Chapter 1 Solutions

NCERT Solutions Class 12 Chemistry Chapter 1 - Solutions

NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions helps students to know how they can progress from a very basic understanding of the concepts of mixtures, to obtaining a quantitative description of how homogeneous substances behave as physical systems. Students who are able to master the NCERT Solutions will also be able to prepare themselves for the CBSE board examinations, as well as prepare themselves for meeting the requirements of high stakes competitive exams such as the JEE Main, JEE Advanced and NEET.

The chapter discusses both the physical and mathematical principles relating to how different materials interact with one another at a molecular level. By using NCERT Solutions, students learn how to calculate the concentration of a solution, and why the addition of salt to water results in an increase in the temperature at which water will boil; explains the logical reasoning behind many chemical phenomena that occur regularly. NCERT Solutions for Class 12 Chemistry will allow students to identify areas in which they need to improve their problem-solving skills and increase their ability to complete high-weightage numerical problems so that they can be successful in their studies.

1.0Class 12 Chemistry Chapter 1: Key Concepts

Class 12 Chemistry Chapter 1 – Solutions explains how substances dissolve in one another and how the properties of a solution can be measured. The chapter covers concentration, solubility, vapour pressure, and the properties of solutions through important laws and numerical problems.

The important concepts in NCERT Class 12 Chemistry Chapter 1 include:

Types of Solutions: Understanding gaseous, liquid, and solid solutions with specific examples such as amalgams and alloys.
Concentration of Solutions: Learning how to express the amount of solute present in a solution using molarity (M), molality (m), mole fraction (x), and parts per million (ppm).
Solubility: Understanding the amount of a substance that can dissolve in a given solvent and the factors that affect the solubility of solids and gases in liquids.
Henry’s Law: Studying the solubility of gases in liquids and how it changes with pressure, with applications such as carbonated drinks and scuba diving.
Vapour Pressure: Understanding how the vapour pressure of a solution changes when a non-volatile solute is added.
Ideal and Non-Ideal Solutions: Understanding the difference between ideal and non-ideal solutions and how their vapour pressure differs from the expected behaviour.
Colligative Properties: Studying properties that depend only on the number of solute particles:
1. Relative lowering of vapour pressure
2. Elevation of boiling point
3. Depression of freezing point
4. Osmotic pressure
Determination of Molar Mass: Understanding how colligative properties can be used to determine the molar mass of a solute.
Abnormal Molar Masses: Understanding how association or dissociation of solute particles affects molar mass calculations using the Van’t Hoff factor (i).

2.0Class 12 Chemistry Chapter 1: Detailed NCERT Textbook Solutions

INTEXT QUESTIONS

Calculate the mass percentage of benzene $(C_{6} H_{6})$ and carbon tetrachloride $(CCl_{4})$ if 22g of benzene is dissolved in 122 g of carbon tetrachloride. Sol. Total mass $= 22 + 122 = 144 g$

$Mass % of benzene = \frac{22 \times 100}{144} = 15.28% &$

Mass % of $CCl_{4} = \frac{122}{144} \times 100 = 84.72%$

2. Calculate the mole fraction of benzene in a solution containing 30% by mass in carbon tetrachloride. Sol. Mass of benzene $= 30 g$ & Molar mass of $C_{6} H_{6}$ $= 78 g / mol$ No. of moles of $C_{6} H_{6} = \frac{w _{2}}{M _{2}} = \frac{30}{78} = 0.385$ Molar mass of $CCl_{4} = 154 g / mol$ No. of moles of $CCl_{4} = \frac{w _{1}}{M _{1}} = \frac{70}{154} = 0.455$ Total moles $= 0.385 + 0.455 = 0.84$

3. Calculate the molarity of each of the following solutions : (a) 30 g of $Co (NO_{3})_{2} \cdot 6 H_{2} O$ in 4.3 L of solution, (b) 30 mL of $0.5 M H_{2} SO_{4}$ diluted to 500 mL .

Sol.

(a) Molar mass of $Co (NO_{3})_{2} \cdot 6 H_{2} O = 291 g / mol$

$Molarity = \frac{w}{m \times V ( lit )} = \frac{30}{291 \times 4.3} = 0.024 M$

(b) $M_{1} V_{1} = M_{2} V_{2}$ ,

$M_{1} = 0.5, V_{1} = 30 mL, M_{2} = ?, V_{2} = 500 mL 0.5 \times 30 = M_{2} \times 500 \Rightarrow M_{2} = 0.03 M$

Calculate the mass of urea ( $NH_{2} CONH_{2})$ required in making 2.5 kg of 0.25 molal aqueous solution. Sol. 0.25 molal aqueous solution means Moles of urea = 0.25, Mass of water $= 1 kg = 1000 g$ 0.25 mole urea $= 0.25 \times 60 = 15$ gram Total mass of solution $= 1000 + 15$ $= 1015 g = 1.015 kg$ $∵ 1.015 kg$ solution contains urea $= 15 g$ $∴ 2.5 kg$ solution contains urea $= \frac{15 \times 2.5}{1.015} = 36.94$ gram
Calculate (a) molality, (b) molarity, (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g/mL. Sol. 20% mass/mass aqueous KI means Mass of $KI = 20 g$ , Mass of water = 80g, Mass of solution $= 100 g$ (a) For molality : Molar mass of $KI = 39 + 127 = 166 g / mol$ Molality $= \frac{w _{2}}{M _{2} \times w _{1} ( kg )}$ $= \frac{20}{166 \times 80 \times 1 0 ^{- 3}} = 1.50 mol kg^{- 1}$ (b) For Molarity :

$Volume of solution = \frac{mass}{density} = \frac{100}{1.202} = 83.2 ml Molarity = \frac{W _{2}}{M _{2} \times V ( lit )} = \frac{20}{166 \times 83.2 \times 1 0 ^{- 3}} = 1.45 mol L^{- 1}$

$& Moles of water = \frac{w _{1}}{M _{1}} = \frac{80}{18} = 4.44$

Total moles $= 0.12 + 4.44 = 4.56$ & Mole fraction of $KI = \frac{0.12}{4.56} = 0.0263$

6. $H_{2} S$ , a toxic gas with rotten egg smell, is used for the qualitative analysis. If the solubility of $H_{2} S$ in water at STP is 0.195 m, calculate Henry's law constant. Mass of solvent (water) $= 1 kg = 1000 g$ . Sol. Solubility $= 0.195 mole$ in one kg of water Mass of solvent $($ water $) = 1 kg = 1000 g$

$& Moles of water = \frac{w _{1}}{M _{1}} = \frac{1000}{18} = 55.55$

Total moles $= 0.195 + 55.55$ Mole fraction of $H_{2} S (X_{H_{2} S})$ in solution

$= \frac{0.195}{0.195 + 55.55} = 0.0035$

Pressure of $H_{2} S$ at $STP = 0.987$ bar. Since partial pressure of the gas is given,

$p_{H_{2} S} = K_{H} \times X_{H_{2} S} or 0.987 = K_{H} \times 0.0035 or K_{H} = 282 bar$

