 5.056 ≈ 0.011
Mole fraction of water (x_water)
= moles of water / total moles
x_water = 5 / 5.056 ≈ 0.989 (or simply 1 – x_glucose)

5. Calculate the molarity:

Mass of the solution = 100 g (as we initially considered)
Volume of the solution = mass / density = 100 g / 1.2 g/mL ≈ 83.33 mL
Convert volume to liters: 83.33 mL * (1 L / 1000 mL) ≈ 0.08333 L
Molarity (M) = moles of glucose / volume of solution (in L)
Molarity (M) = 0.056 moles / 0.08333 L
Molarity (M) ≈ 0.672 M

Summary of results:

Molality: ≈ 0.617 mol/kg
Mole fraction of glucose: ≈ 0.011
Mole fraction of water: ≈ 0.989
Molarity: ≈ 0.672 M

6. How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2C03 and NaHCO3 containing equimolar amounts of both?

Ans :

1. Calculate the moles of each compound in the mixture:

Let ‘x’ be the moles of Na₂CO₃ and also the moles of NaHCO₃ (since they are equimolar).
Molar mass of Na₂CO₃
= (2 * 23) + 12 + (3 * 16)
= 106 g/mol
Mass of Na₂CO₃ = x * 106 g/mol
Molar mass of NaHCO₃ = 23 + 1 + 12 + (3 * 16) = 84 g/mol
Mass of NaHCO₃ = x * 84 g/mol
Total mass of the mixture
= mass of Na₂CO₃ + mass of NaHCO₃
1 g = 106x + 84x
1 = 190x
x = 1/190 moles (This is the number of moles of both Na₂CO₃ and NaHCO₃)

2. Write the balanced chemical equations:

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
NaHCO₃ + HCl → NaCl + H₂O + CO₂

3. Calculate the total moles of HCl required:

From the balanced equations, 1 mole of Na₂CO₃ reacts with 2 moles of HCl, and 1 mole of NaHCO₃ reacts with 1 mole of HCl.
Moles of HCl required for Na₂CO₃ = 2 * (1/190) = 2/190 moles
Moles of HCl required for NaHCO₃ = 1 * (1/190) = 1/190 moles
Total moles of HCl required = (2/190) + (1/190) = 3/190 moles

4. Calculate the volume of 0.1 M HCl required:

Molarity = moles of solute / volume of solution (in L)¹
Volume (in L) = moles of solute / Molarity
Volume (in L) = (3/190 moles) / (0.1 mol/L)
Volume (in L) ≈ 0.1579 L

5. Convert the volume to milliliters:

Volume (in mL) = Volume (in L) * 1000 mL/L
Volume (in mL) ≈ 0.1579 L * 1000 mL/L
Volume (in mL) ≈ 157.9 mL

Therefore, approximately 157.9 mL of 0.1 M HCl are required to react completely with the 1 g mixture.

7. A solution is obtained by mixing 300g of 25% solution and 400g of 40% solutions by mass. Calculate the mass percentage of the resulting solution.

Ans :

Calculate the mass of solute in the first solution:
- Mass of first solution = 300 g
- Percentage of solute in first solution = 25%
- Mass of solute in first solution = (25/100) * 300 g = 75 g
Calculate the mass of solute in the second solution:
- Mass of second solution = 400 g
- Percentage of solute in second solution = 40%
- Mass of solute in second solution = (40/100) * 400 g = 160 g
Calculate the total mass of the resulting solution:
- Total mass = Mass of first solution + Mass of second solution
- Total mass = 300 g + 400 g
- = 700 g
Calculate the total mass of solute in the resulting solution:
- Total mass of solute = Mass of solute in first solution + Mass of solute in second solution
- Total mass of solute = 75 g + 160 g
- = 235 g
Calculate the mass percentage of the resulting solution:
- Mass percentage = (Total mass of solute / Total mass of solution) * 100
- Mass percentage = (235 g / 700 g) * 100
- Mass percentage ≈ 33.57%

Therefore, the mass percentage of the resulting solution is approximately 33.57%.

8. An antifreeze solution is prepared from 222.6 g of ethylene glycol, (C2 H6O2 ) and200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL-1, then what shall be the molarity of the solution?

Ans :

9. A sample of drinking water was found to be severely contaminated with chloroform (CHCl3), supposed to be a carcinogen. The level of contamination was 15 ppm (by mass).

(i) express this in percent by mass.

(ii) determine the molality of chloroform in the water sample.

Ans :

(i) Expressing contamination in percent by mass:

ppm (parts per million) means that for every 1 million parts of the solution, there are 15 parts of chloroform.
To convert ppm to percent by mass, we can use the following proportion:

15 parts chloroform / 1,000,000 parts solution = x parts chloroform / 100 parts solution
Solving for x: x = (15 * 100) / 1,000,000 x = 0.0015%
Therefore, the contamination is 0.0015% by mass.

