CBSE Class 12 Chemistry Solid State Worksheet Set 01

Important Questions for NCERT Class 12 Chemistry Solid State

Question. Gold has a face centred cubic lattice with an edge length of the unit cube of 407 pm. Assuming the closest packing, the diameter of the gold atom is
(a) 576.6 pm
(b) 287.8 pm 
(c) 352.5 pm
(d) 704.9 pm

Answer : B

Question. Which is not correct about the Schottky defects?
(a) Both cations and anions are missing from their lattice sites without affecting the stoichiometry of the compound
(b) Because of presence of holes the lattice energy decreases.
(c) The presence of holes causes the density of the crystal to decrease.
(d) The defect increases the electrical conductivity of the solid due to migration of the ions into the holes.

Answer : B

Question. Schottky defect defines imperfection in the lattice structure of 
(a) solid
(b) gas
(c) liquid
(d) plasma

Answer : A

Question. An AB2 type structure is found in
(a) NaCl
(b) CaF²
(c) Al²O³
(d) N²O

Answer : B

Question. An element (atomic mass 100 g/mol) having bcc structure has unit cell edge 400 pm. The density of element is (No. of atoms in bcc, Z = 2). 
(a) 2.144 g/cm³
(b) 7.289 g/cm³
(c) 5.188 g/cm³
(d) 10.376 g/cm³

Answer : C

Question. What is the coordination number of sodium in Na2O? 
(a) 6
(b) 4
(c) 8
(d) 2

Answer : B

Question. The crystal system of a compound with unit cell dimensions "a = 0.387, b = 0.387 and c = 0.504 nm and α = β = 90° and γ = 120°" is : 
(a) cubic
(b) hexagonal
(c) orthorhombic
(d) rhombohedral

Answer : B

Question. If z is the number of atoms in the unit cell that represents the closest packing sequence ..... ABC ABC ......, the number of tetrahedral voids in the unit cell is equal to :
(a) z
(b) 2z
(c) z/2
(d) z/4

Answer : B

Question. The compound, found in nature in gas phase but ionic in solid state is : 
(a) PCl₅
(b) CCl₄
(c) PCl₃
(d) POCl₃

Answer : A

Question. The Ca²⁺ and F– are located in CaF2 crystal, respectively at face centred cubic lattice points and in 
(a) Tetrahedral voids
(b) Half of tetrahedral voids
(c) Octahedral voids
(d) Half of octahedral voids

Answer : A

Question. The coordination number in hcp is 
(a) 6
(b) 12
(c) 18
(d) 24

Answer : B

Question. The space lattice of graphite is 
(a) Cubic
(b) Tetragonal
(c) Rhombic
(d) Hexagonal

Answer : D

Question. Coordination numbers of Zn²⁺ and S^2– in the crystal structure of wurtzite are 
(a) 4, 4
(b) 6, 6
(c) 8, 4
(d) 8, 8

Answer : A

Question. If AgI crystallises in zinc blende structure with I– ions at lattice points. What fraction of tetrahedral voids is occupied by Ag+ ions?
(a) 25%
(b) 50%
(c) 100%
(d) 75%

Answer : B

Question. Which set of following characteristics for ZnS crystal is correct? 
(a) Coordination number (4 : 4); ccp; Zn²⁺ ion in the alternate tetrahedral voids
(b) Coordination number (6 : 6); hcp; Zn²⁺ ion in all tetrahedral voids.
(c) Coordination number (6 : 4); hcp; Zn²⁺ ion in all octahedral voids
(d) Coordination number (4 : 4); ccp; Zn²⁺ ion in all tetrahedral voids.

Answer : A

Question. Molecules/ions and their magnetic properties are given below. 
Molecule/ion Magnetic property
(i) C6H6 (1) Antiferromagnetic
(ii) CrO2 (2) Ferrimagnetic
(iii) MnO (3) Ferromagnetic
(iv) Fe3O4 (4) Paramagnetic
(v) Fe3+ (5) Diamagnetic
The correctly matched pairs in the above is
(a) i-5, ii-3, iii-2, iv-1, v-4
(b) i-3, ii-5, iii-1, iv-4, v-2
(c) i-5, ii-3, iii-1, iv-2, v-4
(d) i-5, ii-3, iii-1, iv-4, v-2

Answer : C

Question. Which one of the following statements about packing in solids is incorrect ?
(a) Coordination number in bcc mode of packing is 8.
(b) Coordination number in hcp mode of packing is 12.
(c) Void space in hcp mode of packing is 32%.
(d) Void space is ccp mode of packing is 26%.

