1. SOLUTIONS

CBSE Class 12 Chemistry Chapter 1 Solutions

Chapter 1 : Solutions

CBSE Class 12 Chemistry Chapter 1 Solutions Intext Questions and Answers :

Intext Question1.1 Calculate the mass percentage of benzene ( ) and carbon tetrachloride () if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Solution : Mass of Benzene () = 22 g

Mass of Carbon Tetrachloride () = 122 g

Now, calculate the total mass of the solution:

Total Mass of Solution = Mass of Benzene + Mass of Carbon Tetrachloride

Total Mass of Solution = 22 g + 122 g = 144 g

Mass Percentage of Benzene ():

Mass Percentage of Benzene

Mass Percentage of Carbon Tetrachloride ():

Mass Percentage of Carbon Tetrachloride

So, the mass percentage of benzene in the solution is approximately 15.28%, and the mass percentage of carbon tetrachloride in the solution is approximately 84.72%.

Intext Question1.2 Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Solition: Mass Percentage of Benzene () = 30% = 0.30

Mass Percentage of = 100% – 30% = 70% = 0.70

Molar mass of benzene () = 6 × 12 + 6 × 1 = 78 g/mol

Molar mass of carbon tetrachloride () = 1 × 12 + 4 × 35.5 = 154 g/mol

Moles of Benzene ()

Moles of Benzene 0.3842 moles

Moles of

Moles of = 0.4552 moles

Mole Fraction of Benzene

So, the mole fraction of benzene in the solution containing 30% by mass in carbon tetrachloride is approximately 0.458.

Intext Question1.3 Calculate the molarity of each of the following solutions: (a) 30 g of in 4.3 L of solution (b) 30 mL of 0.5 M diluted to 500 mL.

Solution: (a) Molar mass of = 1 × 59 + 2 × 14 + 6 × 16 + 6(1 × 2 + 16) = 292 g/mol

Number of moles moles

The volume of the solution is given as 4.3 L.

Molarity (M)

So, the molarity of the solution containing 30 g of in 4.3 L of solution is 0.024 M.

(b) Here, , , and

We have,

So, after dilution, the molarity of the solution is approximately 0.03 M.

Intext Question1.4 Calculate the mass of urea ( ) required in making 2.5 kg of 0.25 molal aqueous solution.

Solution : We have,

Molality (m)

Moles of solute = Molality × Mass of solvent

Moles of solute = 0.25 x 2.5 = 0.625 moles

Molar mass of urea () = 1 × 14 + 2 × 1 + 12 × 1 + 16 × 1 + 14 × 1 + 2 × 1 = 60 g/mol

Mass of urea (solute) = (0.625 moles) x (60 g/mol)

Mass of urea = 37.5 grams

Intext Question1.5 Calculate (a) molality (b) molarity and (c) mole fraction of if the density of 20% (mass/mass) aqueous is 1.202 g m .

Solution: (a) Molality of : Molar mass of KI = 39 g/mol + 127 g/mol = 166 g/mol

Molality (m) = 20% (mass/mass) = 20 g of KI in 100 g of solution

Mass of KI = 20 g Mass of water = 100 g – 20 g = 80 g

Moles of KI moles

Again, Mass of water = 80 g = 0.080 kg

Molality (m)

(b) Molarity of :

Weight of solution = 100 g ; Density of solution = 1.202 g/mL

Volume of solution

Molarity (M)

Moles of KI moles

Molar mass of water = 18 g/mol

Moles of water moles

Mole fraction of KI

Intext Question1.6 , a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of in water at STP is 0.195 m, calculate Henry’s law constant.

Solution: Here, mole , mole

bar

Applying Henry’s law , we have

Bar

Intext Question1.7 Henry’s law constant for in water is Pa at 298 K. Calculate the quantity of in 500 mL of soda water when packed under 2.5 atm pressure at 298 K.

Solution: We know that, 1 atm = 101325 Pa

Molar mass of = 44 g/mol

Here, Pa ,

Apply Henry’s law , we have

Moles of water mole

We have,

mole

Therefore, the mass of

Intext Question1.8 The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K . Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Solution: Here, mm , mm , mm

According to Raoult's law, the vapor pressure of the mixture () is given by :

We have,

So, the mole fraction of A () = 0.4

and the mole fraction of B () = 1 – 0.4 = 0.6

Mole fraction of A (vapor phase)

Mole fraction of B (vapor phase)

Intext Question1.9 Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea () is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering .

Solution: Here, P° = 23.8 mm

Molar mass of H2O = 18 g/mol

Molar mass of urea () = 14 + 2 × 1 + 12 + 16 + 14 + 2 × 1 = 60 g/mol

Moles of water moles

Moles of urea moles

We have,

So, the vapor pressure of the solution is 23.4 mm Hg.

The relative lowering of vapor pressure using Raoult's law:

Relative Lowering of Vapor Pressure (RLVP)

So, the relative lowering of vapor pressure for this solution 0.017 .

Intext Question1.10 Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.

Solution: Molar mass of sucrose () = 12 × 12 + 22 × 1 + 11 × 16 = 342 g/mol

Here, g , 342 g/mol , (100+273) – (99.63+273) = 0.37 K and 0.52 K kg/mol

We know that ,

So, we need to add 121.67g of sucrose to 500 g of water to raise its boiling point to 100°C.

Intext Question1.11 Calculate the mass of ascorbic acid (Vitamin C, ) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C . K kg .

Solution: Molar mass of sucrose ( ) = 12 × 6 + 8 × 1 + 6 × 16 = 176 g/mol

Here, g , 176 g/mol , 1.5 K and 0.52 K kg/mol

We know that ,

Intext Question1.12 Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.

Solution: Here, 185000 , 1.0 g , T = 273 + 37 = 310 K , V = 450 mL = 0.45 L ,

R = Pa L

We know that ,

CBSE Class 12 Chemistry Chapter 1 Solutions Example Questions and Answers :

Example 1.1 Calculate the mole fraction of ethylene glycol ( ) in a solution containing 20% of by mass.

Solution: Given, Mass of water = 80 g , Mass of = 20 g

Molar mass of water ()= 2×1 + 16 = 18 g/mol

Molar mass of = 2 × 12 + 6 × 1 + 2 × 16 = 62 g/mol

Moles of water mole

Moles of mole

The mole fraction of

Example 1..2 Calculate the molarity of a solution containing 5 g of in 450 mL solution.

Solution: Given, molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

Moles of NaOH mole

Volume of the solution in L = 450/1000 L = 0.45 L

The molarity of a solution

mol/L

Example 1.3 Calculate molality of 2.5 g of ethanoic acid () in 75 g of benzene.

Solution: Molar mass of = 12 + 3 × 1 + 12 + 16 + 16 + 1 = 60 g/mol

Mole of mol

Mass of benzene in kg kg

We know that ,

Molality mol/kg

Example 1.4 If gas is bubbled through water at 293 K, how many millimoles of gas would dissolve in 1 litre of water? Assume that exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for at 293 K is 76.48 kbar.

Solution: Here, k bar = 76480 bar , bar

Apply Henry’s law , we have

i.e.,

Again, 1 litre of water contains 55.5 mol of it and let n be the number of moles of in solution.

So,

[ is very negligible value. ]

milli mole

Example 1.5 Vapour pressure of chloroform () and dichloromethane () at 298 K are 200 mm Hg and 415 mm Hg respectively. (i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of

and 40 g of () at 298 K and, (ii) mole fractions of each component in vapour phase.

Solution: (i) Here, molar mass of = 12 + 1 + 35.5 × 3 = 119.5 g/mol

Molar mass of = 12 + 2 × 1 + 35.5 × 2 = 85 g/mol

mole

Here, mm , mm , mol

mm Hg

(ii) We have ,

mm Hg

We know that,

Example 1.6 The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g ). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?

Solution: Molar mass of benzene () = 12 × 6 +6 × 1 = 78 g/mol

Here, bar , bar , g/mol , g , g

We know that ,

g/mol

Example 1.7 18 g of glucose, , is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? for water is 0.52 K kg .

Solution: Molar mass of C6H12O6 = 12 × 6 + 12 × 1 + 16 × 6 = 180 g/mol

Moles of = 18/180 = 0.1 mole

Given , 0.52 K kg

Molality (m) mole/ kg

We know that , K

Since water boils at 373.15 K at 1.013 bar pressure .

Therefore, the boiling point of solution will be 373.15 + 0.052 = 373.202 K .

Example 1.8 The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. for benzene is 2.53 K kg.

Solution: Given , 354.11 – 353.23 = 0.88 K , 90 g , 1.8 g , 2.53 K kg

We know that ,

g/mol

Therefore, molar mass of the solute is 57.5 g/mol .