(ii) Determining the molality of chloroform:

We’ll assume we have 100 g of the contaminated water sample. This makes the calculation easier due to the percentage we just calculated.
Mass of chloroform = 0.0015% of 100 g = 0.0015 g
Mass of water (solvent) = 100 g – 0.0015 g ≈ 100 g (Since the mass of chloroform is so small compared to the total mass, we can safely approximate). We will use 100g = 0.1kg for molality calculations.
Molar mass of chloroform (CHCl₃) = 12 + 1 + (3 * 35.5) = 119.5 g/mol
Moles of chloroform = mass of chloroform / molar mass of chloroform
Moles of chloroform = 0.0015 g / 119.5 g/mol ≈ 1.255 x 10⁻⁵ moles
Molality (m) = moles of chloroform / mass of water (in kg)
Molality (m) = 1.255 x 10⁻⁵ moles / 0.1 kg
Molality (m) ≈ 1.255 x 10⁻⁴ mol/kg or 0.0001255 m

10. What role does the molecular interaction play in solution of alcohol in water?

Ans : The molecular interactions, particularly hydrogen bonding, are the key to understanding the behavior of alcohol in water. The fact that the hydrogen bonding between water and alcohol is weaker than in the pure substances leads to a positive deviation from Raoult’s law, a non-ideal solution, a lower boiling point, and complete miscibility.

11. Why do gases always tend to be less soluble in liquids as the temperature is raised?

Ans : The decreased solubility of gases in liquids with increasing temperature is due to a combination of factors: the exothermic nature of the dissolution process, Le Chatelier’s principle, and the increased kinetic energy of gas molecules at higher temperatures.

12. State Henry’s law and mention some of its important applications.

Ans :

Henry’s Law:

At a fixed temperature, the quantity of a specific gas that dissolves in a certain type and volume of liquid is directly proportional to the partial pressure of that gas when in equilibrium with the liquid.

In simpler terms:

Higher pressure = More gas dissolves
Lower pressure = Less gas dissolves

Important Applications of Henry’s Law:

Carbonated Beverages:
- Carbon dioxide gas is dissolved in soft drinks under high pressure. When the bottle is opened, the pressure decreases, and the gas comes out of the solution, causing the fizz.
Respiration:
- In the lungs, oxygen from the inhaled air dissolves in the blood due to its partial pressure in the alveoli. Similarly, carbon dioxide from the blood dissolves into the air in the alveoli to be exhaled.
Scuba Diving:
- Divers breathe air at high pressure underwater. This causes more nitrogen to dissolve in their blood. During ascent, if the pressure decreases too quickly, the dissolved nitrogen can form bubbles, leading to decompression sickness (“the bends”).
Manufacturing of Fertilizers:
- Henry’s Law is used in the production of ammonia, a key component of fertilizers. The process involves dissolving gases like nitrogen and hydrogen in liquids under pressure.
Industrial Processes:
- Many industrial processes involve dissolving gases in liquids, and Henry’s Law is used to determine the optimal conditions for these processes.

13. The partial pressure of ethane over a solution containing 6.56 × 10-3 g of ethane is 1 bar. If the solution contains 5.00 × 10-2 g of ethane, then what shall be the partial pressure of the gas?

Ans :

14. According to Raoult’s law, what is meant by positive and negative deviaitions and how is the sign of ∆solH related to positive and negative deviations from Raoult’s law?

Ans :

Raoult’s Law states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent in the solution. In simpler terms, it predicts how the vapor pressure of a liquid (solvent) is lowered when another substance (solute) is dissolved in it.

Deviations from Raoult’s Law

Most solutions are not ideal and show deviations from Raoult’s Law. These deviations can be:

Positive Deviation:
- The vapor pressure of the solution is higher than what Raoult’s Law predicts.
- This indicates that the interactions between solute and solvent particles are weaker than the interactions in the pure substances.
- Solute and solvent molecules escape the solution more easily, leading to higher vapor pressure.
- ΔsolH is positive (heat is absorbed during mixing).
- Example: A mixture of ethanol and water.
Negative Deviation:
- This indicates that the interactions between solute and solvent particles are stronger than the interactions in the pure substances.
- Solute and solvent molecules are held more tightly in the solution, leading to lower vapor pressure.
- ΔsolH is negative (heat is released during mixing).
- Example: A mixture of acetone and chloroform.

Relationship between ΔsolH and Deviations

The sign of ΔsolH (enthalpy of solution) directly relates to the type of deviation from Raoult’s Law:

ΔsolH > 0 (positive): Positive deviation from Raoult’s Law.
ΔsolH < 0 (negative): Negative deviation from Raoult’s Law.

15. An aqueous solution of 2 percent non-volatile solute exerts a pressure of 1·004 bar at the boiling point of the solvent. What is the molecular mass of the solute ?

Ans :

According to law :

16. Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35.0 g of octane?

Ans :

1. Calculate the moles of each component:

Molar mass of heptane (C₇H₁₆) = (7 * 12) + (16 * 1) = 100 g/mol
Moles of heptane = mass / molar mass = 26.0 g / 100 g/mol = 0.26 moles
Molar mass of octane (C₈H₁₈) = (8 * 12) + (18 * 1) = 114 g/mol
Moles of octane = mass / molar mass = 35.0 g / 114 g/mol ≈ 0.307 moles

2. Calculate the mole fraction of each component:

Total moles = moles of heptane + moles of octane = 0.26 + 0.307 ≈ 0.567 moles
Mole fraction of heptane (x_heptane) = moles of heptane / total moles = 0.26 / 0.567 ≈ 0.459
Mole fraction of octane (x_octane) = moles of octane / total moles = 0.307 / 0.567 ≈ 0.541 (or 1 – x_heptane)

3. Apply Raoult’s Law:

P_total = (x_heptane * P°_heptane) + (x_octane * P°_octane)

Where:

P_total is the total vapor pressure of the mixture.
x_heptane and x_octane are the mole fractions of heptane and octane, respectively.
P°_heptane and P°_octane are the vapor pressures of pure heptane and octane, respectively.

