Question:

Benzene and toluene form an 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 80 g of benzene is mixed with 100 g of toluene.

Answer:

Step 1: Calculate the moles of benzene (B) and toluene (T) B = 80 g / 78.11 g/mol = 1.02 mol T = 100 g / 92.14 g/mol = 1.08 mol

Step 2: Calculate the total moles of the solution Total moles = 1.02 + 1.08 = 2.10 mol

Step 3: Calculate the mole fraction of benzene Mole fraction of benzene = 1.02 mol / 2.10 mol = 0.48

Step 4: Calculate the vapour pressure of the solution Vapour pressure = (50.71 mm Hg x 0.48) + (32.06 mm Hg x 0.52) = 41.84 mm Hg

Question:

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL^-1?

Answer:

Step 1: Calculate the mass of nitric acid in the solution. Mass of nitric acid = 68% of 1.504 g/mL = 1.024 g/mL

Step 2: Calculate the moles of nitric acid in the solution. Moles of nitric acid = mass of nitric acid/molar mass of nitric acid = 1.024 g/mL/63.01 g/mol = 0.016 mol/mL

Step 3: Calculate the molarity of nitric acid in the solution. Molarity of nitric acid = moles of nitric acid/volume of solution = 0.016 mol/mL/1 mL = 0.016 M

Question:

Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.

Answer:

(i) Mole fraction: Mole fraction is the ratio of the number of moles of one component of a mixture to the total number of moles of all components in the mixture.

(ii) Molality: Molality is the number of moles of solute per kilogram of solvent.

(iii) Molarity: Molarity is the number of moles of solute per liter of solution.

(iv) Mass percentage: Mass percentage is the mass of solute per 100 grams of solution.

Question:

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.

Answer:

1. Acetic acid, trichloroacetic acid and trifluoroacetic acid are all organic acids.

2. When dissolved in water, each acid will lower the freezing point of the water.

3. The amount of depression in the freezing point of the water is related to the strength of the acid.

4. Acetic acid is the weakest of the three acids, and therefore produces the smallest depression in the freezing point of the water.

5. Trichloroacetic acid is stronger than acetic acid and produces a greater depression in the freezing point of the water.

6. Trifluoroacetic acid is the strongest of the three acids and produces the greatest depression in the freezing point of the water.

7. Therefore, the depression in the freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above.

Question:

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​,pchloroform​, and pacetone​ as a function of xacetone​. The experimental data observed for different compositions of mixture is: 100xxacetone​ 0 11.8 23.4 36.0 50.8 58.2 64.5 72.1 Pacetone​/mm Hg 0 54.9 110.1 202.4 322.7 405.9 454.1 521.1 Pchloroform​/mm Hg 632.8 548.1 469.4 359.7 257.7 193.6 161.2 120.7 Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

Answer:

Step 1: Calculate the total vapour pressure for each composition. Ptotal = Pacetone + Pchloroform

Step 2: Plot ptotal, pchloroform, and pacetone as a function of xacetone.

Step 3: Plot the experimental data on the same graph paper.

Step 4: Determine whether the data has a positive or negative deviation from the ideal solution. To do this, compare the total vapour pressure calculated in Step 1 to the total vapour pressure observed in the experimental data. If the observed total vapour pressure is greater than the calculated total vapour pressure, the deviation is positive. If the observed total vapour pressure is less than the calculated total vapour pressure, the deviation is negative.

Question:

A solution containing 30 gm of non-volatile solute exactly in 90 gm of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 gm of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298 K.

Answer:

(i) To calculate the molar mass of the solute, use the following equation:

M = (30 g/90 g) x (2.8 kPa/2.9 kPa) x 18 g

M = 16.22 g/mol

(ii) To calculate the vapour pressure of water at 298 K, use the following equation:

P = (30 g/108 g) x (2.8 kPa/2.9 kPa) x 2.9 kPa

P = 2.67 kPa

Question:

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

Answer:

1. Freezing point of pure water = 273.15 K

2. Freezing point of 5% cane sugar in water = 271 K

3. Difference in freezing point between 5% cane sugar and pure water = 271 K - 273.15 K = -2.15 K

4. Freezing point of 5% glucose in water = 273.15 K - 2.15 K = 271 K

Question:

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2​SO4​ in 2 litre of water at 25°C, assuming that it is completely dissociated.