Answer : C

Question. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29Å. The radius of sodium atom is approximately : 
(a) 5.72Å
(b) 0.93Å
(c) 1.86Å
(d) 3.22Å

Answer : C

Question. The existence of a substance in more than one solid modifications is known as 
(a) isomorphism
(b) Polymorphism
(c) Amorphism
(d) Allotropy

Answer : B

Question. An element (atomic mass = 100 g / mol) having bcc structure has unit cell edge 400 pm. Then, density of the element is 
(a) 10.376 g/cm³
(b) 5.188 g/cm³
(c) 7.289 g/cm³
(d) 2.144 g/cm³

Answer : B

Question. A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids.
The formula of the compound is
(a)C₃A₂
(b)C₃A₄
(c)C₄A₃
(d)C₂A₃

Answer : B

Question. All the metallic elements like iron, copper; non-metallic elements like sulphur, iodine and compounds like NaCl, ZnS form
(a) amorphous solids
(b) crystalline solids
(c) polycrystalline solis
(d) Both (b) and (c)

Answer : B

Question. Some of the physical properties of crystalline solids like refractive index show different values on measuring along different directions in the same crystals. This property is called
(a) isotropy
(b) cleavage property
(c) anisotropy
(d) None of these

Answer : C

Question. Which primitive unit cell has unequal edge lengths (a≠b≠ c) and all axial angles different from 90°?
(a) Hexagonal
(b) Monoclinic
(c) Tetragonal
(d) Triclinic

Answer : D

Question. Identify the type of crystal system of the following
(A) KNO₃ ;  (B) CaCO₃ ;  (C) CaSO₄ ; (D) CuSO₄  5H₂O  
(a) A-Cubic; B-Triclinic; C-Hexagonal; D-Rhombohedral
(b) A-Tetragonal; B-Monoclinic; C-Triclinic; D-Hexagonal
(c) A-Orthorhombic; B-Trigonal; C-Tetragonal; D-Triclinic
(d) A-Rhombohedral; B-Hexagonal; C-Trigonal; D-Orthorhombic

Answer : C

Question. In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is
(a) A₂ B
(b) A B₂
(c) A₂ B₂
(d) A₂ B₅

Answer : D

Question. The lattice points of a crystal of hydrogen iodide are occupied by
(a) HI molecules
(b) H atoms and I atoms
(c) H⁺ cations and I⁻ anions
(d) H₂ molecules and I₂ molecules

Answer : A

Question. The number of octahedral void(s) per atom present at a cubic close packed structure is
(a) 1
(b) 3
(c) 2
(d) 4

Answer : A

Question. In a crystalline solid, having formula XY₂O₄ , oxide ions are arranged in cubic close packed lattice, while cations X are present in tetrahedral voids and cations Y are present in octahedral voids. The percentage of tetrahedral voids occupied by X is
(a) 12.5%
(b) 25%
(c) 50%
(d) 75%

Answer : A

Question. Which type of magnetic behaviour is exhibited by MgFe₂O₄ is ?
(a) Diamagnetic
(b) Paramagnetic
(c) Ferromagnetic
(d) Ferrimagnetic

Answer : D

Question. Element ‘B ’ forms ccp structure and ‘A ’ occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is
(a) A₂BO₄
(b) AB₂O₄
(c)  A₂B₂O
(d) A₄B₂

Answer : B

Question. Which of the following is correct order of packing efficiency?
(a) hcp = fcc > bcc > sc
(b) sc > bcc > hcp = fcc
(c) bcc > sc > hcp < fcc
(d) fcc = hcp > sc > bcc

Answer : A

Question. A metal has bcc structure and the edge length of its unit cell is 3.04 Å. The volume of the unit cell in cm³ will be
(a) 1.6 × 10⁻²¹ cm³
(b) 2.81 × 10⁻²³ cm³
(c) 6.02 × 10⁻²³ cm³
(d) 6.6 × 10⁻²⁴ cm³

