Page No. 5

1.1. Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.?

Mass of solution = Mass of benzene + Mass of carbon tetrachloride

= 22 g + 122 g

= 144 g

Mass percentage of benzene

Mass percentage of carbon tetrachloride

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

Let the total mass of the solution be 100 g, and the mass of benzene be 30 g.

∴ Mass of carbon tetrachloride = (100 − 30) g

= 70 g

Molar mass of benzene (C6H6) = (6×12 + 6×1) g mol−1

= 78 g mol−1

∴ Number of moles of C6H6

= 0.385 mol

Molar mass of carbon tetrachloride (CCl4) = 1×12 + 4×35.5

= 154 g mol−1

∴ Number of moles of CCl4

= 0.455 mol

Thus, the mole fraction of C6H6 is given as:

= 0.458

∴ Mole fraction of CCl4 = 1 − 0.458 = 0.542

1.3. Calculate the molarity of the following solution:

(a) 30 g of Co(NO3)2⋅6H2O in 4.3 L of solution.

(b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.

(a) Molar mass of Co(NO3)2⋅6H2O

= 58.7 + 2(14 + 48) + 6×18 g mol−1

= 58.7 + 2(62) + 108 g mol−1

= 58.7 + 124 + 108 g mol−1

= 290.7 g mol−1

∴ Number of moles of Co(NO3)2⋅6H2O

= 0.103 mol

Volume of solution = 4.3 L

Molarity of solution

= 0.024 M

(b) Number of moles present in 1000mL of 0.5 M H2SO4 = 0.5 mol

∴ Number of moles present in 30mL of 0.5 M H2SO4

= 0.015 mol

Therefore, molarity

= 0.03 M

1.4. Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.

Moles of urea = 0.25 mol

Mass of solvent (water) = 1 kg = 1000 g

Molar mass of urea (NH2CONH2) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16

= 60 g mol−1

∴ Mass of urea in 1000 g of water = 0.25 mol × 60 g mol−1

= 15 g

Total mass of solution = 1000 + 15 g

= 1015 g

= 1.015 kg

Thus, 1.015 kg of solution contain urea = 15 g

∴ 2.5 kg of solution will require urea

= 36.95 g

= 37 g (approximately)

Hence, the mass of urea required is 37 g.

1.5. Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL−1.

20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.

∴ Mass of solvent (water) = 100 − 20 = 80 g = 0.080 kg

(a) Calculation of morality:

Molar mass of KI = 39 + 127 = 166 g mol−1

Moles of KI

= 0.120 mol

Molality of solution

= 1.5 mol kg−1

(b) Calculation of molarity:

Density of the solution = 1.202 g mL−1

∴ Volume of solution

= 83.2 mL

= 0.0832 L

Molarity of the solution

= 1.44 M

(c) Calculation of the mole fraction of KI:

No. of moles of KI = 0.120 mol

No. of moles of water

= 4.44 mol

Mole fraction of KI

= 0.0263

1.6. H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant.

Solubility of H2S gas = 0.195 m

∴ Moles of H2S = 0.195 mol

Mass of water = 1000 g

No. of moles of water

= 55.55 mol

∴ Mole fraction of H2S gas in the solution (x)

= 0.0035

Pressure at STP = 1 bar

According to Henry’s law,

pH2S = KH × xH2S

= 285.7 bar

1.7. Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.

Given, KH = 1.67 × 108 Pa

PCO2 = 2.5 atm = 2.5×101325 Pa

According to Henry’s law,

PCO2 = KH × CO2

= 1.517 × 10−3

We can write,

= 1.517 × 10−3 …[Since, nCO2 is negligible as compared to to nH2O]

For 500 mL of soda water, the volume of water = 500 mL; the mass of water = 500 g,

We can write,

= 500/18 mol of water

= 27.78 mol of water

i.e., nH2O = 27.78

⇒ nCO2 = 42.14×10−3 mole

Mass of CO2 = 42.14×10−3 × 44 g = 1.854 g

1.8. The vapour pressures 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 the total vapour pressure is 600 mm Hg. Also, find the composition of the vapour phase.

Given,

PA0 = 450 mm Hg

PB0 = 700 mm Hg

PTotal = 600 mm Hg

xA = ?