4. Plug in the values and calculate:

P_total = (0.459 * 105.2 kPa) + (0.541 * 46.8 kPa) P_total = 48.29 + 25.33 P_total ≈ 73.62 kPa

Therefore, the vapor pressure of the mixture is approximately 73.62 kPa.

17. The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Ans :

Calculate Moles of Water

Molar mass of water (H₂O) = 18.015 g/mol
Moles of water = mass of water / molar mass of water
Moles of water = 1000 g / 18.015 g/mol ≈ 55.51 moles

Calculate Mole Fraction of Solute

Mole fraction of solute (x_solute) = moles of solute / (moles of solute + moles of water)
x_solute = 1 mole / (1 mole + 55.51 moles)
x_solute ≈ 0.0177

Apply Raoult’s Law

P_solution = P°_water * (1 – x_solute)

Where:

P_solution is the vapor pressure of the solution
P°_water is the vapor pressure of pure water

Calculate Vapor Pressure of Solution

P_solution = 12.3 kPa * (1 – 0.0177)
P_solution ≈ 12.08 kPa

Therefore, the vapor pressure of the 1 molal solution is approximately 12.08 kPa.

18. Calculate the mass of a non-volatile solute (molecular mass 40 g mol-1) that should be dissolved in 114 g of octane to reduce its pressure to 80%.

Ans :

P_solution = P°_octane * x_octane

Where:

P_solution is the vapor pressure of the solution (80% of P°_octane)
P°_octane is the vapor pressure of pure octane
x_octane is the mole fraction of octane in the solution

2. Express the given information:

P_solution = 0.80 * P°_octane

3. Combine and simplify:

Substitute the expression for P_solution into Raoult’s Law:

0.80 * P°_octane = P°_octane * x_octane

Divide both sides by P°_octane:

0.80 = x_octane

4. Calculate the mole fraction of the solute:

Since x_octane + x_solute = 1, we can find the mole fraction of the solute:

x_solute = 1 – x_octane x_solute = 1 – 0.80 x_solute = 0.20

5. Calculate moles of octane:

Molar mass of octane (C₈H₁₈) = (8 * 12) + (18 * 1) = 114 g/mol
Moles of octane = mass of octane / molar mass of octane
Moles of octane = 114 g / 114 g/mol = 1 mole

6. Calculate moles of solute:

x_solute = moles of solute / (moles of solute + moles of octane)

Let ‘n’ be the moles of solute:

0.20 = n / (n + 1)

Solving for n:

0.20n + 0.20 = n 0.20 = 0.80n n = 0.20 / 0.80 n = 0.25 moles

7. Calculate mass of solute:

Mass of solute = moles of solute * molar mass of solute
Mass of solute = 0.25 moles * 40 g/mol
Mass of solute = 10 g

Therefore, 10 g of the non-volatile solute should be dissolved in 114 g of octane to reduce its vapor pressure to 80%

19. A solution containing 30g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18g of water is then added to the solution and the new of vapour pressure becomes 2.9 kPa at 298 K. Calculate

(i) molar mass of the solute.

(ii) vapour pressure of water at 298 K.

Ans :

(i) Molar mass of the solute:

Set up Raoult’s Law equations for both scenarios:

Scenario 1: P₁ = P° * (n_water1 / (n_water1 + n_solute))
Scenario 2: P₂ = P° * (n_water2 / (n_water2 + n_solute))

Where:

P₁ = 2.8 kPa (vapor pressure in scenario 1)
P₂ = 2.9 kPa (vapor pressure in scenario 2)
P° = vapor pressure of pure water (unknown)
n_water1 = moles of water in scenario 1 (90g / 18 g/mol = 5 moles)
n_water2 = moles of water in scenario 2 (90g + 18g)/18g/mol=6 moles)
n_solute = moles of solute (unknown)

Divide the first equation by the second:

P₁/P₂ = (n_water1*(n_water2 + n_solute)) / (n_water2*(n_water1+ n_solute))

Substituting the values

2.8/2.9 = (5*(6 + n_solute)) / (6*(5 + n_solute))

Solving we get n_solute = 0.5 moles

Calculate the molar mass of the solute:

Molar mass = mass of solute / moles of solute Molar mass = 30 g / 0.5 moles = 60 g/mol

(ii) Vapor pressure of water at 298 K:

Substitute back into either Raoult’s Law equation (using scenario 1):

2.8 kPa = P° * (5 / (5 + 0.5))

Solve for P°:

P° = (2.8 kPa * 5.5) / 5 P° = 3.08 kPa (approximately 3.1 kPa)

Therefore:

(i) The molar mass of the solute is 60 g/mol.
(ii) The vapor pressure of water at 298 K is approximately 3.1 kPa.