Henry's law constant for $CO_{2}$ in water is $1.67 \times 1 0^{8} Pa$ at 298 K. Calculate the quantity of $CO_{2}$ in 500 mL of soda water when packed under 2.5 atmospheric $CO_{2}$ pressure at 298 K. Sol. $K_{H} = 1.67 \times 1 0^{8} Pa$ ,

$P_{CO_{2}} = 2.5 atm = 2.5 \times 1.01325 \times 1 0^{5} = 2.533 \times 1 0^{5} Pa P_{CO_{2}} = K_{H} \times X_{CO_{2}} X_{CO_{2}} = \frac{P _{CO_{2}}}{K _{H}} = \frac{2.533 \times 1 0 ^{5}}{1.67 \times 1 0 ^{8}} = 0.00152 = 1.52 \times 1 0^{- 3}$

$n_{CO_{2}}$ is neglected in denominator in comparison to $n_{H_{2} O}$ . Water $= 500 mL = 500 g$ , So, $n_{H_{2} O} = \frac{500}{18} = 27.78 mol$ $x_{CO_{2}} = \frac{n _{CO_{2}}}{n _{H_{2} O}}$ $\Rightarrow \frac{n _{CO_{2}}}{n _{H_{2} O}} = 1.520 \times 1 0^{- 3}$ or $\frac{n _{CO_{2}}}{27.78} = 1.52 \times 1 0^{- 3}$ or $n_{CO_{2}} = 42.2 \times 1 0^{- 3}$ Mass of $CO_{2} =$ moles × molar mass $= 42.2 \times 1 0^{- 3} \times 44 = 1.85$ gram

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. Sol. $P_{A}^{0} = 450 mm Hg, P_{B}^{0} = 700 mm Hg$ ,

$P_{Total} = 600 mm Hg$

or $P_{s} = P_{B}^{o} + X_{A} (P_{A}^{o} - P_{B}^{o})$ $600 = 700 + X_{A} (450 - 700)$ $\Rightarrow X_{A} = \frac{100}{250} = 0.40$

$∴ X_{B} = 1 - X_{A} = 1 - 0.40 = 0.60$

$P_{A} = P_{A}^{0} X_{A} = 450 \times 0.40 = 180 mm Hg$ ; $P_{B} = P_{B}^{0} X_{B} = 700 \times 0.60 = 420 mm Hg$ Total vapour pressure $= 180 + 420$ $= 600 mmHg$ Mole fraction of A in vapour phase

$= \frac{P _{A}}{P _{total}} = \frac{180}{600} = 0.30$

Mole fraction of B in vapour phase

$= \frac{P _{B}}{P _{total}} = \frac{420}{600} = 0.70$

Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 gram of urea $(NH_{2} CONH_{2})$ is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering. Sol. Solution contains non-volatile solid. Our aim is to calculate

$P_{s} and \frac{P _{1}^{o} - P _{S}}{P _{S}} = \frac{w _{2} \times M _{1}}{M _{2} \times w _{1}}$

$P_{1}^{o} = 23.8 mm Hg, w_{2} = 50 g, w_{1} = 850 g,$

$P_{s} = ?, M_{1} = 18 g / mol for H_{2} O,$

$M_{2} = 60 g / mol$ for urea

$\frac{P _{1}^{o} - P _{S}}{P _{S}} = \frac{w _{2} \times M _{1}}{M _{2} \times w _{1}}$

$\Rightarrow \frac{23.8 - P _{S}}{P _{S}} = \frac{50 \times 18}{60 \times 850} = \frac{3}{170} = 0.0176$ or $P_{s} = 23.4 mm Hg$ Again, relative lowering in V.P. $= \frac{P _{1}^{o} - P _{S}}{P _{S}}$

$= \frac{23.8 - 23.4}{23.4} = 0.017$

Boiling point of water at 750 mm Hg is $99.6 3^{\circ} C$ . How much sucrose is to be added to 500 g of water such that it boils at 100°C ? $K_{b} = 0.52 K kg / mol$ . Sol. $Δ T_{b} = T_{b} - T_{b}^{\circ}$

$= (100 + 273) K - (99.63 + 273) K = 0.37 K$

$w_{1} =$ ?,

$w_{1} = 500 g,$

$M_{2} = 342 g / mol$ for sucrose,

$K_{b} = 0.52 K kg / mol$

$Δ T_{b} = \frac{K _{b} \times w _{2} \times 1000}{M _{2} \times w _{1}} or 0.37$

$= \frac{0.52 \times w _{2} \times 1000}{342 \times 500}$

$\Rightarrow w = 121.67 g$

Calculate the mass of ascorbic acid (vitamin C, $C_{6} H_{8} O_{6}$ ) to be dissolved in 75g acetic acid to lower its melting point by 1.5°C. $K_{f} = 3.9 Kkg / mol$ . Sol. $Δ T_{f} = \frac{K _{f} \times w _{2} \times 1000}{M _{2} \times w _{1}}$ Formula for lowering in freezing point or melting point is same.

$Δ T_{f} = 1.5 K, w = ?, w_{1} = 75 g,$

$M_{2}$ (ascorbic acid)

$= 176 g / mol 1.5 = \frac{3.9 \times w _{2} \times 1000}{176 \times 75}$

$\Rightarrow w_{2} = 5.08 g$

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

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0g of polymer of molar mass 185000 in 450 mL of water at 37°C. Sol. Osmotic pressure

$(Π) = \frac{n _{2}}{V} RT = \frac{w _{2}}{M _{2}} RT$

Volume of water,

$V = 450 ml = 0.45 L$

Temperature $(T) = (37 + 273) K$

$= 310 K π = \frac{1}{18500} mol \times \frac{1}{0.45 L} \times 8.314 \times 1 0^{3} PaLK^{- 1} mol^{- 1} \times 310 K = 30.98 \approx 31 Pa (approximately)$

3.0NCERT EXERCISE

Define the term solution. How many types of solutions are formed? Write briefly about each type with an example. Sol. Homogeneous mixtures of two or more than two components are known as solutions.

There are three types of solutions.

(i) Gaseous solution : The solution in which the solvent is a gas in called a gaseous solution. In these solutions, the solute may be liquid, solid, or gas. For example, a mixture of oxygen and nitrogen gas is a gaseous solution. (ii) Liquid solution : The solution in which the solvent is a liquid is known as a liquid solution. The solute in these solutions may be gas, liquid, or solid. For example, a solution of ethanol in water is a liquid solution. (iii) Solid solution : The solution in which the solvent is a solid known as a solid solution. The solute may be gas, liquid or solid. For example, a solution of copper in gold is a solid solution.

2. Give an example of solid solutions in which the solute is a gas. Sol. A solution of hydrogen in palladium is a solid solution in which the solute is a gas.

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

Sol.