Applying Raoult's law,

PA = xA × PA0

PB = xB × PB0 = (1-xA)PB0

PTotal = PA + PB

= xA × PA0 + (1-xA)PB0

= PB0 + (PA0-PB0)xA

Substituting the given values, we get

600 = 700 + (450 − 700)xA or 250x= 100

⇒ xA = 100/250 = 0.40

Thus, the composition of the liquid mixture will be

xA (mole fraction of A) = 0.40

xB (mole fraction of B) = 1 − 0.40 = 0.60

Calculation of composition in the vapour phase:

PA = xA×PA0

= 0.40 × 450 mm Hg

= 180 mm Hg

PB = xB× PB0

= 0.60 × 700 mm Hg

= 420 mm Hg

Mole fraction of A in the vapour phase

= 0.30

Mole fraction of B in the vapour phase = 1 − 0.30 = 0.70

1.9. Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Given, P0 = 23.8 mm Hg

w1 = 850 g, M1 (water) = 18 g mol−1

w2 = 50 g, M2 (urea) = 60 g mol−1

Applying Raoult's law,

Thus, the relative lowering of vapour pressure = 0.017

∴ Î”P = 0.017 × 23.8

⇒ P0 − Ps = 0.017 × 23.8

⇒ Ps = 23.8 − (0.017 × 23.8)

⇒ Ps = 23.4 mm Hg

Thus, the vapour pressure of water in the solution is 23.4 mm Hg.

1.10. 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?

Molal elevation constant for water is 0.52 K kg mol−1.

Here, elevation of boiling point Î”Tb = (100 + 273) − (99.63 + 273)

= 0.37 K

Mass of water, w1 = 500 g

Molar mass of sucrose (C12H22O11), M2 = 11×12 + 22×1 + 11×16

= 342 g mol−1

Molal elevation constant, Kb = 0.52 K kg mol−1

We know that,

= 121.67 g (approximately)

= 122 g

Hence, 122 g of sucrose is to be added.

1.11. Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf = 3.9 K kg mol−1.

Lowering in melting point (Î”Tf) = 1.5°C

Mass of solvent (CH3COOH), w1 = 75 g

Mass of solute, w2 = ?

Molar mass of solvent (CH3COOH), M1 = 60 g mol−1

Molar mass of solute (C6H8O6), M2 = 6 × 12 + 8 × 1 + 6 × 16

= 72 + 8 + 96

= 176 g mol−1

Kf = 3.9 K kg mol−1

Applying the formula,

= 5.08 g (approx)

Hence, 5.08 g of ascorbic acid is needed to be dissolved.

1.12. 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.

Given,

Number of moles of solute dissolved (n)

V = 450 mL = 0.45 L,

T = 37°C = (37 + 273) K = 310 K

R = 8.314 k Pa L K−1 mol−1 = 8.314 × 103 Pa L K−1 mol−1

= 30.96 Pa

= 31 Pa (approximately)

#### Exercises

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

Homogeneous mixtures of two or more than two components are known as solutions.

A homogeneous mixture means that its composition and properties are uniform throughout the mixture. The substances that make up the solution are called the components of the solution. Based on the total number of components present in a solution, it is called a binary solution (two components), ternary solution (three components), quaternary solution (four components), etc.

The components of a binary solution are usually called the solute and the solvent. Generally, the component present in the largest quantity is called the solvent, while the other component present in the smallest quantity is called the solute. The solvent determines the physical state in which the solution exists. In other words, the solute is the substance that dissolves and the solvent is the substance in which the solute dissolves.

For example, if some crystals of sugar are put in a beaker filled with water, they dissolve in the water and form a solution. In this case, sugar is the solute, and water is the solvent. The solution's particles have a molecular size of about 1000 pm, and no physical technique, such as filtration, decantation, centrifugation, etc., can separate its various components.

Types of Solutions

Depending upon the physical state of the solute and solvent, solutions may be classified into the following types:

 Type of Solution Solute Solvent Common Examples Gaseous Solutions Gas Gas mixture of oxygen and nitrogen gases Liquid Gas Chloroform mixed with nitrogen gas Solid Gas Camphor in nitrogen gas Liquid Solutions Gas Liquid Oxygen dissolved in water Liquid Liquid Ethanol dissolved in water Solid Liquid Glucose dissolved in water Solid Solutions Gas Solid Solution of hydrogen in palladium Liquid Solid Amalgam of mercury with sodium Solid Solid Copper dissolved in gold

Three of the nine types of solutions listed above are very common: solid in liquid, gas in liquid, and liquid in liquid. All three types of solutions have liquid as the solvent. Solutions that have water as the solvent are called aqueous solutions, while solutions that do not have water as the solvent are called non-aqueous solutions. Examples of common non-aqueous solvents are ether, benzene, carbon tetrachloride, etc.