20. A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Ans :

Calculate the molality of the cane sugar solution:

5% cane sugar solution means 5g of cane sugar in 95g of water.
Molar mass of cane sugar (C₁₂H₂₂O₁₁) = (1212)+(221)+(11*16) = 342 g/mol
Moles of cane sugar = 5g / 342 g/mol ≈ 0.0146 moles
Mass of water in kg = 95g / 1000 = 0.095 kg
Molality of cane sugar solution = moles of cane sugar / mass of water (in kg) = 0.0146 moles / 0.095 kg ≈ 0.154 mol/kg

Calculate the freezing point depression for cane sugar:

ΔTf (cane sugar) = Freezing point of pure water – Freezing point of cane sugar solution
ΔTf (cane sugar) = 273.15 K – 271 K = 2.15 K

Relate ΔTf to molality using the freezing point depression constant (Kf):

ΔTf = Kf * m
Since the solvent is water in both cases, the Kf will be the same. We can use this to relate the two solutions.

Calculate the molality of the glucose solution:

5% glucose solution means 5g of glucose in 95g of water.
Molar mass of glucose (C₆H₁₂O₆) = (612)+(121)+(6*16) = 180 g/mol
Moles of glucose = 5g / 180 g/mol ≈ 0.0278 moles
Molality of glucose solution = moles of glucose / mass of water (in kg) = 0.0278 moles / 0.095 kg ≈ 0.293 mol/kg

Calculate the freezing point depression for glucose:

We can set up a proportion since Kf is the same for both:

ΔTf(cane sugar) / m(cane sugar) = ΔTf(glucose) / m(glucose)

2.15K / 0.154 m = ΔTf(glucose) / 0.293 m

ΔTf(glucose) = (2.15K * 0.293 m) / 0.154 m ≈ 4.11 K

Calculate the freezing point of the glucose solution:

Freezing point of glucose solution = Freezing point of pure water – ΔTf(glucose)
Freezing point of glucose solution = 273.15 K – 4.11 K ≈ 269.04 K

Therefore, the freezing point of the 5% glucose solution is approximately 269.04 K.

21. Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol-1. Calculate atomic masses of A and B.

Ans :

1. Use the formula for freezing point depression:

ΔTf = Kf * m

Where:

ΔTf is the depression in freezing point
Kf is the molar depression constant

2. Calculate the molality of each solution:

For AB₂: 2.3 K = 5.1 K kg/mol * m_AB2 m_AB2 = 2.3 K / 5.1 K kg/mol ≈ 0.451 mol/kg
For AB₄: 1.3 K = 5.1 K kg/mol * m_AB4 m_AB4 = 1.3 K / 5.1 K kg/mol ≈ 0.255 mol/kg

3. Calculate the molar mass of each compound:

Molality (m) = (mass of solute / molar mass of solute) / mass of solvent (in kg)

For AB₂: 0.451 mol/kg = (1 g / M_AB2) / (20 g / 1000 g/kg) (Note: we convert grams of benzene to kg) M_AB2 = (1g * 1000g/kg) / (0.451 mol/kg * 20g) ≈ 111 g/mol
For AB₄: 0. M_AB4 = (1g* 1000g/kg) / (0.255 mol/kg * 20g) ≈ 196 g/mol

4. Set up equations to solve for the atomic masses:

Let ‘a’ be the atomic mass of A and ‘b’ be the atomic mass of B.

Equation 1: a + 2b = 111 (Molar mass of AB₂)
Equation 2: a + 4b = 196 (Molar mass of AB₄)

5. Solve the system of equations:

Subtract Equation 1 from Equation 2:

2b = 85 b = 42.5

Substitute the value of b back into Equation 1:

a + 2(42.5) = 111 a + 85 = 111 a = 26

Therefore, the atomic mass of A is approximately 26, and the atomic mass of B is approximately 42.5.

22. At 300 K, 36 g glucose present per litre in its solution has osmotic pressure of 4·98 bar. If the osmotic pressure of the solution is 1·52 bar at the same temperature, what would be its concentration?

Ans :

23. Suggest the most important type of intermolecular attractive interaction in the following pairs:

(i) n-hexane and n-octane

(ii) I2 and CCl4.

(iii) NaCl04 and water

(iv) methanol and acetone

(v) acetonitrile (CH3CN) and acetone (C3H60)

Ans :

(i) n-hexane and n-octane:

London Dispersion Forces (LDFs) or Van der Waals forces: Both n-hexane and n-octane are nonpolar hydrocarbons. The only significant intermolecular forces present are weak London Dispersion Forces, which arise from temporary fluctuations in electron distribution.

(ii) I₂ and CCl₄:

London Dispersion Forces (LDFs) or Van der Waals forces:
Iodine (I₂) is a nonpolar molecule, and carbon tetrachloride (CCl₄) is also nonpolar due to its symmetrical structure. The dominant intermolecular forces are London Dispersion Forces.

(iii) NaClO₄ and water:

Ion-Dipole Interactions: Sodium perchlorate (NaClO₄) is an ionic compound. When dissolved in water, it dissociates into Na⁺ cations and ClO₄⁻ anions. These ions interact strongly with the polar water molecules through ion-dipole interactions. Hydrogen bonding also plays a role between water molecules.

(iv) Methanol and acetone:

Hydrogen Bonding and Dipole-Dipole Interactions: Methanol (CH₃OH) has an -OH group, which allows it to form hydrogen bonds. Acetone (C₃H₆O) has a polar carbonyl (C=O) group, making it a polar molecule. The primary interactions are dipole-dipole between acetone molecules and hydrogen bonding between methanol molecules and between methanol and acetone.