(i) Mole fraction : 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. (ii) Molality : Molality (m) is defined as the number of moles of the solute per kilogram of the solvent. It is expressed as : (iii) Molarity : Molarity (M) is defined as the number of moles of the solute dissolved in one Litre of solution. It is expressed as : (iv) Mass percentage : The mass percentage of a component of a solution is defined as the mass of the solute in grams present in 100g of the solution. It is expressed as : 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 $1.504 g mL^{- 1}$ ? Sol. 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 is dissolved in 100 g of the solution. Molar mass of nitric acid ( $HNO_{3}$ )

$= 1 \times 1 + 14 + 3 \times 16 = 63 g mol^{- 1}$

Then, number of moles of $HNO_{3}$

$= \frac{68}{63} mol = 1.079 mol$

Given,

$Density of solution = 1.504 g mL^{- 1} ∴ Volume of 100 g solution = \frac{100}{1.504} mL = 66.49 mL = 66.49 \times 1 0^{- 3} L$

Molarity of solution

$= \frac{1.079}{66.49 \times 1 0 ^{- 3} L} = 16.23 M$

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? Sol. 10% w/w solution of glucose in water means that 10 g of glucose in present in 100g of the solution i.e., 10 g of glucose is present in $(100 - 10) g = 90 g$ of water.

Molar mass of glucose $(C_{6} H_{12} O_{6})$

$= 6 \times 12 + 12 \times 1 + 6 \times 16 = 180 g mol^{- 1} Then, number of moles of glucose = \frac{10}{180} mol = 0.056 mol & Molality of solution = \frac{0.056}{0.09 kg} = 0.62 m$

Number of moles of water $= \frac{90 g}{18 gmol ^{- 1}} = 5 mol$ & Mole fraction of glucose $(x_{g}) = \frac{0.056}{0.056 + 5}$

$= 0.011$

And, mole fraction of water $x_{w} = 1 - x_{g}$

$= 1 - 0.011 = 0.989$

If the density of the solution is $1.2 g mL^{- 1}$ , then the volume of the 100 g solution can be given as :

$V = \frac{100 g}{1.2 g mL ^{- 1}} = 83.33 mL = 83.33 \times 1 0^{- 3} L & Molarity of the solution = \frac{0.056 mol}{83.33 \times 1 0 ^{- 3} L} = 0.67 M$

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of $Na_{2} CO_{3}$ and $NaHCO_{3}$ containing equimolar amounts of both ? Sol. Let the amount of $Na_{2} CO_{3}$ in the mixture be x g. Then, the amount of $NaHCO_{3}$ in the mixture is $(1 - x) g$ . Molar mass of $Na_{2} CO_{3}$

$= 2 \times 23 + 1 \times 12 + 3 \times 16 = 106 g mol^{- 1}$

Molar mass of $NaHCO_{3}$

$= 1 \times 23 + 1 \times 1 \times 12 + 3 \times 16 = 84 g mol^{- 1}$

$∴ Number of moles NaHCO_{3} = \frac{1 - x}{84} mol$

According to the question, $\frac{x}{106} = \frac{1 - x}{84}$

$\Rightarrow \Rightarrow \Rightarrow 84 x = 106 - 106 x 190 x = 106 x = 0.5579.$

Therefore, number of moles of $Na_{2} CO_{3}$

$= \frac{0.5579}{106} mol = 0.0053 mol$

And, number of moles of $NaHCO_{3}$

$= \frac{1 - 0.5579}{84} = 0.0053 mol$

HCl reacts with $Na_{2} CO_{3}$ and $NaHCO_{3}$ according to the following equation.

$2 mol 2 HCl + 1 mol Na_{2} CO_{3} ⟶ 2 NaCl + H_{2} O + CO_{2}$

$1 mol HCl + 1 mol NaHCO ⟶ NaCl + H_{2} O + CO_{2}$

1 mol of $Na_{2} CO_{3}$ reacts with 2 mol of HCl . Therefore, 0.0053 mol of $Na_{2} CO_{3}$ reacts with $HCl = 2 \times 0.0053 mol = 0.0106 mol$ . Similarly, 1 mol of $NaHCO_{3}$ reacts with 1 mol of HCl. Therefore, 0.0053 mol of $NaHCO_{3}$ reacts with 0.0053 mol of HCl . Total moles of HCl required

$= (0.0106 + 0.0053) mol = 0.0159 mol$

0.1 mol of HCl is preset in 1000 mL of the solution. Therefore 0.0159 mol of HCl is present in $\frac{1000 \times 0.0159}{0.1} = 159 mL$ of the solution Hence, 159 mL of 0.1 M of HCl is required to react completely with 1 g of mixture of $Na_{2} CO_{3}$ and $NaHCO_{3}$ containing equimolar amounts of both.

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. Sol. Total amount of solute present in the mixture is

$300 \times \frac{25}{100} + 400 \times \frac{40}{100} = 75 + 160 = 235 g$

Total amount of solution $= 300 + 400 = 700 g$ Therefore, mass percentage (w/w) of the solute in the resulting solution

$= \frac{235}{700} \times 100% = 33.57%$

And, mass percentage (w/w) of the solvent in the resulting solution $= (100 - 33.57%) = 66.43%$

An antifreeze solution is prepared from 222.6 g of enthylene glycol $(C_{2} H_{6} O_{2})$ 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? Sol. Molar mass of ethylene glycol $[C_{2} H_{6} O_{2}]$

$= 2 \times 12 + 6 \times 1 + 2 \times 16 = 62 gmol^{- 1}$

Mass of solvent $(H_{2} O) = 200 g = 0.200 kg$ Number of moles of ethylene glycol

$= \frac{222.6 g}{62 gmol ^{- 1}} = 3.59 mol$

Therefore, molality of the solution

$= \frac{3.59 mol}{0.200 kg} = 17.95 m$

Total mass of the solution

$= (222.6 + 200) g = 422.6 g$

Given, Density of the solution $= 1.072 g mL^{- 1}$

$∴ Volume of the solution = \frac{422.6 g}{1.072 g mL ^{- 1}} = 394.22 mL = 0.3942 \times 1 0^{- 3} L \Rightarrow Molarity of the solution = \frac{3.59 mol}{0.39422 \times 1 0 ^{- 3} L} = 9.10 M$

A sample of drinking water was found to be severely contaminated with chloroform $(CHCl_{3})$ 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.

Sol.

(i) 15 ppm (by mass) means 15 parts per million $(1 0^{6})$ of the solution. Therefore, percent by mass

$= \frac{15}{1 0 ^{6}} \times 100% = 1.5 \times 1 0^{- 3} %$

(ii) Molar mass of chloroform $(CHCl_{3})$

$= 1 \times 12 + 1 \times 1 + 3 \times 35.5 = 119.5 g mol^{- 1}$

Now, according to the question, 15 g of chloroform is present in $1 0^{6} g$ of the solution. i.e., 15 g chloroform is present in $(1 0^{6} - 15) \approx 1 0^{6} g$ of water. ∴ Molality of the solution

$= \frac{\frac{15}{119.5} mol}{1 0 ^{6} \times 1 0 ^{- 3} kg} = 1.25 \times 1 0^{- 4} m$

What role does the molecular interaction play in a solution of alcohol and water? Sol. In pure alcohol and water, the molecules are held tightly by a strong hydrogen bonding. The interaction between the molecules of alcohol and water is weaker than alcohol-alcohol and water-water interactions. As a result, when alcohol and water are mixed, the intermolecular interactions become weaker and the molecules can easily escape. This increase the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.
Why do gases always tend to be less soluble in liquids as the temperature is raised ? Sol. Solubility of gases in liquids decreases with an increase in temperature. This is because dissolution of gases in liquids is an exothermic process.

