Class 12 Chemistry Chapter 1 Solution Complete Answers

Welcome to the dedicated Answers and Solutions page for Class 12 Chemistry Chapter 1 Solution Complete Ans. Mastering this chapter requires more than just memorizing formulas—it is about understanding how different components interact on a molecular level. Whether you are calculating the exact concentration of a solute, predicting how a solution will behave under different temperatures, or tackling complex numerical problems on colligative properties, having clear, step-by-step answers is key to building your confidence.

Class 12 Chemistry Chapter 1 Solution Complete Ans- Textbook Answer

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

A solution is a homogeneous mixture of two or more substances. The component present in larger amount is called the solvent, while the component present in smaller amount is called the solute.

Depending on the physical states of solute and solvent, nine types of solutions are possible.

Solute	Solvent	Type of Solution	Example
Gas	Gas	Gaseous solution	Air
Liquid	Gas	Gaseous solution	Chloroform mixed with nitrogen gas
Solid	Gas	Gaseous solution	Camphor in nitrogen gas
Gas	Liquid	Liquid solution	Oxygen in water
Liquid	Liquid	Liquid solution	Ethanol in water
Solid	Liquid	Liquid solution	Salt in water
Gas	Solid	Solid solution	Hydrogen in palladium
Liquid	Solid	Solid solution	Amalgam of mercury with sodium
Solid	Solid	Solid solution	Brass (zinc in copper)

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

Hydrogen dissolved in palladium is an example of a solid solution in which the solute is a gas.

1.3 Define the following terms:

(i) Mole fraction
Mole fraction of a component is the ratio of the number of moles of that component to the total number of moles of all components present in the solution.

$X_A=\frac{n_A}{n_A+n_B}$ XA=nA+nBnA

where $X_A$ XA is mole fraction of component A.

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

$m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$ m=mass of solvent in kgmoles of solute

(iii) Molarity
Molarity is the number of moles of solute present in 1 litre of solution.

$M=\frac{\text{moles of solute}}{\text{volume of solution in litre}}$ M=volume of solution in litremoles of solute

(iv) Mass percentage
Mass percentage is the mass of solute present in 100 g of solution.

$\text{Mass \%}=\frac{\text{mass of solute}}{\text{mass of solution}}\times100$ Mass %=mass of solutionmass of solute×100

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?

Given:
Mass percentage of HNO₃ = 68%
Density of solution = 1.504 g mL⁻¹

Mass of 1 litre solution: $= 1.504 \times 1000$ $= 1504 \text{ g}$

Mass of HNO₃ in 1 litre solution: $= \frac{68}{100} \times 1504$ $= 1022.72 \text{ g}$

Molar mass of HNO₃ = 63 g mol⁻¹

Number of moles of HNO₃: $= \frac{1022.72}{63}$ =631022.72 $= 16.23 \text{ mol}$

Since these moles are present in 1 litre solution,

$M=\frac{16.23\ \text{mol}}{1\ \text{L}}=16.23\ M$

Therefore, the molarity of the nitric acid solution is 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?

10% w/w glucose solution means:

10 g glucose is present in 100 g solution.
Mass of water = 100 – 10 = 90 g

Molar mass of glucose (C₆H₁₂O₆) = 180 g mol⁻¹

Moles of glucose: $= \frac{10}{180}$ $= 0.0556 \text{ mol}$

Mass of water = 90 g = 0.09 kg

Molality

$m=\frac{0.0556}{0.09}=0.617\ m$

Moles of water: $= \frac{90}{18}$ $= 5 \text{ mol}$

Total moles: $= 5 + 0.0556$ $= 5.0556$

Mole fraction of glucose

$X_{glucose}=\frac{0.0556}{5.0556}=0.011$

Mole fraction of water $X_{water} = 1 – 0.011$ $= 0.989$

Density of solution = 1.2 g mL⁻¹

Volume of 100 g solution: $= \frac{100}{1.2}$ $= 83.33 \text{ mL}$ $= 0.08333 \text{ L}$