(v) Acetonitrile (CH₃CN) and acetone (C₃H₆O):

Dipole-Dipole Interactions: Both acetonitrile and acetone are polar molecules. Acetonitrile has a polar C≡N group, and acetone has a polar C=O group. The most important intermolecular forces are dipole-dipole interactions between these polar groups. Although acetone can act as a hydrogen bond acceptor, it cannot act as a hydrogen bond donor, so hydrogen bonding between the two molecules is not a significant factor.

24. Based on solute solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.

Ans :

n-octane: A nonpolar hydrocarbon solvent. “Like dissolves like” is the key principle.

KCl: Ionic compound. Strong ion-ion forces within KCl are far stronger than any interactions with nonpolar n-octane. Insoluble.

CH₃OH: Polar molecule (methanol). Has some slight solubility due to the small nonpolar part of the molecule, but the polar -OH group makes it prefer polar solvents. Low solubility in n-octane.

CH₃CN: Polar molecule (acetonitrile). More polar than methanol. Even less interaction with nonpolar n-octane. Very low solubility in n-octane.

Cyclohexane: Nonpolar hydrocarbon, like n-octane. Only weak London Dispersion Forces. Completely miscible with n-octane (most soluble).

25. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

(i) phenol

(ii) toluene

(iii) formic acid

(iv) ethylene glycol

(v) chloroform

(vi) pentanol

Ans :

(i) Phenol (C₆H₅OH): Partially soluble. Phenol has a polar -OH group that can form hydrogen bonds with water, increasing its solubility. However, the large nonpolar aromatic ring makes it significantly less polar overall, limiting its solubility.

(ii) Toluene (C₆H₅CH₃): Insoluble. Toluene is a nonpolar hydrocarbon. It cannot form hydrogen bonds with water, and the weak London Dispersion Forces aren’t strong enough to overcome the strong hydrogen bonds between water molecules.

(iii) Formic Acid (HCOOH): Highly soluble. Formic acid is a small, polar molecule that can form hydrogen bonds with water through its -COOH group. It also ionizes slightly in water, further enhancing its solubility.

(iv) Ethylene Glycol (HOCH₂CH₂OH): Highly soluble. Ethylene glycol has two -OH groups, allowing it to form extensive hydrogen bonds with water. This makes it highly miscible in water.

(v) Chloroform (CHCl₃): Insoluble. Chloroform is a relatively nonpolar molecule. While it has some slight polarity, it cannot form strong hydrogen bonds with water, making it essentially insoluble.

(vi) Pentanol (C₅H₁₁OH): Partially soluble. Pentanol has a polar -OH group, which contributes to some solubility in water. However, the larger nonpolar hydrocarbon chain (C₅H₁₁) decreases its overall polarity, limiting its solubility. Compared to smaller alcohols, pentanol is less soluble in water.

26. If the density of lake water is 1·25 g mL-1, and it contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake.

Ans :

Calculate the moles of Na⁺ ions:

The problem states there are 92 g of Na⁺ ions per kg of water.
The molar mass of Na⁺ is approximately 23 g/mol.
Moles of Na⁺ = mass of Na⁺ / molar mass of Na⁺
Moles of Na⁺ = 92 g / 23 g/mol = 4 moles

Calculate the molality:

Molality (m) = moles of Na⁺ / mass of water (in kg)
We are given that 92 g of Na+ ions are present per 1 kg of water
Molality (m) = 4 moles / 1 kg = 4 m

27. If the solubility product of CuS is 6 x 10-16, calculate the maximum molarity of CuS in aqueous solution.

Ans :

1. Write the equilibrium equation for the dissolution of CuS:

CuS(s) ⇌ Cu²⁺(aq) + S²⁻(aq)

2. Define the solubility product (Ksp):

Ksp = [Cu²⁺][S²⁻]

where [Cu²⁺] and [S²⁻] are the concentrations of Cu²⁺ and S²⁻ ions in the saturated solution.

3. Relate the ion concentrations to the molar solubility (s):

Since CuS dissolves in a 1:1 ratio, the concentration of Cu²⁺ and S²⁻ ions will be equal to the molar solubility (s) of CuS.

[Cu²⁺] = s [S²⁻] = s

4. Substitute into the Ksp expression:

Ksp = (s)(s) = s²

5. Solve for the molar solubility (s):

s = √Ksp = √(6 x 10⁻¹⁶) ≈ 2.45 x 10⁻⁸ M

Therefore, the maximum molarity of CuS in aqueous solution is approximately 2.45 x 10⁻⁸ M.

28. Calculate the mass percentage of aspirin (C9H8O4 in acetonitrile (CH3CN) when 6.5g of CHO is dissolved in 450 g of CH3CN.

Ans :

1. Calculate the total mass of the solution:

Mass of aspirin (solute) = 6.5 g
Mass of acetonitrile (solvent) = 450 g
Total mass of the solution = mass of aspirin + mass of acetonitrile
Total mass = 6.5 g + 450 g = 456.5 g

2. Calculate the mass percentage of aspirin:

Mass percentage of aspirin = (mass of aspirin / total mass of solution) * 100%
Mass percentage = (6.5 g / 456.5 g) * 100%
Mass percentage ≈ 1.42%

Therefore, the mass percentage of aspirin in the acetonitrile solution is approximately 1.42%.

29. Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 x 10-3 m aqueous solution required for the above dose.

Ans :

Convert the dose to grams:

1.5 mg = 1.5 x 10⁻³ g

Understand molality:

A 1.5 x 10⁻³ m solution means there are 1.5 x 10⁻³ moles of nalorphene per 1 kg (1000 g) of water.