The types of solutions are explained below:

1. Solid Solution

2. Liquid Solution

3. Gaseous Solution

1. Liquid solution: These solutions are formed by mixing solids or gases in liquids or by mixing two liquids. Some solid substances also form liquid solutions when mixed. For instance, mixing equimolar amounts of sodium and potassium metals at room temperature results in a liquid solution. Oxygen dissolved in water in sufficient quantity protects aquatic organisms in ponds, rivers and seas.
1. When both the components are completely miscible: In this case, both the liquids are of a similar nature, i.e., either both are polar (e.g., ethyl, alcohol and water) or nonpolar (e.g., benzene and hexane).
2. When both the components are almost miscible: Here, one liquid is polar, and the other is nonpolar in nature, e.g. benzene and water, oil and water, etc.
3. When both the components are partially miscible: If the intermolecular attraction A-A in liquid A is different from the intermolecular attraction B-B in liquid B, but the A-B attraction is of intermediate order, then both the liquids are limitedly miscible with each other. For example, ether and water are partially mixed.
2. Solid solution: A solid solution is a solution in which the solvent is a solid. The solute may be gas, liquid or solid. For example, gold and copper form solid solutions because gold atoms replace copper atoms in copper crystals, and similarly, copper atoms can replace gold atoms in gold crystals. Alloys of two or more metals are solid solutions.
Solid solutions can be divided into two classes:
1. Substitutable solid solutions: In these solutions, atoms, molecules or ions of one substance replace the particles of the other substance in the crystal lattice. Brass, copper and zinc are common examples of substitutable solid solutions.
2. Interstitial solid solutions: In these solutions, atoms of one type occupy the vacancies or gaps present in the lattice of atoms of the other substance. A common example of an interstitial solid solution is tungsten carbide (WC).
3. Gaseous solution: A gaseous solution is a solution in which the solvent is a gas. In these solutions, the solute may be liquid, solid, or gaseous. For example, a mixture of oxygen and nitrogen gas is a gaseous solution.

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

In case a solid solution is formed between two substances (one having very large particles and the other having very small particles), an interstitial solid solution will be formed. For example, a solution of hydrogen in palladium is a solid solution in which the solute is a gas.

1.3. Define the following term:

(i) Mole fraction

(ii) Molality (m)

(iii) Molarity (M)

(iv) Mass Percentage

(i) The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component to the total number of moles of all the components in the mixture i.e.

Mole fraction of a component = Number of moles of the component/Total number of moles of all component

Mole fraction is denoted by ‘x’.

If in a binary solution, the number of moles of the solute and the solvent are nA and nB respectively, then the mole fraction of the solute in the solution is given by,

Similarly, the mole fraction of the solvent in the solution is given

The sum of the mole fractions of all the components in a solution is always 1, i.e.

Therefore, if the mole fraction of one component of a binary solution is known, then the mole fraction of the other component can be determined.

For example, for a binary solution, the mole fraction xA is related to xB in the following way:

xA = 1 – xB

⇒ xB = 1 – xA

Mole fractions do not depend on the temperature of the solution.

(ii) Molality is a property of a solution and is defined as the number of moles of solute per kilogram of solvent. The SI unit for molality is mol/kg.

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

(iii) Molarity (M) is defined as the number of moles of the solute dissolved in one Litre (or one cubic decimetre) of solution.

It is expressed as:

Molarity (M) = Moles of solute/Volume of Solution in Litre

For example, 0.25 mol L–1 (or 0.25 M) solution of NaOH means that 0.25 mol of NaOH has been dissolved in one litre (or one cubic decimetre).

(iv) The mass percentage of a component of a solution is defined as the mass of the solute in gram present in 100 g of the solution. It is expressed as:

For example, if a solution is described as 10% glucose in water by mass, it means that 10 g of glucose is dissolved in 90 g of water, resulting in a 100 g solution. Concentration, described by mass percentage, is commonly used in industrial chemical applications. For example, a commercial bleaching solution contains a 3.62 mass percentage of sodium hypochlorite in water.

1.4. 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 g mL−1?

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. This means that 68 g of nitric acid are dissolved in 100 g of the solution.

Molar mass of nitric acid (HNO3) = 1×1 + 1×14 + 3×16 = 63 g mol−1

Then, the number of moles of HNO3 = 68/63 mol

= 1.079 mol

Given,

Density of solution = 1.504 g mL−1

∴ Volume of solution

= 66.5 mL

= 0.0665 L

Molarity of Solution

= 16.23 M

1.5. 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 mL−1, then what shall be the molarity of the solution?

Let mass of solution = 100 g

∴ Mass of glucose = 10 g

Mass of water = 100 – 10 = 90 g = 0.09 kg

No. of moles in 10 g glucose = 10/180

= 0.0555 mol

No. of moles in 90 g H2O = 90/18

= 5 moles

= 83.33 mL

= 0.0833 L

= 0.617 m

= 0.01

∴ x (H2O) = 1 − 0.01 = 0.99

= 0.67 M

1.6. How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?

Let the amount of Na2CO3 in the mixture be x g.