Molarity

$M=\frac{0.0556}{0.08333}=0.667\ M$

Therefore:

Molality = 0.617 m
Mole fraction of glucose = 0.011
Mole fraction of water = 0.989
Molarity = 0.667 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 moles of Na₂CO₃ = moles of NaHCO₃ = x

Molar mass of Na₂CO₃ = 106 g mol⁻¹
Molar mass of NaHCO₃ = 84 g mol⁻¹

Total mass: $106x + 84x = 1$ $190x = 1$ $x = \frac{1}{190}$ $= 0.00526 \text{ mol}$

Reactions: $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$ $NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2$

HCl required: $= 2(0.00526) + 0.00526$ $= 0.01578 \text{ mol}$

Using molarity formula:

$V=\frac{n}{M}=\frac{0.01578}{0.1}=0.1578\ L$ $= 157.8 \text{ mL}$

Therefore, volume of HCl required = 157.8 mL.

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.

Mass of solute in first solution: $= \frac{25}{100} \times 300$ $= 75 \text{ g}$

Mass of solute in second solution: $= \frac{40}{100} \times 400$ $= 160 \text{ g}$

Total mass of solute: $= 75 + 160$ $= 235 \text{ g}$

Total mass of solution: $= 300 + 400$ $= 700 \text{ g}$

Mass percentage:

$\text{Mass \%}=\frac{235}{700}\times100=33.57\%$

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

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: $= (2 \times 12) + (6 \times 1) + (2 \times 16)$ $= 62 \text{ g mol}^{-1}$

Moles of ethylene glycol: $= \frac{222.6}{62}$ $= 3.59 \text{ mol}$

Mass of water = 200 g = 0.2 kg

Molality

$m=\frac{3.59}{0.2}=17.95\ m$

Total mass of solution: $= 222.6 + 200$ $= 422.6 \text{ g}$

Density = 1.072 g mL⁻¹

Volume of solution: $= \frac{422.6}{1.072}$ $= 394.2 \text{ mL}$ $= 0.3942 \text{ L}$

Molarity

$M=\frac{3.59}{0.3942}=9.11\ M$

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

15 ppm means: $15 \text{ parts of solute in }10^6 \text{ parts of solution}$

Mass percent:

$\text{Mass \%}=\frac{15}{10^6}\times100=1.5\times10^{-3}\%$

Therefore, percentage by mass = 1.5 × 10⁻³ %

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

15 ppm means 15 g CHCl₃ in $10^6$

Molar mass of CHCl₃: $= 12 + 1 + (35.5 \times 3)$ $= 119.5 \text{ g mol}^{-1}$

Moles of CHCl₃: $= \frac{15}{119.5}$ $= 0.1255 \text{ mol}$

Mass of water: $= 10^6 \text{ g} = 1000 \text{ kg}$

Molality:

$m=\frac{0.1255}{1000}=1.255\times10^{-4}\ m$

Therefore, molality = 1.255 × 10⁻⁴ m

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

In alcohol and water solution, strong hydrogen bonding exists between alcohol molecules and water molecules. These interactions are different from those present in pure alcohol and pure water. Due to these intermolecular attractions, the solution shows negative deviation from Raoult’s law and volume contraction occurs on mixing.

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

The dissolution of gases in liquids is generally an exothermic process. When temperature increases, according to Le Chatelier’s principle, the equilibrium shifts in the direction that absorbs heat, causing dissolved gas molecules to escape from the solution. Therefore, the solubility of gases decreases with increase in temperature.

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

Henry’s law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

$p=K_Hx$ p=KHx

where:

$p$ p = partial pressure of gas
$x$ x = mole fraction of gas
$K_H$ KH = Henry’s law constant

Applications of Henry’s law:

In the manufacture of soft drinks and soda water, carbon dioxide is dissolved under high pressure.
Deep-sea divers use oxygen diluted with helium to avoid bends.
At high altitudes, lower oxygen pressure causes breathing difficulties.

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?