$Gas + Liquid ⟶ Solution + Heat$

Therefore, when the temperature is increased, heat is supplied and the equilibrium shifts backwards, thereby decreasing the solubility of gases.

State Henry's law and mention some important applications? Sol. Henry's law states that partial pressure of a gas in the vapour phase is proportional to the mole fraction of the gas in the solution. If p is the partial pressure of the gas in the vapour phase and x is the mole fraction of the gas, then Henry's law can be expressed as : $P_{g} = K_{H} X_{g}$ where, $K_{H}$ is Henry's law constant Some important applications of Henry's law are mentioned below. (i) Bottles are sealed under high pressure to increase the solubility of $CO_{2}$ in soft drinks and soda water. (ii) The oxygen tanks used by scuba divers are filled with air and diluted with helium to avoid BENDS (Blockage of cappillaries). (iii) The concentration of oxygen is low in the blood and tissue of people living at high altitudes such as climbers. Low-blood oxygen cause climbers to become weak and disables them from thinking clearly. These are symptoms of ANOXIA.
The partial pressure of ethane over a solution containing $6.56 \times 1 0^{- 3} g$ of ethane is 1 bar. If the solution contains $5.00 \times 1 0^{- 2} g$ of ethane, then what shall be the partial pressure of the gas? Sol. Molar mass of ethane $(C_{2} H_{6})$

$= 2 \times 12 + 6 \times 1 = 30 g mol^{- 1}$

∴ Number of moles present in $6.56 \times 1 0^{- 3} g$

$of ethane = \frac{6.56 \times 1 0 ^{- 3}}{30} = 0.218 \times 1 0^{- 3} mol = 2.18 \times 1 0^{- 4} mol$

Let the number of moles of the solvent be x. According to Henry's law, $p_{g} = K_{H} x_{g}$

$\Rightarrow 1 bar = K_{H} \cdot \frac{2.18 \times 1 0 ^{- 4}}{2.18 \times 1 0 ^{- 4} + x} \Rightarrow 1 bar = K_{H} \frac{2.18 \times 1 0 ^{- 4}}{x}$

$(Since x ≫ 2.18 \times 1 0^{- 4}) \Rightarrow K_{H} = \frac{x _{g}}{2.18 \times 1 0 ^{- 4}} bar$

Number of moles present in $5.00 \times 1 0^{- 2} g$ of ethane $= \frac{5.00 \times 1 0 ^{- 2}}{30} mol = 1.67 \times 1 0^{- 3} mol$ According to Henry's law,

$P_{g} = K_{H} X_{g} = \frac{x}{2.18 \times 1 0 ^{- 4}} \times \frac{1.67 \times 1 0 ^{- 3}}{( 1.67 \times 1 0 ^{- 3} ) + x} = \frac{x}{2.18 \times 1 0 ^{- 4}} \times \frac{1.67 \times 1 0 ^{- 3}}{x}$

(Since, $x ≫ 1.67 \times 1 0^{- 3}$ ) $= 7.64$ bar Hence, partial pressure of the gas shall be 7.64 bar.

14. 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? Sol. According to Raoult's law, the partial vapour pressure of each volatile component in any solution is directly proportional to its mole fraction. The solutions which obey Raoult's law over the entire range of concentration are known as ideal solutions. The solutions that do not obey Raoult's law (non - ideal solutions) have vapour pressure either higher or lower than that predicted by Raoult's law. If the vapour pressure is higher, then the solution is said to exhibit positive deviation, and if it is lower, then the solution is said to exhibit negative deviation from Raoult's law.

Vapour pressure of a two - component solution showing positive deviation from Raoult's law.

Vapour pressure of a two - component solution showing negative deviation from Raoult's law

In case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solutions is zero.

$Δ_{sol} H = 0$

In the case of solutions showing positive deviations, absorption of heat takes place.

$Δ_{sol} H = Positive$

In the case of solutions showing negative deviations, evolution of heat takes place.

$Δ_{sol} H = Negative$

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? Sol. Here, Vapour pressure of the solution at normal boiling point $(P_{1}) = 1.004$ bar Vapour pressure of pure water at normal boiling point $(P_{1}^{0}) = 1.013$ bar Mass of solute, $(w_{2}) = 2 g;$ Mass of solvent (water), $(w_{1}) = 98 g$ Molar mass of solvent (water), $(M_{1})$

$= 18 g mol^{- 1}$

According to Raoult's law,

$\frac{P _{1}^{0} - P _{S}}{P _{S}} = \frac{w _{2} \times M _{1}}{M _{2} \times w _{1}} \Rightarrow \frac{1.013 - 1.004}{1.004} = \frac{2 \times 18}{M _{2} \times 98}$

$\Rightarrow M_{2}$ (molar mass of solute) = 41 g/mol

16. Heptane and octane form an ideal solution. At 373 K, the vapour pressure 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 ? Sol. Vapour pressure of heptane $(P_{h}^{0}) = 105.2 kPa$ Vapour pressure of octane $(P_{O}^{0}) = 46.8 kPa$ Molar mass of heptane $(C_{7} H_{16})$

$= 7 \times 12 + 16 \times 1 = 100 g mol^{- 1}$

∴ Number of moles of heptane $= \frac{26}{100}$ $= 0.26 mol$ Molar mass of octane $(C_{8} H_{18}) = 8 \times 12 + 18 \times 1$

$= 114 g mol^{- 1}$

∴ Number of moles of octane $= \frac{35}{114} mol$

$= 0.31 mol$

Mole fraction of heptane, $X_{h} = \frac{0.26}{0.26 + 0.31}$

$= 0.456$

And, mole fraction of octane,

$X_{O} = 1 - 0.456 = 0.544$

Now, partial pressure of heptane,

$P_{h} = P_{h}^{0} X_{h} = 105.2 \times 0.456 = 47.97 kPa$

Partial pressure of octane, $P_{O} = P_{O}^{0} X_{O}$

$= 46.8 \times 0.544 = 25.46 kPa$

Hence, vapour pressure of solution,

$P_{total} = P_{h} + P_{O} = 47.97 + 25.46 = 73.43 kPa$

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. Sol. 1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water). Molar mass of water $= 18 g mol^{- 1}$ ∴ Number of moles present in 1000 g of water $= \frac{1000}{18} = 55.56 mol$ Therefore, mole fraction of the solute

$X_{2} = \frac{1}{1 + 55.56} = 0.0176$

Vapour pressure of water,

$P_{1}^{0} = 12.3 kPa \frac{12.3 - P _{S}}{P _{S}} = 0.0176$

$\Rightarrow 12.3 = 1.0176 P_{S}$

$P_{S} = \frac{12.3}{1.0176}$

$\Rightarrow P_{S} = 12.08 kPa$ (approximately)

Hence, the vapour pressure of the solution is 12.08 kPa.