Then, the amount of NaHCO3 in the mixture is (1 − x) g.

Molar mass of Na2CO3 = 2×23 + 1×12 + 3×16

= 106 g mol−1

∴ Number of moles Na2CO3 = x/106 mol

Molar mass of NaHCO3 = 1×23 + 1×1×12 + 3×16

= 84 g mol−1

According to the question,

⇒ 84x = 106 − 106x

⇒ 190x = 106

⇒ x = 0.558

Therefore, number of moles of Na2CO3 = 0.558/106 mol

= 0.00526 mol

And,

= 0.00526 mol

HCl reacts with Na2CO3 and NaHCO3 according to the following equation:

1 mol of Na2CO3 reacts with 2 mol of HCl.

Therefore, 0.00526 mol of Na2CO3 reacts with 2 × 0.00526 mol = 0.01052 mol.

Similarly, 1 mol of NaHCO3 reacts with 1 mol of HCl.

Therefore, 0.00526 mol of NaHCO3 reacts with 0.00526 mol of HCl.

Total moles of HCl required = (0.01052 + 0.00526) mol

= 0.01578 mol

In 0.1 M of HCl,

0.1 mol of HCl is preset in 1000 mL of the solution.

Therefore,

= 157.8 mL of the solution

Hence, 157.8 mL of 0.1 M of HCl is required to react completely with a 1 g mixture of Na2CO3 and NaHCO3, containing equimolar amounts of both.

1.7. 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.

The total amount of solute present in the mixture is given by,

= 75 + 160

= 235 g

Total amount of solution = 300 + 400 = 700 g

Therefore, mass percentage (w/w) of the solute in the resulting solution

= 33.57%

And, mass percentage (w/w) of the solvent in the resulting solution,

= (100 − 33.57)%

= 66.43%

1.8. 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?

Molar mass of ethylene glycol [C2H4(OH)2] = 2 × 12 + 6 × 1 + 2 × 16

= 62 g mol−1

= 3.59 mol

Therefore,

= 17.95 m

Total mass of the solution = (222.6 + 200) g

= 422.6 g

Given,

Density of the solution = 1.072 g mL−1

= 394.2 mL

= 0.3942 L

= 9.11 M

1.9. 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.

15 ppm (by mass) means 15 parts per million (106) of the solution.

(i) Therefore,

= 1.5 × 10−4

(ii) Molar mass of chloroform (CHCl3) = 1 × 12 + 1 × 1 + 3 × 35.5

= 119.5 g mol−1

Now, according to the question,

15 g of chloroform is present in 106 g of the solution.

i.e., 15 g of chloroform is present in (106 − 15) ≈ 106 g of water.

= 1.25 × 10−4 m

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

Alcohol and water both have a strong tendency to form intermolecular hydrogen bonds. On mixing the two, a solution is formed as a result of the formation of H-bonds between alcohol and H2O molecules, but these interactions are weaker and less extensive than those in pure H2O. Thus, they show a positive deviation from ideal behaviour. As a result of this, the solution of alcohol and water will have a higher vapour pressure and a lower boiling point than that of water and alcohol.

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

The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas in equilibrium with the liquid at that temperature. The dissolution of a gas in a liquid is an exothermic process, that is, it is accompanied by the evolution of heat. Thus,

Gas + Solvent ⇋ Solution + Heat

Applying Le Chatelier’s principle, the increase of temperature would shift the equilibrium in the backward direction, that is, solubility would decrease. Therefore, gases always tend to be less soluble in liquids as the temperature is raised.

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

This law states that ‘the solubility of a gas in a liquid is directly proportional to the pressure of the gas.’

or

‘The partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution’.

It is expressed as:

p = KHx

where,

KH = Henry’s law constant

p = Partial pressure of the gas in vapour phase

x = Mole fraction of the gas

Applications of Henry’s law

• To increase the solubility of CO2 in soda water and soft drinks, the bottle is sealed under high pressure.
• To avoid the toxic effects of high concentration of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen).
• At high altitudes, low blood oxygen causes climbers to become weak and make them unable to think clearly, which are symptoms of a condition known as anoxia.

1.13. 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?

Molar mass of ethane (C2H6) = 2×12 + 6×1

= 30 g mol−1

∴ Number of moles present in 6.56 × 10−3 g of ethane

= 2.187 × 10−3 mol

Let the number of moles of the solvent be x.

According to Henry’s law,

p = KHx

Number of moles present in 5.00 × 10−2 g of ethane

= 1.67 × 10−3 mol

According to Henry’s law,

p = KHx

Hence, the partial pressure of the gas shall be 0.764 bar.