Calculate the molar mass of nalorphene (C₁₉H₂₁NO₃):

Molar mass = (19 * 12) + (21 * 1) + 14 + (3 * 16) = 228 + 21 + 14 + 48 = 311 g/mol

Calculate the moles of nalorphene in the given dose:

Moles = mass / molar mass = (1.5 x 10⁻³ g) / 311 g/mol ≈ 4.82 x 10⁻⁶ moles

Calculate the mass of water in the solution containing the required dose:

Using the molality definition:

Molality = moles of nalorphene / mass of water (in kg)

Mass of water (in kg) = moles of nalorphene / molality

Mass of water (in kg) = (4.82 x 10⁻⁶ moles) / (1.5 x 10⁻³ mol/kg) ≈ 0.00321 kg

= 3.21 g

Calculate the total mass of the solution:

Total mass of solution = mass of nalorphene + mass of water

Total mass ≈ 1.5 x 10⁻³ g + 3.21 g ≈ 3.21 g (since the mass of nalorphene is negligible compared to the mass of water)

Therefore, approximately 3.21 g of the 1.5 x 10⁻³ m aqueous nalorphene solution is required for the 1.5 mg dose.

30. Calculate the amount of benzoic acid (C5H5COOH) required for preparing 250 mL of 0· 15 M solution in methanol.

Ans :

Calculate the moles of benzoic acid required:

Molarity (M) = moles of solute / volume of solution (in Liters)

Volume of solution
= 250 mL = 0.250 L
Molarity = 0.15 M

Moles of benzoic acid = Molarity * Volume of solution Moles = 0.15 mol/L * 0.250 L = 0.0375 moles

Calculate the molar mass of benzoic acid (C₆H₅COOH):

Molar mass = (7 * 12) + (6 * 1) + (2 * 16) = 84 + 6 + 32 = 122 g/mol

Calculate the mass of benzoic acid required:

Mass = moles * molar mass Mass

= 0.0375 moles * 122 g/mol

= 4.575 g

Therefore, 4.575 g of benzoic acid is required to prepare 250 mL of a 0.15 M solution in methanol.

31. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.

Ans :

Acetic Acid (CH₃COOH): Acetic acid is a weak acid. It only partially dissociates in water, meaning it forms relatively few ions in solution. Therefore, it causes a smaller depression in freezing point compared to the other two.

Trichloroacetic Acid (Cl₃CCOOH): The presence of three chlorine atoms (which are electron-withdrawing groups) makes trichloroacetic acid a stronger acid than acetic acid. It dissociates to a greater extent, producing more ions in solution. This leads to a greater depression in freezing point compared to acetic acid.

Trifluoroacetic Acid (F₃CCOOH): The three fluorine atoms (also electron-withdrawing groups, and even more electronegative than chlorine) make trifluoroacetic acid an even stronger acid than trichloroacetic acid. It dissociates almost completely in water (it’s considered a strong acid), producing the most ions in solution of the three. This results in the largest depression in freezing point.

32. Calculate the depression in the freezing point of water when 10g of CH3CH2CHClCOOH is added to 250g of water. Ka = 1.4 x 1o-3 Kg = 1.86 K kg mol-1.

Ans :

1. Calculate the moles of CH₃CH₂CHClCOOH (2-chlorobutanoic acid):

Molar mass of CH₃CH₂CHClCOOH = (4 * 12) + (8 * 1) + 35.5 + (2 * 16) = 48 + 8 + 35.5 + 32 = 123.5 g/mol
Moles of CH₃CH₂CHClCOOH = mass / molar mass = 10 g / 123.5 g/mol ≈ 0.081 moles

2. Calculate the molality (m) of the solution:

Molality (m) = moles of solute / mass of solvent (in kg)
Mass of water = 250 g
= 0.250 kg
m = 0.081 moles / 0.250 kg ≈ 0.324 mol/kg

3. Determine the van’t Hoff factor (i):

Since 2-chlorobutanoic acid is a weak acid, it will partially dissociate in water. We need to use the given Ka value to determine the degree of dissociation (α) and then calculate the van’t Hoff factor (i).

CH₃CH₂CHClCOOH ⇌ CH₃CH₂CHClCOO⁻ + H⁺

Ka = [CH₃CH₂CHClCOO⁻][H⁺] / [CH₃CH₂CHClCOOH]

Let ‘c’ be the initial concentration of the acid and α be the degree of dissociation. Then at equilibrium:

[CH₃CH₂CHClCOO⁻] = cα [H⁺] = cα [CH₃CH₂CHClCOOH] = c(1-α)

Ka = (cα)(cα) / c(1-α) = cα² / (1-α)

Since Ka is small, we can assume 1-α ≈ 1. Then Ka ≈ cα².

α = sqrt(Ka / c)

= sqrt(1.4 x 10⁻³ / 0.324)

≈ 0.066

The van’t Hoff factor (i) is related to α by i

= 1 + α (since the acid dissociates into 2 ions).

i = 1 + 0.066 ≈ 1.066

4. Calculate the depression in freezing point (ΔTf):

ΔTf = i * Kf * m

Where Kf is the freezing point depression constant for water (1.86 K kg/mol).