Answer:

1. Calculate the molarity of the K2SO4 solution: Molarity = (25 mg / (2 L * 158.12 g/mol)) = 0.78 mM

2. Calculate the osmotic pressure of the solution: Osmotic Pressure = (2 * 0.78 mM * 0.0821 L atm/mol K * 298 K) = 4.3 atm

Question:

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.

Answer:

1. Calculate the molarity of the solution: Molarity = Osmotic Pressure (atm) x Gas Constant (R) x Temperature (K) / (i x 1000) Molarity = 0.75 x 0.082 x 300 / (2.47 x 1000) Molarity = 0.0075 mol/L

2. Calculate the amount of CaCl₂ needed to produce the required molarity: Amount of CaCl₂ = Molarity (mol/L) x Volume (L) Amount of CaCl₂ = 0.0075 mol/L x 2.5 L Amount of CaCl₂ = 0.01875 mol

3. Convert the amount of CaCl₂ from moles to grams: Grams of CaCl₂ = Amount of CaCl₂ (mol) x Molar Mass of CaCl₂ (g/mol) Grams of CaCl₂ = 0.01875 mol x 111.99 g/mol Grams of CaCl₂ = 2.09 g

Question:

19.5 g of CH₂​FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

Answer:

1. Calculate the molality of the solution: Molality = (mass of solute/mass of solvent)*1000 Molality = (19.5 g/500 g)*1000 = 39 mol/kg

2. Calculate the depression in freezing point of water: ΔTf = (iKfm) Where, i = van’t Hoff factor, Kf = Freezing point depression constant, m = molality

3. Rearrange the equation to find van’t Hoff factor: i = (ΔTf/Kfm) i = (1.00°C/(1.86°C/mmol)*39 mol/kg) i = 2.1

4. Calculate the dissociation constant of fluoroacetic acid: K = (iKfm)/ΔTf K = (2.11.86°C/mmol*39 mol/kg)/1.00°C K = 79.7 (mol/kg)

Question:

Calculate the amount of benzoic acid (C₆​H₅​COOH) required for preparing 250 mL of 0.15 M solution in methanol.

Answer:

Step 1: Calculate the moles of benzoic acid (C₆H₅COOH) required.

Molarity (M) = moles of solute/volume of solution (L)

Moles of benzoic acid = molarity x volume of solution

Moles of benzoic acid = 0.15 M x 0.250 L

Moles of benzoic acid = 0.0375 moles

Step 2: Calculate the mass of benzoic acid (C₆H₅COOH) required.

Molar mass of benzoic acid = 122.12 g/mol

Mass of benzoic acid = moles x molar mass

Mass of benzoic acid = 0.0375 moles x 122.12 g/mol

Mass of benzoic acid = 4.57 g

Question:

If the solubility product of CuS is 6×10¹⁶, calculate the maximum molarity of CuS in aqueous solution.

Answer:

Step 1: Identify the solubility product (Ksp) of CuS, which is 6×10¹⁶.

Step 2: Calculate the molar solubility of CuS by using the equation Ksp = [Cu₂⁺][S₂^-].

Step 3: Calculate the maximum molarity of CuS by dividing the molar solubility by the number of moles of CuS in the solution (1 mole of CuS).

Question:

What role does the molecular interaction play in a solution of alcohol and water?

Answer:

1. The molecular interaction between alcohol and water is an example of a type of chemical reaction known as an ‘intermolecular force.’

2. This type of force is responsible for the attraction between molecules of different substances, such as alcohol and water.

3. In a solution of alcohol and water, the molecular interaction creates a mixture of molecules in which the molecules of alcohol and water are distributed evenly throughout the solution.

4. This allows the molecules of alcohol and water to interact with each other, resulting in a solution that is a homogeneous mixture.

5. The molecular interaction also affects the physical properties of the solution, such as its boiling point and viscosity.

6. Additionally, the molecular interaction between alcohol and water can affect the solubility of other substances in the solution, as well as the rate of chemical reactions that take place in the solution.

Question:

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂​CO₃​ and NaHCO₃​ containing equimolar amounts of both?

Answer:

Step 1: Calculate the moles of Na₂CO₃ and NaHCO₃ in 1g of the mixture.

Na₂CO₃: 1 g/105.99 g/mol = 0.00945 mol NaHCO₃: 1 g/84.01 g/mol = 0.01190 mol

Step 2: Calculate the moles of HCl required to react completely with the mixture.