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%. Sol. Molar mass of solute, $M_{2} = 40 g mol^{- 1}$ Mass of octane, $w_{1} = 114 g$ Molar mass of octane, $(C_{8} H_{18})$ ,

$M_{1} = 8 \times 12 + 18 \times 1 = 114 g mol^{- 1}$

Applying the relation, $\frac{p _{1}^{0} - p _{s}}{p _{s}} = \frac{w _{2} \times M _{1}}{M _{2} \times w _{1}}$

$\Rightarrow \frac{100 - 80}{80} = \frac{w _{2} \times 114}{40 \times 114}$

$\Rightarrow \frac{20}{80} = \frac{w _{2}}{40}$

$\Rightarrow w_{2} = 10 g$

Hence, the required mass of the solute is 10 g.

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. Sol. (i) Let, the molar mass of the solute be $Mg mol^{- 1}$ Now, the no. of moles of solvent (water),

$n_{1} = \frac{90 g}{18 g mol ^{- 1}} = 5 mol$

And, the no. of moles of solute,

$n_{2} = \frac{30 g}{M mol ^{- 1}} = \frac{30}{M} mol P_{S} = 2.8 kPa$

Applying the relation : $\frac{P _{1}^{0} - P _{S}}{P _{S}} = \frac{n _{2}}{n _{1}}$

$\Rightarrow \frac{P _{1}^{0} - 2.8}{2.8} = \frac{\frac{30}{M}}{5}$

(ii) After the addition of 18 g of water :

$n_{1} = \frac{90 + 18 g}{18} = 6 mol P_{1} = 2.9 kPa$

Again, applying the relation : $\frac{P _{1}^{0} - P _{S}}{P _{S}} = \frac{n _{2}}{n _{1}}$

$\Rightarrow \frac{P _{1}^{0} - 2.9}{2.9} = \frac{\frac{30}{M}}{6}$

Dividing equation (i) by (ii), we have :

$\frac{P _{1}^{o} - 2.8}{2.8} \times \frac{2.9}{P _{1}^{o} - 2.9} = \frac{30}{5 M} \times \frac{6 M}{30}$

$\Rightarrow 14.5 (P_{1}^{o} - 2.8) = 16.8 (P_{1}^{o} - 2.9)$ $\Rightarrow 8.12 = 2.3 P_{A}^{o}$ $\Rightarrow$ Vapour pressure of water $(P_{1}^{o}) = 3.53 kPa$ Put value of $P_{1}^{o}$ in eq. (i)

$\Rightarrow \frac{3.53 - 2.8}{2.8} = \frac{30}{5 M}$

⇒ Molar mass of solute (M)

$= \frac{30 \times 2.8}{0.73 \times 5} = \frac{84}{3.65} = 23 u$

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. Sol. Here, $Δ T_{f} = (273.15 - 271) K = 2.15 K$ Molar mass of sugar $(C_{12} H_{22} O_{11})$

$= 12 \times 12 + 22 \times 1 + 11 \times 16 = 342 g mol^{- 1}$

5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in $(100 - 5) g = 95 g$ of water. Now, number of moles of cane sugar

$= \frac{5}{342} mol = 0.0146 mol$

Therefore, molality of the solution,

$m = \frac{0.0146 mol}{0.095 kg} = 0.1537 mol kg^{- 1} Δ T_{f} = K_{f} \times m \Rightarrow K_{f} = \frac{Δ T _{f}}{m} = \frac{2.15 K}{0.1537 mol kg ^{- 1}} = 13.99 K kg mol^{- 1}$

Molar mass of glucose $(C_{6} H_{12} O_{6})$

$= 6 \times 12 + 12 \times 1 + 6 \times 16 = 180 g mol^{- 1} .$

5% glucose in water means 5g of glucose is present in $(100 - 5) = 95 g$ of water. ∴ Number of moles of glucose

$= \frac{5}{180} mol = 0.0278 mol$

Therefore, molality of the solution,

$m = \frac{0.0278 mol}{0.095 kg} = 0.2926 mol kg^{- 1} Δ T_{f} = K_{f} \times m = 13.99 K kg mol^{- 1} \times 0.2926 mol kg^{- 1} = 4.09 K (approximately)$

Hence, the freezing point of 5% glucose solution is $(273.15 - 4.09) K = 269.06 K$ .

21. Two element A and B form compounds having formula $AB_{2}$ and $AB_{4}$ . When dissolved in 20 g of benzene $(C_{6} H_{6}), 1 g$ of $AB_{2}$ lowers the freezing point by 2.3 K whereas 1.0 g of $AB_{4}$ 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. Sol. We know that $M_{2} = \frac{1000 \times w _{2} \times k _{f}}{Δ T _{f} \times w _{1}}$ Then, $M_{AB_{2}} = \frac{1000 \times 1 \times 5.1}{2.3 \times 20} = 110.87 g mol^{- 1}$

$& M_{AB_{4}} = \frac{1000 \times 1 \times 5.1}{1.3 \times 20} = 196.15 g mol^{- 1}$

Now, we have the molar masses of $AB_{2}$ and $AB_{4}$ as $110.87 g mol^{- 1}$ and $196.15 g mol^{- 1}$ respectively. Let the atomic masses of A and B are x and y respectively. Now, we can write :

$x + 2 y = 110.87 x + 4 y = 196.15$

Subtracting equation (1) from (2), we have

$2 y = 85.28 \Rightarrow y = 42.64 u$

Putting the value of 'y' in equation (1), we have

$x + 2 \times 42.64 = 110.87 \Rightarrow x = 25.59 u$

Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.

22. At 300 K, 36g 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? Sol. Here, $T = 300 K; π = 1.52$ bar ;

$R = 0.083 bar L K^{- 1} mol^{- 1}$

$= π = CRT \Rightarrow C = \frac{π}{RT} \frac{1.52}{0.083 barL.K ^{- 1} mol ^{- 1} \times 300 K} = 0.061 mol$

Since the volume of the solution is 1 L , the concentration of the solution would be 0.061 M.

Suggest the most important type of intermolecular attractive interaction in the following pairs. (i) n - hexane and n-octane (ii) $I_{2}$ and $CCl_{4}$ (iii) $NaClO_{4}$ and water $(H_{2} O)$ (iv) methanol and acetone (v) acetonitrile $(CH_{3} CN)$ and acetone $(C_{3} H_{6} O)$ . Sol. (i) Van der Wall's forces of attraction. (ii) Van der Wall's forces of attraction. (iii) Ion-dipole interaction. (iv) Dipole - dipole interaction. (v) Dipole - dipole interaction.
Based on solute- solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, $KCl, CH_{3} OH, CH_{3} CN$ . Sol. n-octane is non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in the n-octane. The order of increasing polartiy is : Cyclohexane $< CH_{3} CN < CH_{3} OH < KCl$ Therefore, the order of increasing solubility is : $KCl < CH_{3} OH < CH_{3} CN <$ Cyclohexane.
Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water ? (i) phenol (ii) toluene (iii) formic acid (iv) ethlene glycol (v) chloroform (vi) pentanol. Sol. (i) Phenol $(C_{6} H_{5} OH)$ has the polar group -OH and non-polar group $- C_{6} H_{5}$ . Thus, phenol is partially soluble in water. (ii) Toluene $(C_{6} H_{5} - CH_{3})$ has no polar group. Thus, toluene is insoluble in water.