ΔTf = 1.066 * 1.86 K kg/mol * 0.324 mol/kg ≈ 0.647 K

Therefore, the depression in the freezing point of water is approximately 0.65 K.

33. 19.5g of CH2FCOOH is dissolved in 500g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’s Hoff factor and dissociation constant of fluoroacetic acid.

Ans :

1. Calculate the moles of CH₂FCOOH (Fluoroacetic Acid):

Molar mass of CH₂FCOOH = (2 * 12) + (3 * 1) + 19 + (2 * 16) = 24 + 3 + 19 + 32 = 78 g/mol
Moles of CH₂FCOOH = mass / molar mass = 19.5 g / 78 g/mol = 0.25 moles

2. Calculate the molality (m) of the solution:

Molality (m) = moles of solute / mass of solvent (in kg)
Mass of water = 500 g
= 0.5 kg
m = 0.25 moles / 0.5 kg = 0.5 mol/kg

3. Calculate the expected freezing point depression (ΔTf) assuming no dissociation:

ΔTf = Kf * m where Kf is the freezing point depression constant for water (1.86 K kg/mol).
ΔTf = 1.86 K kg/mol * 0.5 mol/kg = 0.93 K

4. Calculate the van’t Hoff factor (i):

The observed freezing point depression is given as 1.0°C. The van’t Hoff factor relates the observed depression to the expected depression.

i = Observed ΔTf / Expected ΔTf
i = 1.0°C / 0.93 K ≈ 1.075

5. Calculate the degree of dissociation (α):

For a weak acid that dissociates into two ions (HA ⇌ H⁺ + A⁻), the van’t Hoff factor is related to the degree of dissociation (α) as follows:

i = 1 + α

α = i – 1
α = 1.075 – 1 = 0.075

6. Calculate the dissociation constant (Ka):

We can set up an ICE table and use the expression for Ka. Initial: c 0 0 Change: -cα +cα +cα Equilibrium: c(1-α) cα cα Ka = [H⁺][A⁻] / [HA] = (cα)(cα) / c(1-α) = cα² / (1-α)

Where ‘c’ is the initial concentration or molality and in this case c is 0.5 m, α is the degree of dissociation.

Since α is small, we can make the approximation that 1 – α ≈ 1. Then we get:

Ka = cα²

Ka = (0.5) * (0.075)² ≈ 0.0028 or 2.8 x 10⁻³

Therefore:

Van’t Hoff factor (i) ≈ 1.075
Dissociation constant (Ka) ≈ 2.8 x 10⁻³

34. Vapour pressure of water at 293 K is 17·535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Ans :

35. Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Ans :

P = k_H * m

Where:

P is the partial pressure of the gas above the solution
k_H is Henry’s law constant (for molality)
m is the molality of the gas in the solution

2. Plug in the given values:

P = 760 mm Hg
k_H = 4.27 x 10⁵ mm Hg

760 mm Hg = (4.27 x 10⁵ mm Hg) * m

3. Solve for molality (m):

m = 760 mm Hg / (4.27 x 10⁵ mm Hg) m ≈ 0.00178 mol/kg

Therefore, the solubility of methane in benzene at 298 K under 760 mm Hg is approximately 0.00178 mol/kg. This means that approximately 0.00178 moles of methane will dissolve in 1 kg of benzene under the given conditions.

36. 100g of liquid A (molar mass 140 g mol-1) was dissolved in 1000g of liquid B (molar mass 180g mol-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

Ans :

1. Calculate the moles of each liquid:

Moles of liquid A = mass of A / molar mass of A = 100 g / 140 g/mol ≈ 0.714 moles
Moles of liquid B = mass of B / molar mass of B = 1000 g / 180 g/mol ≈ 5.556 moles

2. Calculate the mole fraction of each liquid:

Total moles = moles of A + moles of B = 0.714 + 5.556 ≈ 6.27 moles
Mole fraction of A (x_A) = moles of A / total moles = 0.714 / 6.27 ≈ 0.114
Mole fraction of B (x_B) = moles of B / total moles = 5.556 / 6.27 ≈ 0.886 (or 1 – x_A)

3. Apply Raoult’s Law:

Raoult’s Law states that the partial pressure of each component in an ideal solution is equal to the mole fraction of that component multiplied by its vapor pressure in the pure state.

P_total = P_A + P_B
P_total = (x_A * P°_A) + (x_B * P°_B)

Where:

P_total is the total vapor pressure of the solution
P_A and P_B are the partial pressures of A and B, respectively
P°_A and P°_B are the vapor pressures of pure A and pure B, respectively

4. Plug in the known values and solve for P°_A:

We know:

P_total = 475 torr
P°_B = 500 torr
x_A = 0.114
x_B = 0.886

475 torr = (0.114 * P°_A) + (0.886 * 500 torr) 475 torr = 0.114 * P°_A + 443 torr 32 torr = 0.114 * P°_A P°_A ≈ 280.7 torr

37. Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot Ptotal, Pchlroform and Pacetone as a function of χacetone. The experimental data observed for different compositions of mixtures is:

NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions 6

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

Ans :

38. Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80g of benzene is mixed with 100g of toluene.