Example 1.9 45 g of ethylene glycol ( ) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of the solution.

Solution: Here, 1.86 K kg/mol

Molar mass of = 12 ×2 + 1 × 6 + 16 × 2 = 62 g/mol

Moles of = mole

Mass of water = 600/1000 = 0.6 kg

Molality of mol/kg

(a) The freezing point depression,

(b) The Freezing point of the aqueous solution = 273.15 K – 2.2 K = 270.95 K

Example 1.10 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg . Find the molar mass of the solute.

Solution: Given , 0.40 K , 50 g , 1.00 g , 5.12 K kg

We know that ,

g/mol .

Thus, molar mass of the solute = 256 g/mol

Example 1.11 200 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be bar. Calculate the molar mass of the protein .

Solution : Given, V = 200 = 200/1000 L = 0.2 L , 1.26 g

T = 300 K , R = 0.083 bar , bar

We know that ,

g/mol

Example 1.12 2g of benzoic acid ( ) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg . What is the percentage association of acid if it forms dimer in solution ?

Solution : Given , 1.62 K , 25 g , 2 g , 4.9 K kg

We know that ,

g/mol

Thus, experimental molar mass of benzoic acid in benzene is 241.97 g/mol .

Molar mass of = 12 × 6 + 1 × 5 + 12 + 16 + 16 + 1 = 122 g/mol .

Let x be the degree of association of the solute then we would have (1 – x ) mol of benzoic acid left in unassociated form and correspondingly x/2 as associated moles of benzoic acid at equilibrium.

So,

We know that ,

A/Q ,

Therefore, degree of association of benzoic acid in benzene is 99.2 %.

Example 1.13 0.6 mL of acetic acid ( ), having density 1.06 g m , is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205°C. Calculate the van’t Hoff factor and the dissociation constant of acid.

Solution: Molar mass of = 12 + 3 × 1 + 12 + 16 + 16 + 1 = 60 g/mol

Number of moles of acetic acid =

mole

Molality (m) mol/kg

1.86 K kg/mol

We know that ,

and

Let x is the degree of dissociation of acetic acid, then we would have n(1 – x) moles of undissociated acetic acid, nx moles of and nx moles of ions .

Thus total moles of particles =

A/Q,

Thus degree of dissociation of acetic acid = 0.041

Then,

CBSE Class 12 Chemistry Chapter 1 Solutions Exercise Questions and Answers :

Q2.1 : Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Answer : A solution is a homogeneous mixture where one or more substances (solutes) are dissolved in another substance (solvent). The solute particles become evenly distributed throughout the solvent, resulting in a well-mixed and uniform solution.

There are three main types of solutions:

Liquid Solutions: In liquid solutions, both the solute and solvent are in a liquid state. A common example is saltwater, where salt (solute) is dissolved in water (solvent).

Solid Solutions: In solid solutions, the solute is in a solid state and is uniformly distributed within the solid solvent. A classic example is alloys, such as brass, which is a mixture of copper (solute) and zinc (solvent).

Gas Solutions: In gas solutions, the solute is in a gaseous state and is dissolved in a gas solvent. One example is air, which is a mixture of various gases like oxygen, nitrogen, and carbon dioxide.

Q2.2 : Give an example of a solid solution in which the solute is a gas.

Answer : One example of a solid solution where the solute is a gas is the combination of hydrogen gas () and palladium (Pd). When hydrogen gas is absorbed into the crystal lattice of palladium, it forms a solid solution known as "palladium hydride."

In this solid solution, the hydrogen gas molecules become evenly dispersed within the palladium's metal structure. Palladium hydride can store large amounts of hydrogen, making it useful for hydrogen storage applications in industries like hydrogen fuel cells and hydrogen storage tanks.

Q2.3 : Define the following terms:

(i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.

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

i.e., Mole fraction of a component

It is used to express the concentration of a component in a solution and is represented by the symbol "x" Mole fraction values range from 0 to 1.

(ii) Molality: Molality is a measure of the concentration of a solute in a solution.

It is defined as the number of moles of solute per kilogram of solvent. Molality is denoted by the symbol "m" and is independent of temperature.

(iii) Molarity: Molarity is also a measure of the concentration of a solute in a solution.

It is defined as the number of moles of solute per liter of solution. Molarity is denoted by the symbol "M" and can vary with temperature.