2HCl + Na₂CO₃ –> 2NaCl + CO₂ + H₂O 1 mol HCl = 1 mol Na₂CO₃

Therefore, 0.00945 mol HCl is required to react completely with Na₂CO₃.

2HCl + NaHCO₃ –> 2NaCl + H₂O + CO₂ 1 mol HCl = 1 mol NaHCO₃

Therefore, 0.01190 mol HCl is required to react completely with NaHCO₃.

Step 3: Calculate the mL of 0.1 M HCl required to react completely with the mixture.

0.00945 mol HCl x 1000 mL/1 L x 0.1 mol/L = 9.45 mL 0.01190 mol HCl x 1000 mL/1 L x 0.1 mol/L = 11.90 mL

Therefore, 9.45 mL + 11.90 mL = 21.35 mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both.

Question:

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

Answer:

Step 1: Calculate the total mass of the mixture.

Total mass = 300 g + 400 g = 700 g

Step 2: Calculate the total amount of solute in the mixture.

Total amount of solute in the mixture = (300 g x 0.25) + (400 g x 0.40) = 270 g + 160 g = 430 g

Step 3: Calculate the mass percentage of the resulting solution.

Mass percentage of the resulting solution = (430 g / 700 g) x 100 = 61.43%

Question:

Suggest the most important type of intermolecular attractive interaction in the following pairs. (i) n-hexane and n-octane (ii) I2​ and CCl4​ (iii) NaClO4​ and water (iv) methanol and acetone (v) acetonitrile (CH₃​CN) and acetone (C3​H6​O)

Answer:

(i) London dispersion forces (ii) Dipole-dipole interactions (iii) Ion-dipole interactions (iv) Hydrogen bonding (v) Dipole-dipole interactions

Question:

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

Answer:

Step 1: Vapour pressure of water at 300 K is 12.3 kPa.

Step 2: Vapour pressure of a non-volatile solute in a 1 molal solution is calculated using the Raoult’s Law.

Step 3: According to Raoult’s Law, the vapour pressure of the solution is equal to the sum of the vapour pressure of the solvent and the vapour pressure of the solute.

Step 4: Therefore, the vapour pressure of the solution is equal to 12.3 kPa + (1 molal x vapour pressure of the solute).

Question:

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

Answer:

(i) Phenol: Highly soluble (ii) Toluene: Partially soluble (iii) Formic acid: Highly soluble (iv) Ethylene glycol: Highly soluble (v) Chloroform: Insoluble (vi) Pentanol: Partially soluble

Question:

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

Answer:

1. A solid solution in which the solute is a gas is a solid solution that contains a gas dissolved in a solid.

2. An example of this is solid carbon dioxide, commonly known as dry ice. Dry ice is made up of solid carbon dioxide molecules dissolved in a solid state.

Question:

100 g of liquid A (molar mass 140 g mol^-1) was dissolved in 1000 g of liquid B (molar mass 180 g 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.

Answer:

1. Calculate the moles of liquid A: 100 g of liquid A / 140 g mol^-1 = 0.714 moles of liquid A

2. Calculate the moles of liquid B: 1000 g of liquid B / 180 g mol^-1 = 5.56 moles of liquid B

3. Calculate the vapour pressure of pure liquid A: Vapour pressure of pure liquid A = 500 torr

4. Calculate the total vapour pressure of the solution: Total vapour pressure of the solution = 475 torr

5. Calculate the vapour pressure of liquid A in the solution: Vapour pressure of liquid A in the solution = (0.714 moles of liquid A / (0.714 moles of liquid A + 5.56 moles of liquid B)) x 475 torr = 45.5 torr

Question:

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.

Answer:

(i) 15 ppm (by mass) = 15 x 10^-6 = 0.000015

Percent by mass = 0.000015 x 100 = 0.0015%

(ii) Molality of chloroform = (mass of solute (CHCl3) / mass of solvent (water)) x 1000

Molality of chloroform = (0.000015 / 1) x 1000 = 0.015 m

Question:

If the density of some lake water is 1.25g mL^-1 and contains 92 g of Na^+ ions per kg of water, calculate the molality of Na^+ ions in the lake.

Answer:

Step 1: Calculate the mass of 1 kg of water.

1 kg = 1000 g

Step 2: Calculate the moles of Na+ ions in 1 kg of water.

Moles of Na+ ions = 92 g Na+/23 g Na+ = 4 moles

Step 3: Calculate the molality of Na+ ions in the lake.