(iii) Formic acid (HCOOH) has the polar group -OH and can from H-bond with water. Thus, formic acid is highly soluble in water. (iv) Ethylene glycol has polar -OH group and can form H-bond. Thus, it is highly soluble in water. (v) Chloroform is insoluble in water. (vi) Pentanol $(C_{5} H_{11} OH)$ has polar -OH group, but it also contains a very bulky non-polar $- C_{5} H_{11}$ group. Thus, pentanol is partially soluble in water.

26. If the density of some lake water is $1.25 g mL^{- 1}$ and contains 92 g of $Na^{+}$ ions per kg of water, calculate the molarity of $Na^{+}$ ions in the lake. Sol. No. of moles of $Na^{+}$ ion

$= \frac{92 g}{23 g / mol} = 4 mol$

Mass of water $= 1 kg$

$Volume = \frac{mass}{density} = \frac{1000 g}{1.25 g ml} = 0.8 L$

Molarity of $Na^{+}$ ion $= \frac{4 mol}{0.8 L}$ $= 5 M$ or $5 mol / L$

27. If the solubility product of CuS is $6 \times 1 0^{- 16}$ , calculate the maximum molarity of CuS in aqueous solution. Sol. Solubility product of $CuS, K_{sp} = 6 \times 1 0^{- 16}$ Let's be the solubility of CuS is $mol L^{- 1}$ .

$CuS ⇌ Cu_{S}^{2 +} + S_{S}^{2 -}$

Now,

$K_{sp} = [Cu^{2 +}] [S^{2 +}] = s \times s = s^{2}$

Then, we have,

$K_{sp} = s^{2} = 6 \times 1 0^{- 16} \Rightarrow s = 6 \times 1 0^{- 16} = 2.45 \times 1 0^{- 8} mol L^{- 1}$

Hence, the maximum molarity of CuS in an aqueous solution is $2.45 \times 1 0^{- 8} mol L^{- 1}$ .

Calculate the mass percentage of aspirin $(C_{9} H_{8} O_{4})$ in acetonitrile $(CH_{3} CN)$ when 6.5 g of $C_{9} H_{8} O_{4}$ is dissolved in 450 g of $CH_{3} CN$ . Sol. 6.5 g of aspirin $(C_{9} H_{8} O_{4})$ is dissolved in 450 g of acetonitrile $(CH_{3} CN)$ Then, total mass of the solution

$= (6.5 + 450) g = 456.5 g$

Therefore, mass percentage of $C_{9} H_{8} O_{4}$

$= \frac{6.5}{456.5} \times 100% = 1.424%$

Nalorphene $(C_{19} H_{21} NO_{3})$ , 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 $1.5 \times 1 0^{- 3} m$ aqueous solution required for the above dose. Sol. $1.5 \times 1 0^{- 3} m$ solution means $1.5 \times 1 0^{- 3} mol$ of nalorphene is dissolved in 1 kg of water. Molar mass of Nalorphene

$\Rightarrow \Rightarrow 19 \times 12 + 21 + 14 + 48 = 311 g / mol 1.5 \times 1 0^{- 3} \times 311 g = 0.466 g = 466 mg Mass of solution = 1000 g + 0.466 g = 1000.466 g$

For 1.5 mg of nalorphene solution required

$= \frac{1000.466 \times 1.5 \times 1 0 ^{- 3}}{466} = 3.22 g$

Calculate the amount of benzoic acid $(C_{6} H_{5} COOH)$ required for preparing 250 mL . of 0.15 M solution in methanol. Sol. 0.15 M solution of benzoic acid in methanol means, 1000 mL of solution contains 0.15 mol of benzoic acid. Therefore, 250 mL of solution contains

$= \frac{0.15 \times 250}{1000} mol of benzoic acid = 0.0375 mol of benzoic acid$

Molar mass of benzoic acid $(C_{6} H_{5} COOH)$

$= 7 \times 12 + 6 \times 1 + 2 \times 16 = 122 g mol^{- 1}$

Hence, required benzoic acid

$= 0.0375 mol \times 122 g mol^{- 1} = 4.575 g$

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. Sol.

Among H, Cl, and F, H is least elecronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H . Thus, trifluoroacetic acid can easily lose $\mathrm{H}^{+}$ions i.e. trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point. Hence, the depression in the freezing point increases in the order. Acetic acid < trichloroacetic acid < trifluoroacetic acid
Calculate the depression in the freezing point of water when 10 g, CH3CH2CHClCOOH is added to 250 g of water, Ka = 1.4 × 10–3, Kf = 1.86 K kg mol–1

Sol. Molar mass of CH3CH2CHClCOOH

$= 15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1 = 122.5 g mol^{- 1}$

No. of moles present in

$10 g CH_{3} CH_{2} CHClCOOH = \frac{10 g}{122.5 g mol ^{- 1}} = 0.0816 mol$

It is given that $10 g CH_{3} CH_{2} CHClCOOH$ is added to 250 g of water.

$Molality of the solution, = \frac{0.0186}{250} \times 1000 = 0.3265 mol kg^{- 1}$

Let a be the degree of dissociation of $CH_{3} CH_{2} CHClCOOH$ . $CH_{3} CH_{2} CHClCOOH$ undergoes dissociation according to the following equation : $CH_{3} CH_{2} CHClCOOH \leftrightarrow CH_{3} CH_{2} CHClCOO^{-} + H^{+}$ Initial conc. $C mol L^{- 1} 00$ At equilibrium $C (1 - α) C α C α$

$∴ K_{a} = \frac{C α \cdot C α}{C ( 1 - α )} = \frac{C ^{2}}{( 1 - α )}$

Since a is very small with respect to $1, 1 - α \approx 1$ Now,

$\Rightarrow \Rightarrow K_{a} = C^{2} α = \frac{K _{a}}{C} = \frac{1.4 \times 1 0 ^{- 3}}{0.3264} = 0.0655 (K_{a} = 1.4 \times 1 0^{- 3})$

Again,

$CH_{3} CH_{2} CHClCOOH \leftrightarrow Initial moles At equilibrium CH_{3} CH_{2} CHClCOO^{-} + H^{+} 1 1 - α 0 α 0 α$

Total mass of equilibrium

$= 1 - α + α + α = 1 + α ∴ i = \frac{1 + α}{1} = 1 + α = 1 + 0.0655 = 1.0655$

Hence, the depression in the freezing point of water is given as : $Δ T_{f} = i . K_{f} m$