Answer : B

Question. An atom forms face centred cubic crystal with density d = 8.92 g / mL and edge length a = 3.6 ×10⁻⁸ cm. The molecular mass of atom in amu is
(a) 98 amu
(b) 63 amu
(c) 32 amu
(d) 93 amu

Answer : B

Question. Sodium metal crystallises in a body centred cubic lattice with a unit cell edge of 4.29 A ° . The radius of sodium atom is approximately
(a) 1.86 Å
(b) 3.22 Å
(c) 5.72 Å
(d) 0.93 Å

Answer : A

Question. Iron exhibits bcc structure at room temperature.
Above 900°C , it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is
(a) 3√3/4√2
(b) 4√3/3√2
(c) √3/√2
(d) 1/2

Answer : A

Question. The cubic unit cell of Al (molar mass 27 g mol⁻¹ ) has an edge length of 405 pm. Its density is 2.7 g cm^–3 .
The cubic unit cell is
(a) face centred
(b) body centred
(c) primitive
(d) edge centred

Answer : A

Question. Which of the following compounds is likely to show both Frenkel and Schottky defects in its crystalline form?
(a) AgBr
(b) CsCl
(c) KBr
(d) ZnS

Answer : A

Question. Electrical conductivity of semiconductors increases with increase in
(a) temperature
(b) pressure
(c) volume
(d) None of these

Answer : A

Question. Which of the following is the ratio of packing density of fcc, bcc and simple cubic structures ?
(a) 0.92 : 0.70 : 1
(b) 0.70 : 0.92 : 1
(c) 1 : 0.92 : 0.70
(d) 1 : 0.70 : 0.92

Answer : C

Question. Which of the following has the highest value of energy gap?
(a) Aluminium
(b) Silver
(c) Germanium
(d) Diamond

Answer : D

Question. Which one is not a ferroelectric compound?
(a) KH₂ PO₄
(b) K₄ [Fe(CN)₆ ]  
(c) Rochelle salt
(d) BaTiO₃

Answer : B

1 MARK QUESTIONS

Question. What is molarity?
Answer: Molarity represents the total moles of solute that are dissolved per litre (or \( 1\text{ dm}^3 \)) of the final solution.

Question. What do you understand by saying that molality of a solution is 0.2?
Answer: This statement indicates that \( 0.2\text{ mol} \) of solute is present inside \( 1\text{ kg} \) of the respective solvent.

Question. Why is the vapour pressure of a liquid remains constant at constant temperature?
Answer: The rate of evaporation equals the rate of condensation when a system reaches dynamic equilibrium. Consequently, a liquid maintains a steady vapour pressure if the temperature remains unchanged.

Question. Define Azeotropes?
Answer: Solutions that boil at a constant temperature without any change in composition are referred to as azeotropes.

Question. Which substance is usually added into water in the car radiator to act as antifreeze?
Answer: Automobile radiators typically use water mixed with ethylene glycol to serve as an effective agent for lowering the freezing point.

Question. Which liquids form ideal solution?
Answer: Liquid components that possess comparable molecular configurations and similar polar characters tend to form ideal solutions.

Question. Which property of solution depend only upon the number of moles of solute dissolved and not on the nature of the solute?
Answer: These are known as colligative properties.

Question. Write one example each of solid in gas and liquid in gas solution?
Answer: An example of a solid-in-gas mixture is camphor dispersed in nitrogen gas (\( \text{N}_2 \)). For a liquid-in-gas mixture, a common example is chloroform combined with nitrogen gas.

Question. What is molal elevation constant or ebullioscopic constant?
Answer: The ebullioscopic constant, or molal elevation constant, represents the increase in boiling point observed when the concentration of the solution is exactly one molal.

Question. Define van’t Hoff factor.
Answer: The van't Hoff factor is defined as the observed value of a colligative property divided by its calculated theoretical value.

Question. Two liquids A and B boil at 120 C and 160 C respectively. Which of them has higher vapour pressure at 70 C?
Answer: A substance with a lower boiling temperature is more volatile. Thus, liquid A will exhibit a higher vapour pressure when measured at \( 70^\circ\text{C} \).

Question. What happens when blood cells are placed in pure water?
Answer: Solvent molecules travel into the blood cells via osmosis through their semi-permeable membranes. This causes the cells to expand and potentially rupture.