Molality = 4 moles/1000 g = 0.004 moles/kg

Question:

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 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?

Answer:

Step 1: Convert the masses of ethylene glycol and water to moles.

Moles of ethylene glycol = 222.6 g / (62.07 g/mol) = 3.60 mol

Moles of water = 200 g / (18.02 g/mol) = 11.11 mol

Step 2: Calculate the molality of the solution.

Molality = (3.60 mol ethylene glycol) / (11.11 mol water + 3.60 mol ethylene glycol) = 0.24 m

Step 3: Calculate the molarity of the solution.

Molarity = (3.60 mol ethylene glycol + 11.11 mol water) / (1.072 g/mL x 1000 mL/L) = 3.31 M

Question:

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.

Answer:

1. Calculate the mole fraction of glucose (X) in the solution:

X = (25 g of glucose / (25 g of glucose + 450 g of water)) x 100 = (25/475) x 100 = 5.26%

2. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water using Raoult’s Law:

Vapour Pressure of Water = (X x P°) + ((1-X) x P°) = (5.26% x 17.535 mm Hg) + (94.74% x 17.535 mm Hg) = 17.25 mm Hg

Question:

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

Answer:

Step 1: Calculate the molar mass of octane, which is 114 g/mol.

Step 2: Find the vapour pressure of octane at 100%.

Step 3: Calculate the vapour pressure of octane at 80%. This can be done by multiplying the vapour pressure of octane at 100% by 0.8.

Step 4: Calculate the amount of solute needed to reduce the vapour pressure of octane to 80%. This can be done by dividing the vapour pressure of octane at 80% by the molar mass of the solute (40 g/mol).

Step 5: Calculate the mass of the solute needed. This can be done by multiplying the amount of solute needed by its molar mass (40 g/mol).

Question:

Calculate the mass percentage of aspirin (C9​H8​O4​) in acetonitrile (CH₃​CN) when 6.5 g of C9​H8​O4​ is dissolved in 450 g of CH₃​CN.

Answer:

Step 1: Calculate the molar mass of aspirin (C9​H8​O4​). Molar mass of aspirin = (9 x 12.01) + (8 x 1.008) + (4 x 16.00) = 180.09 g/mol

Step 2: Calculate the moles of aspirin. Moles of aspirin = 6.5 g / 180.09 g/mol = 0.036 mol

Step 3: Calculate the moles of acetonitrile (CH₃CN). Moles of acetonitrile = 450 g / 41.05 g/mol = 10.98 mol

Step 4: Calculate the total moles of the solution. Total moles = 0.036 mol + 10.98 mol = 11.02 mol

Step 5: Calculate the mass percentage of aspirin. Mass percentage of aspirin = (0.036 mol / 11.02 mol) x 100% = 0.3258%

Question:

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

Answer:

1. Calculate the moles of ethane in the solution containing 6.56×10^-3g of ethane:

Moles of ethane = (6.56×10^-3g) / (30.07g/mol) = 0.00216 moles

2. Calculate the partial pressure of ethane over the solution containing 6.56×10^-3g of ethane:

Partial pressure of ethane = 1 bar

3. Calculate the moles of ethane in the solution containing 5.00×10^-2g of ethane:

Moles of ethane = (5.00×10^-2g) / (30.07g/mol) = 0.01662 moles

4. Calculate the partial pressure of ethane over the solution containing 5.00×10^-2g of ethane:

Partial pressure of ethane = (1 bar) x (0.01662 moles / 0.00216 moles) = 7.6 bar

Question:

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

Answer:

1. Use Henry’s Law Constant: K = 4.27×105 mm Hg

2. Solve for the solubility of methane in benzene at 298 K: Solubility = K × Pressure Solubility = (4.27×105 mm Hg) × (760 mm Hg) Solubility = 3.25×107 mm Hg

Question:

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 g of octane?