$= 1.0655 \times 1.86 kg mol^{- 1} \times 0.364 mol kg^{- 1} = 0.65 K$

19.5 g of $CH_{2} FCOOH$ is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid. Sol. It is given that :

$w_{1} = 500 g, w_{2} = 19.5 g K_{f} = 1.86 K kg mol^{- 1}, Δ T_{f} = 1 K$

We know that : $M_{2} = \frac{K _{f} \times w _{2} \times 1000}{Δ T _{f} \times w _{1}}$

$= \frac{1.86 K kg mol ^{- 1} \times 19.5 g \times 1000 g kg ^{- 1}}{500 g \times 1 K} = 72.54 mol^{- 1}$

Therefore, observed molar mass of $CH_{2} FCOOH$ , $(M_{2})_{obs} = 72.54 mol$ The calculated molar mass of $CH_{2} FCOOH$ is :

$(M_{2})_{cal} = 14 + 19 + 12 + 16 + 16 + 1 = 78 g mol^{- 1}$

Therefore, van't Hoff factor,

$i = \frac{( M _{2} ) _{cal}}{( M _{2} ) _{obs}} = \frac{78 g mol ^{- 1}}{72.54 g mol ^{- 1}} = 1.0753$

Let a be the degree of dissociation of $CH_{2} FCOOH$

$CH_{2} FCOOH ⇌ CH_{2} FCOO^{-} + H^{+}$

Initial conc. $C mol L^{- 1} 000$ At equilibrium $C (1 - α) C α C α$

$Total = C (1 + α)$

$∴ i = \frac{C ( 1 + α )}{C} = 1 + α$

$\Rightarrow α = 1.0753 - 1 = 0.0753 = 7.53%$

Now, the value of $K_{a}$ is given as :

$K_{a} = \frac{[ CH _{2} FCOO ^{-} ] [ H ^{+} ]}{[ CH _{2} FCOOH ]} = \frac{C α \cdot C α}{C ( 1 - α )} = \frac{C ^{2}}{1 - α}$

Taking the volume of the solution as 500 mL . We have the concentration : 19.5 M

$weight = 19.5 g$

$Concentration = \frac{\frac{19.5}{78}}{500} \times 1000 = 0.5 M$

Therefore, $K_{a} = \frac{C α ^{2}}{1 - α}$

$= \frac{0.5 \times ( 0.0753 ) ^{2}}{1 - 0.0753} = \frac{0.5 \times 0.00567}{0.9247}$

$= 0.00307 (approximately) = 3.07 \times 1 0^{- 3}$

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. Sol. Vapour pressure of water, $P_{1}^{0} = 17.535 mm$ of Hg Mass of glucose, $w_{2} = 25 g$ & Mass of water, $w_{1} = 450 g$ Molar mass of glucose $(C_{6} H_{12} O_{6})$ , $M_{2} = 6 \times 12 + 12 \times 1 + 6 \times 16 = 180 g mol^{- 1}$ Molar mass of water, $M_{1} = 18 g mol^{- 1}$ Then, number of moles of glucose,

$n_{2} = \frac{25}{180 g mol ^{- 1}} = 0.139 mol$

And, number of moles of water,

$n_{1} = \frac{450 g}{18 g mol ^{- 1}} = 25 mol$

$\Rightarrow \frac{P _{1}^{o} - P _{s}}{P _{s}} = \frac{n _{2}}{n _{1}}$ $\Rightarrow \frac{P _{1}^{o} - P _{s}}{P _{s}} = \frac{0.139}{25}$ $\Rightarrow 17.535 - P_{s} = 0.0055 P_{s}$ $\Rightarrow P_{s} = 17.44 mm$ of Hg So, the vapour pressure of water is 17.44 mm of Hg.

35. Henry's law constant for the molality of methane in benzene at 298 K is $4.27 \times 1 0^{5} mm$ Hg. Calculate the solubility of methane in benzene at 298 K 760 mm Hg . Sol. Here, $P = 760 mm Hg$ ;

$k_{H} = 4.27 \times 1 0^{5} mm Hg$

According to Henry's law,

$P_{g} = K_{H} X_{g} \Rightarrow x_{g} = \frac{p _{g}}{K _{H}} = \frac{760 mm Hg}{4.27 \times 1 0 ^{5} mm Hg} = 178 \times 1 0^{- 5} (approx)$

Hence, the mole fraction of methane in benzene is $178 \times 1 0^{- 5}$ .

36. 100g 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. Sol. Number of moles of liquids A, $n_{A} = \frac{100}{140} mol$

$= 0.714 mol$

Number of moles of liquids $B, n_{B} = \frac{1000}{180} mol$

$= 5.55 mol$

Then, mole fraction of A,

$x_{A} = \frac{n _{A}}{n _{A} + n _{B}} = \frac{0.714}{0.714 + 5.55} = 0.114$

And, mole fraction of B,

$x_{B} = 1 - 0.114 = 0.886$

Vapour pressure of pure liquid B,

$p_{B}^{0} = 500 torr$

Therefore, vapour pressure of liquid B in the solution,

$p_{B} = p_{B}^{0} x_{B} = 500 \times 0.886 = 443 torr$

Total vapour pressure of the solution,

$p_{total} = 475 torr$

$∴$ Vapour pressure of liquid A in the solution,

$p_{A} = p_{total} - p_{B} = 475 - 443 = 32 torr$

Now, $p_{A} = p_{A}^{0} x_{A}$ $\Rightarrow p_{A}^{0} = \frac{p _{A}}{x _{A}} = \frac{32}{0.114} = 280.7$ Hence, the vapour pressure of pure liquid A is 280.7 torr.

37. Vapour pressure 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 $p_{total} p_{chloroform}$ ' and $p_{acetone}$ as a function of $x_{acetone}$ .

The experimental data observed for different compositions of mixture is :

$100 \times x_{acetone}$	Pacetone /mm Hg	Pchloroform /mm Hg
0	0	632.8
11.8	54.9	548.1
23.4	110.1	469.4
36	202.4	359.7
50.8	322.7	257.7
58.2	405.9	193.6
64.5	454.1	161.2
72.1	521.1	120.7

Sol. From the question, we have the following data

100 × $X_{acetone}$	Pacetone /mm Hg	Pchloroform /mm Hg	$P_{total}$ /(mm Hg)
0	0	632.8	632.8
11.8	54.9	548.1	603
23.4	110.1	469.4	579.5
36	202.4	359.7	562.1
50.8	322.7	257.7	580.4
58.2	405.9	193.6	615.3
64.5	454.1	161.2	615.3
72.1	521.1	120.7	614.8

It can be observed from the graph that the plot for the $p_{total}$ of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.