(iv) Mass percentage: Mass percentage (also known as weight percentage) is a way to express the concentration of a component in a mixture.

i.e., Mass % of a component

It is the ratio of the mass of the solute to the total mass of the solution, multiplied by 100 to express it as a percentage. Mass percentage is represented by the symbol "%(w/w)" and is often used in various fields, such as chemistry and pharmacy.

Q2.4 : 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 ?

Solution : Let us assume we have 100 grams of the 68% nitric acid solution.

Mass of pure nitric acid = 68% of 100 grams = 0.68 × 100 grams = 68 grams

Molar mass of (nitric acid) = 1 + 14 + 48 = 63 g/mol

We have, Moles of nitric acid

Moles of nitric acid moles

Given density of the solution = 1.504 g/mL

Volume of 100 grams of the solution

Volume of the solution in liters

Molarity (M)

So, the molarity of the 68% nitric acid solution is 16.23 M.

2.5 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 , then what shall be the molarity of the solution?

Solution: The molar mass of glucose () = 180 g/mol

Number of moles of glucose (solute)

The mass of water (solvent) is 90 g.

Molality (m)

Total moles in the solution = moles of glucose + moles of water

Mole fraction of glucose () = moles of glucose / total moles

Mole fraction of water () = moles of water / total moles

Molarity (M): Molarity is defined as the number of moles of solute per liter of solution.

Volume of solution = mass of solution / density

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

Molarity (M)

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

Solution: The balanced chemical equation for the reaction between HCl and (sodium carbonate) is:

1 g mixture = 0.5 g + 0.5 g

Molar mass of = 23 × 2 + 12 + 3 × 16.00 = 106 g/mol

Molar mass of = 23 + 1 + 12 + 3 × 16 = 84 g/mol

Moles of

Moles of HCl required = 2 × moles of + moles of

Moles of HCl required = 2 × 0.00471 mol + 0.00595 mol = 0.01537 mol

Molarity (M) = moles / volume (in L)

Volume (in L) = moles / molarity

Volume (in L)

Volume (in mL)

So, 153.7 mL of 0.1 M HCl are required to react completely with the 1 g mixture of and containing equimolar amounts of both .

2.7 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.

Solution: In the 25% solution, the mass of solute is 25% of 300 g = 0.25 × 300 g = 75 g.

In the 40% solution, the mass of solute is 40% of 400 g = 0.40 × 400 g = 160 g.

Calculate the total mass of solute in the mixture:

Total mass of solute = 75 g + 160 g = 235 g.

Total mass of resulting solution = Mass of 25% solution + Mass of 40% solution

= 300 g + 400 g = 700 g.

Mass percentage of solute

The mass percentage of the resulting solution = (100 – 33.57)% = 66.43 %.

2.8 An antifreeze solution is prepared from 222.6 g of ethylene glycol ( ) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072, then what shall be the molarity of the solution?

Solution: The molar mass of ethylene glycol () = 62 g/mol.

Moles of ethylene glycol

Molality (m) = Moles of solute / Mass of solvent (in kg)

Since water is the solvent, its mass is 200 g = 0.2 kg.

Molality (m)

Molarity (M) = Moles of solute / Volume of solution (in L)

Mass of the solution = Mass of ethylene glycol + Mass of water

= 222.6 g + 200 g = 422.6 g

Density of the solution = 1.072 g/mL

Volume of the solution = Mass of solution / Density of solution

Molarity (M)

So, the molality of the solution is 17.95 mol/kg, and the molarity of the solution is 9.10 M.

2.9 A sample of drinking water was found to be severely contaminated with chloroform () 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.

Solution: (i) 1 ppm (part per million) is equivalent to 0.0001%

(since 1/1,000,000 = 0.000001, and multiplying by 100 gives percentage).

So, 15 ppm = 15 × 0.0001% = 0.0015%

(ii) Given that the level of contamination is 15 ppm by mass, we can assume that 15 g of chloroform is present in 1 million grams (1,000,000 g) of water.

Molar mass of chloroform () = 12 + 1 + 3 × 35.5 = 119.5 g/mol

Number of moles of chloroform

Mass of water = 1,000,000 g (1 million grams)

Molality (m) = Moles of solute / Mass of solvent (in kg)

Molality (m)

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

Answer : In the case of alcohol and water, their molecular interactions, like hydrogen bonding, reduce the vapor pressure compared to what Raoult's law predicts for ideal solutions. This is why alcohol-water mixtures evaporate more slowly than expected. The strong interactions between alcohol and water molecules cause them to stick together more, making it harder for them to escape into the vapor phase. This altered vapor pressure behavior is a direct result of the molecular interactions present in the solution.