Answer:

1. Calculate the mole fraction of heptane in the mixture: Moles of heptane = 26.0 g ÷ (7.0 g/mol) = 3.71 moles Moles of octane = 35 g ÷ (8.0 g/mol) = 4.38 moles Mole fraction of heptane = 3.71 moles ÷ (3.71 moles + 4.38 moles) = 0.457

2. Calculate the mole fraction of octane in the mixture: Mole fraction of octane = 4.38 moles ÷ (3.71 moles + 4.38 moles) = 0.543

3. Calculate the vapour pressure of the mixture: Vapour pressure of mixture = (Mole fraction of heptane x Vapour pressure of heptane) + (Mole fraction of octane x Vapour pressure of octane) Vapour pressure of mixture = (0.457 x 105.2 kPa) + (0.543 x 46.8 kPa) = 73.2 kPa

Question:

Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH₃​OH, CH₃​CN

Answer:

1. CH₃CN - The solute-solvent interactions between the carbon and nitrogen atoms of CH₃CN and the carbon atoms of n-octane are strong, resulting in a high solubility in n-octane.

2. CH₃OH - The solute-solvent interactions between the oxygen atom of CH₃OH and the carbon atoms of n-octane are weaker than those of CH₃CN, resulting in a lower solubility in n-octane.

3. Cyclohexane - The solute-solvent interactions between the carbon atoms of cyclohexane and the carbon atoms of n-octane are weaker than those of CH₃OH, resulting in an even lower solubility in n-octane.

4. KCl - The solute-solvent interactions between the chloride ions of KCl and the carbon atoms of n-octane are the weakest, resulting in the lowest solubility in n-octane.

Therefore, the order of increasing solubility in n-octane is CH₃CN > CH₃OH > Cyclohexane > KCl.

Question:

A solution of glucose in water is labelled as 10% w/v, what would be the molality and mole fraction of each component in the solution? If the density of the solution is 1.2 g mL^-1, then what shall be the molarity of the solution?

Answer:

Molality:

1. Calculate the mass of glucose in 1 kg of the solution. Mass of glucose = 10% × 1 kg = 0.1 kg

2. Calculate the moles of glucose in the solution. Moles of glucose = 0.1 kg/180 g/mol = 0.556 mol

3. Calculate the molality of glucose in the solution. Molality of glucose = 0.556 mol/1 kg = 0.556 mol/kg

Mole fraction:

1. Calculate the moles of glucose in the solution. Moles of glucose = 0.1 kg/180 g/mol = 0.556 mol

2. Calculate the moles of water in the solution. Moles of water = 1 kg - 0.556 mol = 0.444 mol

3. Calculate the mole fraction of glucose in the solution. Mole fraction of glucose = 0.556 mol/1.000 mol = 0.556

Molarity:

1. Calculate the mass of glucose in 1 L of the solution. Mass of glucose = 10% × 1 L × 1.2 g mL^-1 = 12 g

2. Calculate the moles of glucose in the solution. Moles of glucose = 12 g/180 g/mol = 0.067 mol

3. Calculate the molarity of glucose in the solution. Molarity of glucose = 0.067 mol/1 L = 0.067 mol/L

Question:

State Henry’s law and mention some important applications?

Answer:

Henry’s Law: At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.

Important Applications:

1. It is used to explain the behavior of gases dissolved in liquids.
2. It is used to calculate the solubility of gases in liquids.
3. It is used to explain the behavior of gases in solution in the atmosphere.
4. It is used to calculate the solubility of gases in solids.
5. It is used to calculate the solubility of gases in ionic solutions.
6. It is used to explain the behavior of gases in the oceans.
7. It is used to calculate the solubility of gases in non-aqueous solutions.
8. It is used to calculate the solubility of gases in supercritical fluids.

Question:

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

Answer:

1. Gases dissolve in liquids because of the forces of attraction between the molecules of the gas and the molecules of the liquid.

2. As the temperature is raised, the molecules of the gas move faster and have more energy, causing them to interact less with the molecules of the liquid, thus reducing their solubility.

3. This is why gases tend to be less soluble in liquids as the temperature is raised.

Question:

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

Answer:

Answer:

1. Calculate the partial pressure of oxygen (PO₂) and nitrogen (PN₂) in the air: PO₂ = 0.20 × 10 atm = 2 atm PN₂ = 0.79 × 10 atm = 7.9 atm

2. Calculate the solubility of oxygen and nitrogen in water at 298 K: SO₂ = 3.30 × 107 mm × 2 atm = 6.60 × 107 mm SN₂ = 6.51 × 107 mm × 7.9 atm = 51.68 × 107 mm

3. Calculate the composition of oxygen and nitrogen in water: CO₂ = SO₂ / (SO₂ + SN₂) = 6.60 × 107 mm / (6.60 × 107 mm + 51.68 × 107 mm) = 0.11 CN2 = SN₂ / (SO₂ + SN₂) = 51.68 × 107 mm / (6.60 × 107 mm + 51.68 × 107 mm) = 0.89

Question:

What is meant by positive and negative deviations from Raoult’s law and how is the sign of Δ^mix​H related to positive and negative deviations from Raoult’s law?