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. Sol. Molar mass of benzene $(C_{6} H_{6})$

$= 6 \times 12 + 6 \times 1 = 78 g mol^{- 1} Molar mass of toluene = 7 \times 12 + 8 \times 1 = 92 g mol^{- 1}$

Now, no. of moles present in 80 g of benzene

$(n_{b}) = \frac{80}{78} mol = 1.026 mol$

And, no. of moles present in 100 g of toluene

$(n_{t}) = \frac{100}{92} mol = 1.087 mol$

Mole fraction of benzene,

$x_{b} = \frac{1.026}{1.026 + 1.087} = 0.486$

And, mole fraction of toluene,

$x_{t} = 1 - 0.486 = 0.514$

It is given that vapour pressure of pure benzene, $p_{b}^{0} = 50.71 mm$ of Hg and vapour pressure of pure toluene, $p_{t}^{0} = 32.06 mm$ of Hg Therefore, partial vapour pressure of benzene,

$p_{b} = x_{b} \times p_{b}^{0} = 0.486 \times 50.71 = 24.64 mm of Hg$

And, partial vapour pressure of toluene,

$p_{t} = x_{t} \times p_{t}^{o} = 0.514 \times 32.06 = 16.47 mm of Hg$

Hence, mole fraction of benzene in vapour phase is given by :

$\frac{p _{b}}{p _{b} + p _{t}} = \frac{24.64}{24.64 + 16.47} = \frac{24.64}{41.11} = 0.599 = 0.6$

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 are $3.30 \times 1 0^{7} mm$ and $6.51 \times 1 0^{7} mm$ respectively. Calculate the composition of these gases in water.

Sol. Percentage of oxygen $(O_{2})$ in air $= 20%$ Percentage of nitrogen $(N_{2})$ in air $= 79%$ Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is, $(10 \times 760) mm Hg = 7600 mm Hg$ Partial pressure of oxygen,

$p_{o_{2}} = \frac{20}{100} \times 7600 mm Hg = 1520 mm Hg$

Partial pressure of nitrogen,

$p_{N_{2}} = \frac{79}{100} \times 7600 mm Hg = 6004 mm of Hg$

Now, according to Henry's law : $p = K_{H} x$ For oxygen : $p_{O_{2}} = K_{H} \cdot x_{O_{2}}$ (Given $K_{H} = 3.30 \times 1 0^{7} mm$ of Hg)

$\Rightarrow x_{O_{2}} = \frac{p _{O_{2}}}{K _{H}} = \frac{1520 mm of Hg}{3.30 \times 1 0 ^{7} mm of Hg} = 4.60 \times 1 0^{- 5} .$

For nitrogen, $p_{N_{2}} = K_{H} \cdot x_{N_{2}}$ (Given $K_{H} = 6.51 \times 1 0^{7} mm$ of Hg)

$\Rightarrow x_{N_{2}} = \frac{p _{N_{2}}}{K _{H}} = \frac{6004 mm Hg}{6.51 \times 1 0 ^{7} mm Hg} = 9.22 \times 1 0^{- 5}$

Hence, the mole fractions of oxygen and nitrogen in water are $4.60 \times 1 0^{- 5}$ and $9.22 \times 1 0^{- 5}$ respectively.

40. Determine the amount of $CaCl_{2} (i = 2.47)$ dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C. Sol. We know that, $π = i \frac{n _{2}}{V} RT$

$\Rightarrow π = i \frac{w _{2}}{M _{2} V} RT \Rightarrow w_{2} = \frac{π M _{2} V}{iRT}$

$π = 0.75 atm; V = 2.5 L; i = 2.47; T = (27 + 273) K = 300 K R = 0.082 L atm K^{- 1} mol^{- 1}$

Molar mass of $CaCl_{2} (M_{2})$

$= 1 \times 40 + 2 \times 35.5 = 111 g mol^{- 1} Therefore, w_{2} = \frac{0.75 \times 111 \times 2.5}{2.47 \times 0.0821 \times 300} = 3.42 g$

Hence, the required amount of $CaCl_{2}$ is 3.42 g .

41. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of $K_{2} SO_{4}$ in 2 liter of water at 25°C, assuming that it is completely dissociated. Sol. When $K_{2} SO_{4}$ is dissolved in water, $K^{+}$ and $SO_{4}^{2 -}$ ions are produced.

$K_{2} SO_{4} ⟶ 2 K^{+} + SO_{4}^{2 -}$

Total number of ions produced $= 3$

$\Rightarrow i = 3$

Given,

$w_{2} = 25 mg = 0.025 g; V = 2 L; T = 2 5^{\circ} C = (25 + 273) K = 298 K R = 0.0821 L atm K^{- 1} mol^{- 1}$

Molar mass of $K_{2} SO_{4} (M_{2})$

$= (2 \times 39) + (1 \times 32) + (4 \times 16) = 174 g mol^{- 1} π = i \frac{n _{2}}{V} RT$

$\Rightarrow i \frac{w _{2}}{M _{2}} \frac{1}{v} RT$

$= 3 \times \frac{0.025}{174} \times \frac{1}{2} \times 0.0821 \times 298 = 5.27 \times 1 0^{- 3} atm$

4.0NCERT Solutions Class 12 Chemistry | Chapter-wise Links

Find NCERT Solutions for Class 12 Chemistry with chapter-wise links to understand concepts, solve textbook questions, and prepare effectively for exams.

Chapter Number	Chapterwise NCERT Solutions
Chapter 2	Electrochemistry
Chapter 3	Chemical Kinetics
Chapter 4	D- and F-Block Elements
Chapter 5	Coordination Compounds
Chapter 6	Haloalkanes and Haloarenes
Chapter 7	Alcohols, Phenols and Ethers
Chapter 8	Aldehydes, Ketones and Carboxylic Acids
Chapter 9	Amines
Chapter 10	Biomolecules

5.0Key Features of NCERT Solutions for Class 12 Chemistry Chapter 1

Step-by-Step Numericals: Easy solutions for molarity, molality, and unit conversion questions.
Clear Graphs: Simple explanations of vapour pressure-composition graphs for ideal and non-ideal solutions.
Important Formulas: Helps students understand and apply the Van’t Hoff factor in colligative property questions.
Easy Concepts: Clear explanations of topics such as azeotropes and reverse osmosis (RO).
Exam Preparation: Step-by-step answers help students understand the formula, calculation, and final answer while solving NCERT questions.

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Frequently Asked Questions

Why is NCERT Solutions for Class 12 Chemistry Chapter 1 important?

Is the chapter Solutions important for the competitive exams like JEE and NEET?

Are these NCERT Solutions as per the latest CBSE syllabus?

What are the important topics in Class 12 Chemistry Chapter 1 Solutions?

How can NCERT Solutions for Class 12 Chemistry Chapter 1 help with numerical questions?

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NCERT Solutions Class 12 Chemistry Chapter 1 - Solutions

1.0Class 12 Chemistry Chapter 1: Key Concepts

2.0Class 12 Chemistry Chapter 1: Detailed NCERT Textbook Solutions

INTEXT QUESTIONS

3.0NCERT EXERCISE

4.0NCERT Solutions Class 12 Chemistry | Chapter-wise Links

5.0Key Features of NCERT Solutions for Class 12 Chemistry Chapter 1

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