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

Answer : As temperature increases, according to Le Chatelier's principle, the equilibrium shifts in a way that reduces the overall effect of the temperature change. In the case of gases dissolving in liquids, this means that with higher temperature, the equilibrium tends to favor the gas escaping from the liquid and forming bubbles. So, gases become less soluble in liquids as the temperature rises because the equilibrium shifts to counteract the temperature increase by reducing gas solubility.

2.12 State Henry’s law and mention some important applications.

Answer : Henry's Law: Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid, at a constant temperature.

Applications of Henry's Law :

(i) Henry's Law explains how carbon dioxide dissolves in sodas and sparkling drinks under pressure. When the bottle is opened, the lower pressure allows the gas to come out of solution, forming bubbles and giving the beverage its characteristic fizz.

(ii) Scuba divers use Henry's Law to understand how nitrogen and other gases dissolve in their blood and tissues at different depths. Ascending too quickly can lead to decompression sickness ("the bends") due to the rapid release of dissolved gases.

(iii) In medical oxygen masks, Henry's Law helps deliver more oxygen to patients by increasing the pressure of oxygen gas above the liquid oxygen solution, promoting better oxygen absorption.

(iv) It explains how gases like oxygen and carbon dioxide move between the atmosphere and oceans. When the air has more dissolved gases, they're released into the atmosphere; when the air has fewer gases, they dissolve into the water.

(v) Henry's Law plays a role in brewing beer and other fermented beverages. During fermentation, gases like carbon dioxide are produced and dissolved in the liquid, affecting the final taste and texture.

(vi) The vapor pressure of gasoline, determined by Henry's Law, is crucial for fuel efficiency and emissions control in vehicles. It affects how easily gasoline evaporates and contributes to air pollution.

2.13 The partial pressure of ethane over a solution containing g of ethane is 1 bar. If the solution contains g of ethane, then what shall be the partial pressure of the gas?

Solution : Using Dalton's law of partial pressures, the partial pressure of a gas is directly proportional to the amount of that gas present.

We know that ,

We have,

And

The partial pressure of the gas is 7.62 bar.

2.14 What is meant by positive and negative deviations from Raoult's law and how is the sign of ∆mixH related to positive and negative deviations from Raoult's law?

Answer : Positive Deviation: When the actual vapor pressure of a solution is higher than expected according to Raoult's law, it's a positive deviation. This happens when the molecular interactions between components are weaker, leading to more vaporization.

Negative Deviation: When the actual vapor pressure of a solution is lower than expected by Raoult's law, it's a negative deviation. Stronger molecular interactions cause this, reducing vaporization.

∆mix H Sign:

Positive : Generally corresponds to a positive deviation. This indicates that breaking the existing interactions and forming new ones requires more energy, causing the components to behave more independently and vaporize more easily.

Negative : Usually related to a negative deviation. It means breaking existing interactions and forming new ones releases energy, leading to stronger interactions and reduced vaporization.

2.15 An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Solution : Here, bar , bar , g/mol , g , g

We know that,

Therefore, the molar mass of the solute is 41.346 g .

2.16 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 g of octane?

Solution : Here, kPa , kPa

Molar mass of heptane () = 7 × 12 + 16 × 1 = 100 g/mol

Moles of heptanes () = 26 / 100 = 0.26 mol

Molar mass of octane () = 8 × 12 + 18 × 1 = 114 g/mol

Moles of octane () = 35 / 114 = 0.31 mol

Total moles of the mixture ( ) = 0.26 mol + 0.31 mol = 0.57 mol

Mole fraction of heptane ()= moles of heptane / total moles = 0.259 mol / 0.565 mol = 0.46

Mole fraction of octane () = moles of octane / total moles = 0.306 mol / 0.565 mol = 0.54

Applying Raoult's law , We have

KPa

and KPa

2.17 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.

Solution:

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

Solution :

2.19 A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g 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.

Solution :

2.20 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.

Solution :

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

Solution :

2.22 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?

Solution :

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

(i) n-hexane and n-octane

(ii) and

(iii) and water

(iv) methanol and acetone

(v) acetonitrile () and acetone ().