Answer:

Positive and negative deviations from Raoult’s law refer to how closely the vapor pressure of a solution follows Raoult’s law. Raoult’s law states that the vapor pressure of a solution is equal to the vapor pressure of the pure components multiplied by the mole fraction of each component. Positive deviations occur when the vapor pressure of the solution is higher than what is predicted by Raoult’s law, and negative deviations occur when the vapor pressure of the solution is lower than what is predicted.

The sign of Δ^mixH (the enthalpy of mixing) is related to positive and negative deviations from Raoult’s law. Positive deviations from Raoult’s law occur when Δ^mixH is negative, and negative deviations from Raoult’s law occur when Δ^mixH is positive.

Question:

Two elements A and B form compounds having formula AB₂ and AB₂. When dissolved in 20 g of benzene (C₆​H₆),1 g of AB₂ lowers the freezing point by 2.3 K whereas 1.0 g of AB₂ 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.

Answer:

1. Determine the moles of AB₂ and AB₂ that were dissolved in the benzene:

Moles of AB₂ = 1 g / (molar mass of AB₂)

Moles of AB₂ = 1 g / (molar mass of AB₂)

2. Calculate the total moles of AB₂ and AB₂ that were dissolved in the benzene:

Total moles = (moles of AB₂) + (moles of AB₂)

3. Calculate the total mass of AB₂ and AB₂ that were dissolved in the benzene:

Total mass = (total moles) x (molar mass of AB₂ + molar mass of AB₂)

4. Calculate the molar depression constant (K kg mol^-1) of benzene:

Molar depression constant = (change in freezing point) / (total mass of AB₂ and AB₂)

5. Calculate the atomic masses of A and B:

Atomic mass of A = (molar mass of AB₂) / 2

Atomic mass of B = (molar mass of AB₂) / 4

Question:

Calculate the depression in the freezing point of water when 10 g of CH₃​CH₂​CHClCOOH is added to 250 g of water. [Ka=1.4×10^-3,Kf​=1.86 K kg mol^-1]

Answer:

1. Calculate the molar mass of CH₃​CH₂​CHClCOOH: Molar mass of CH₃​CH₂​CHClCOOH = (12.011 + 1.008 + 35.453 + 12.011 + 1.008 + 16.00 + 1.008 + 16.00) g/mol = 93.489 g/mol

2. Calculate the number of moles of CH₃​CH₂​CHClCOOH added to the water: Number of moles = 10g/93.489 g/mol = 0.1068 mol

3. Calculate the molality of the solution: Molality = 0.1068 mol/0.250 kg = 0.4272 mol/kg

4. Calculate the depression in the freezing point of water: Depression in the freezing point = Kf​ x molality = 1.86 K kg mol^-1 x 0.4272 mol/kg = 0.79 K

Question:

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

Answer:

Definition of Solution: A solution is a homogeneous mixture of two or more substances, where one substance is dissolved into the other.

Types of Solutions: There are two main types of solutions:

1. Homogeneous Solutions: These are solutions where the components are uniformly distributed throughout the mixture. Examples of homogeneous solutions include salt water, sugar water, and alcohol in water.

2. Heterogeneous Solutions: These are solutions where the components are not uniformly distributed throughout the mixture. Examples of heterogeneous solutions include oil and water, sand and water, and sugar and water.

Question:

At 300 K,36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

Answer:

Step 1: Calculate the molar mass of glucose (C₆H₁₂O₆). Molar mass of glucose = 180 g/mol

Step 2: Calculate the amount of glucose present in the solution. Amount of glucose present = 36 g

Step 3: Calculate the number of moles of glucose present in the solution. Number of moles of glucose = 36 g/180 g/mol = 0.2 mol

Step 4: Calculate the initial osmotic pressure of the solution. Initial osmotic pressure = 4.98 bar

Step 5: Calculate the final osmotic pressure of the solution. Final osmotic pressure = 1.52 bar

Step 6: Calculate the final concentration of the solution. Final concentration = (0.2 mol/L) x (1.52 bar/4.98 bar) = 0.08 mol/L