According to Henry’s law: $p \propto x$

Since mole fraction is proportional to mass for same solvent, $\frac{p_1}{p_2}=\frac{m_1}{m_2}$

Given: $p_1=1 \text{ bar}$ $m_1=6.56\times10^{-3}\text{ g}$ $m_2=5.00\times10^{-2}\text{ g}$ $p_2=\frac{m_2\times p_1}{m_1}$ $=\frac{5.00\times10^{-2}}{6.56\times10^{-3}}$ $=7.62 \text{ bar}$

Therefore, the partial pressure of ethane is 7.62 bar.

1.14 What is meant by positive and negative deviations from Raoult’s law and how is the sign of ΔmixH related to positive and negative deviations from Raoult’s law?

Positive deviation from Raoult’s law:
When the vapour pressure of the solution is greater than predicted by Raoult’s law, the solution shows positive deviation. This occurs when intermolecular forces between unlike molecules are weaker than those between like molecules.

For positive deviation: $\Delta H_{mix} > 0$

Heat is absorbed during mixing.

Negative deviation from Raoult’s law:
When the vapour pressure of the solution is lower than predicted by Raoult’s law, the solution shows negative deviation. This occurs when intermolecular forces between unlike molecules are stronger than those between like molecules.

For negative deviation: $\Delta H_{mix} < 0$

Heat is evolved during mixing.

1.15 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?

At normal boiling point of water: $P^0 = 1.013 \text{ bar}$

Pressure lowering: $\Delta P = 1.013 – 1.004$ $=0.009 \text{ bar}$

Relative lowering:

$\frac{\Delta P}{P^0}=\frac{0.009}{1.013}=0.00888$

For dilute solutions: $\frac{\Delta P}{P^0}=\frac{n_2}{n_1}$

Take 100 g solution:

Solute = 2 g
Water = 98 g

Moles of water: $=\frac{98}{18}=5.44$

Let molar mass of solute = M

Moles of solute: $=\frac{2}{M}$ $0.00888=\frac{2/M}{5.44}$ $M=\frac{2}{0.00888\times5.44}$ $=41.4\text{ g mol}^{-1}$

Therefore, molar mass of solute = 41.4 g mol⁻¹

1.16 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?

Molar mass of heptane = 100 g mol⁻¹
Molar mass of octane = 114 g mol⁻¹

Moles of heptane: $=\frac{26}{100}=0.26$

Moles of octane: $=\frac{35}{114}=0.307$

Total moles: $=0.567$

Mole fractions: $X_{heptane}=\frac{0.26}{0.567}=0.458$ $X_{octane}=\frac{0.307}{0.567}=0.542$

Using Raoult’s law: $P=(X_1P_1^0)+(X_2P_2^0)$ $=(0.458\times105.2)+(0.542\times46.8)$ $=48.18+25.37$ $=73.55\text{ kPa}$

Therefore, vapour pressure of the solution = 73.55 kPa

1.17 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.

1 molal solution means:

1 mole solute in 1000 g water

Moles of water: $=\frac{1000}{18}=55.56$

Mole fraction of water: $X_{water}=\frac{55.56}{55.56+1}$ $=0.9823$

Using Raoult’s law: $P=X_{water}P^0$ $=0.9823\times12.3$ $=12.08\text{ kPa}$

1.18 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%.

1.19 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.

1.20 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.

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

1.22 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?

1.23 Suggest the most important type of intermolecular attractive interaction in the following pairs.
(i) n-hexane and n-octane
(ii) I2 and CCl4
(iii) NaClO4 and water
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H6O).

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

1.25 Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol
(ii) toluene
(iv) ethylene glycol
(v) chloroform
(iii) formic acid
(vi) pentanol.

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

1.27 If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of CuS in aqueous solution.
1.28 Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.

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

1.30 Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.

1.31 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.

1.32 Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250g of water. Ka= 1.4 × 10–3, Kf = 1.86K kg mol–1.

1.33 19.5 g of CH2FCOOH 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 d issociation constant of fluoroacetic acid.
1.34 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. 1.35 Henry’s law constant for the molality of methane in benzene at 298 K is
4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

1.36 100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol1). 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