Solution : The most important types of intermolecular attractive interactions for each pair:

(i) n-hexane and n-octane:

London dispersion forces (van der Waals forces)

(ii) I₂ and CCl₄:

London dispersion forces (van der Waals forces)

(iii) NaClO₄ and water:

Ion-dipole interactions

(iv) Methanol and acetone:

Dipole-dipole interactions

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

Dipole-dipole interactions

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

Solution :

Solution : The order of increasing solubility in n-octane and explanation are :

KCl (Potassium Chloride):

KCl is an ionic compound with strong ion-ion interactions. Since n-octane is nonpolar and lacks the ability to interact with ions, KCl has very low solubility in n-octane.

CH₃OH (Methanol):

Methanol is a polar molecule with dipole-dipole interactions. Although n-octane is nonpolar, it can still dissolve some polar compounds to a limited extent. Therefore, methanol has higher solubility in n-octane compared to KCl.

CH₃CN (Acetonitrile):

Acetonitrile is also a polar molecule with dipole-dipole interactions. Its greater polarity allows it to be more soluble in n-octane than methanol, which means it has higher solubility in n-octane than methanol.

Cyclohexane:

Cyclohexane is a nonpolar molecule like n-octane, and both can mix well due to London dispersion forces. Therefore, cyclohexane has the highest solubility in n-octane among the given substances.

So, the order of increasing solubility in n-octane is: KCl < CH₃OH < CH₃CN < Cyclohexane

2.25 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.

Answer : The given compounds as insoluble, partially soluble, or highly soluble in water are :

Compound	Solubility in Water
Phenol	Partially Soluble
Toluene	Insoluble
Formic acid	Highly Soluble
Ethylene glycol	Highly Soluble
Chloroform	Insoluble
Pentanol	Partially Soluble

Explanation:

Phenol contains a polar hydroxyl (OH) group, which makes it partially soluble in water due to hydrogen bonding.

Toluene is a nonpolar hydrocarbon and is insoluble in water because it lacks the polarity needed for water solubility.

Formic acid is a polar molecule with a carboxyl group (COOH), and it is highly soluble in water due to hydrogen bonding.

Ethylene glycol is a polar molecule with two hydroxyl (OH) groups, making it highly soluble in water due to multiple hydrogen bonding sites.

Chloroform is a nonpolar compound and is insoluble in water due to its lack of polarity.

Pentanol is partially soluble in water because it contains a polar hydroxyl (OH) group, allowing for some degree of hydrogen bonding with water molecules.

2.26 If the density of some lake water is 1.25 and contains 92 g of ions per kg of water, calculate the molarity of ions in the lake.

Solution :

2.27 If the solubility product of CuS is , calculate the maximum molarity of CuS in aqueous solution.

Solution :

2.28 Calculate the mass percentage of aspirin ( ) in acetonitrile ( ) when 6.5 g of ) is dissolved in 450 g of .

Solution :

2.29 Nalorphene ( ), 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 m aqueous solution required for the above dose.

Solution :

2.30 Calculate the amount of benzoic acid () required for preparing 250 mL of 0.15 M solution in methanol.

Solution :

2.31 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.

Solution : The depression in freezing point of water increases in the order: acetic acid < trichloroacetic acid < trifluoroacetic acid. This trend is due to the number of solute particles. Acetic acid dissociates partially into two ions ( and), trichloroacetic acid dissociates into three ions ( , and), and trifluoroacetic acid dissociates into four ions ( , and ). More solute particles result in stronger colligative properties, leading to a greater depression in the freezing point of the solvent (water).

2.32 Calculate the depression in the freezing point of water when 10 g of is added to 250 g of water. , .

Solution:

2.33 19.5 g of 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.

Solution:

2.34 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.

Solution :

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

Solution :

2.36 100 g of liquid A (molar mass 140 ) was dissolved in 1000 g of liquid B (molar mass 180 ). 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.

Solution :

2.37 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 , and as a function of . The experimental data observed for different compositions of mixture is:

100 ×	0	11.8	23.4	36.0	50.8	58.2	64.5	72.1
	0	54.9	110.1	202.4	322.7	405.9	454.1	521.1
	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.

Solution :

2.38 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 80 g of benzene is mixed with 100 g of toluene.

Solution :

2.39 The air is a mixture of a number of gases. The major components are oxygen and nitrogen with 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 the Henry’s law constants for oxygen and nitrogen at 298 K are mm and mm respectively, calculate the composition of these gases in water.

Solution :

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

Solution :

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

Solution :