Ans :

1. Calculate the moles of benzene and toluene:

Molar mass of benzene (C₆H₆) = (6 * 12) + (6 * 1) = 78 g/mol
Moles of benzene = mass of benzene / molar mass of benzene = 80 g / 78 g/mol ≈ 1.026 moles
Molar mass of toluene (C₇H₈) = (7 * 12) + (8 * 1) = 92 g/mol
Moles of toluene = mass of toluene / molar mass of toluene = 100 g / 92 g/mol ≈ 1.087 moles

2. Calculate the mole fraction of benzene :

Total moles = moles of benzene + moles of toluene = 1.026 + 1.087 ≈ 2.113 moles
x_benzene = moles of benzene / total moles = 1.026 / 2.113 ≈ 0.486

3. Apply Raoult’s Law to calculate the partial pressure of benzene (P_benzene):

Raoult’s Law: P_benzene = x_benzene * P°_benzene

Where P°_benzene is the vapor pressure of pure benzene.

P_benzene = 0.486 * 50.71 mm Hg ≈ 24.65 mm Hg

4. Similarly, apply Raoult’s Law to find the partial pressure of toluene (P_toluene):

Mole fraction of toluene (x_toluene) = 1 – x_benzene = 1 – 0.486 ≈ 0.514
P_toluene = x_toluene * P°_toluene = 0.514 * 32.06 mm Hg ≈ 16.48 mm Hg

5. Calculate the total vapor pressure (P_total):

P_total = P_benzene + P_toluene

= 24.65 mm Hg + 16.48 mm Hg

≈ 41.13 mm Hg

6. Calculate the mole fraction of benzene in the vapor phase (y_benzene):

y_benzene = P_benzene / P_total y_benzene

= 24.65 mm Hg / 41.13 mm Hg

≈ 0.600

Therefore, the mole fraction of benzene in the vapor phase is approximately 0.600.

39. The air is a mixture of a number of gases. The major components are oxygen and nitrogen with an approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if Henry’s law constants for oxygen and nitrogen are 3.30 x 107 mm and 6.51 x 107 mm respectively, calculate the composition of these gases in water.

Ans :

1. Calculate the partial pressures of oxygen and nitrogen:

Total pressure = 10 atm
Partial pressure of oxygen = 20% of 10 atm = 0.20 * 10 atm = 2 atm
Partial pressure of nitrogen = 79% of 10 atm = 0.79 * 10 atm = 7.9 atm

2. Convert partial pressures to mm Hg:

1 atm = 760 mm Hg
Partial pressure of oxygen = 2 atm * 760 mm Hg/atm = 1520 mm Hg
Partial pressure of nitrogen = 7.9 atm * 760 mm Hg/atm = 6004 mm Hg

3. Apply Henry’s Law to calculate the molality of each gas:

Henry’s Law: P = k_H * m

Where:

P is the partial pressure of the gas
k_H is Henry’s law constant
m is the molality of the gas
Molality of oxygen = Partial pressure of oxygen / Henry’s law constant for oxygen
Molality of oxygen = 1520 mm Hg / 3.30 x 10⁷ mm Hg ≈ 4.61 x 10⁻⁵ mol/kg
Molality of nitrogen = Partial pressure of nitrogen / Henry’s law constant for nitrogen
Molality of nitrogen = 6004 mm Hg / 6.51 x 10⁷ mm Hg ≈ 9.22 x 10⁻⁵ mol/kg

4. Calculate the total molality of dissolved gases:

Total molality = Molality of oxygen + Molality of nitrogen
Total molality ≈ 4.61 x 10⁻⁵ mol/kg + 9.22 x 10⁻⁵ mol/kg ≈ 1.383 x 10⁻⁴ mol/kg

5. Calculate the mole fraction of each gas in the water:

Mole fraction of oxygen = Molality of oxygen / Total molality
Mole fraction of oxygen ≈ (4.61 x 10⁻⁵ mol/kg) / (1.383 x 10⁻⁴ mol/kg) ≈ 0.333
Mole fraction of nitrogen = Molality of nitrogen / Total molality
Mole fraction of nitrogen ≈ (9.22 x 10⁻⁵ mol/kg) / (1.383 x 10⁻⁴ mol/kg) ≈ 0.667

Therefore, the composition of the gases in water is approximately 33.3% oxygen and 66.7% nitrogen by mole fraction.

40. Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.

Ans :

1. Convert the temperature to Kelvin:

T(K) = T(°C) + 273.15 T(K) = 27 + 273.15 = 300.15 K

2. Use the osmotic pressure formula:

π = iCRT

Where:

π is the osmotic pressure
i is the van’t Hoff factor
T is the temperature in Kelvin

3. Rearrange the formula to solve for C:

C = π / (iRT)

4. Plug in the known values:

C = 0.75 atm / (2.47 * 0.0821 L atm / (mol K) * 300.15 K) C ≈ 0.0123 mol/L

5. Calculate the moles of CaCl₂:

Moles = Molarity * Volume Moles

= 0.0123 mol/L * 2.5 L ≈ 0.0308 moles

6. Calculate the molar mass of CaCl₂:

Molar mass = 40.08 (Ca) + 2 * 35.45 (Cl)

= 110.98 g/mol

7. Calculate the mass of CaCl₂:

Mass = Moles * Molar mass Mass

= 0.0308 moles * 110.98 g/mol

≈ 3.42 g

Therefore, approximately 3.42 grams of CaCl₂ should be dissolved in 2.5 liters of water to achieve an osmotic pressure of 0.75 atm at 27°C.

41. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25°C, assuming that it is completely dissociated.

Ans :

Step I. Calculation of Van’t Hoff factor (i)