Question. What is the effect of temperature on the molality of a solution?
Answer: Temperature variations do not affect this concentration unit because mass remains independent of temperature.

Question. Write Henry’s law.
Answer: According to this law, the amount of gas dissolved in a liquid medium varies directly with its partial pressure above the solution at a specific temperature.

Question. What is an antifreeze?
Answer: This term describes any additive mixed with water to depress its freezing limit, such as ethylene glycol.

Question. Why cutting onions taken from the fridge is more comfortable than cutting onions lying at room temperature?
Answer: Lower temperatures result in a reduced vapour pressure, which decreases the release of the volatile irritants that cause crying.

Question. What will be the van’t Hoff factor for O.1 M ideal solution?
Answer: For an ideal solution, the van't Hoff factor (\( i \)) equals 1 since the solute particles neither associate nor dissociate in the mixture.

Question. What is the optimum concentration of fluoride ions for cleaning of tooth?
Answer: To safely clean teeth, the ideal concentration of fluoride ions is \( 1.5\text{ ppm} \). Concentrations exceeding this threshold can be toxic, whereas lower amounts are generally ineffective.

Question. What role does the molecular interaction play in the solution of alcohol and water?
Answer: The mixture exhibits a positive deviation from Raoult's law due to weaker intermolecular attractions between different molecules.

Question. Henry law constant for two gases are 21.5 and 49.5 atm, which gas is more soluble .
Answer: The Henry's law constant (\( K_H \)) shares an inverse relationship with the solubility of a gas in a liquid. Therefore, the gas with the lower \( K_H \) value (21.5 atm) is more soluble.

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Short Answer Type Questions (2 Marks)

Question. State Raoult’s law. Prove that it is a special case of Henry’ law?
Answer: Raoult’s law states that partial pressure of a volatile component of a solution is directly proportional to its mole fraction. It is a special case of Henry’ law because it becomes the same when \( K_H \) (Henry constant) is equal to pressure of pure solvent.

Question. List two conditions that ideal solutions must satisfy.
Answer:
a. \( \Delta H_{\text{mixing}} \) and \( \Delta V_{\text{mixing}} \) of ideal solutions should be zero.
b. They should obey Raoult’s law over the entire range of concentration.

Question. Explain ideal and non-ideal solutions with respect to intermolecular interactions in a binary solution of A and B.
Answer: For the given binary solution of A and B, it would be ideal if A-B interactions are equal to A-A and B-B interactions and it would be non-ideal if they are different to each other. The deviation from ideal behavior will be positive if A-B interactions are weaker as compared to A-A and B-B. The deviation will be negative if A-B interactions are stronger as compared to A-A and B-B.

Question.
a. What are minimum boiling and maximum boiling azeotropes?
b. Can azeotropes be separated by fractional distillation?
Answer:
(i) Minimum boiling azeotropes are the non-ideal solutions showing positive deviation while maximum boiling azeotropes are those which show negative deviation. Because of positive deviation their vapour pressures are comparatively higher and so they boil at lower temperatures while in case of negative deviation, the vapour pressures are lesser and so higher temperature are required for boiling them.
(ii) No, azeotropes can’t be separated by fractional distillation.

Question.
a. When a non-volatile solute is added to solvent,there is increase in boiling point of solution.Explain.
b. Define ebullioscopic constant and give its units.
Answer:
(i) When a non-volatile solute is added to a volatile solvent the vapour pressure of pure solvent decreases because a part of the surface is occupied by non-volatile solute which can’t volatilise. As a result, the vapour pressure of solution decreases and hence, the solution requires a comparatively higher temperature to boil causing an elevation of boiling point.
(ii) Ebullioscopic constant is defined as the elevation in boiling point of a solution of a non-volatile solute when its molality is unity. Its units are \( \text{K kg mol}^{-1} \).

Question. How did Van’t Hoff explain the abnormal molecular masses of electrolytes like KCl in water and non-electrolytes like benzoic acid in benzene.
Answer: The molecular mass of KCl in aqueous medium has been observed to be almost half than expected and it has been explained as dissociation of KCl into \( \text{K}^+ \) ions and \( \text{Cl}^- \) ions when actual no. of particles become double and so become the colligative properties but since molecular mass is always inversely proportional to colligative property it becomes almost half.
In case of benzoic acid in benzene, association of molecules take place when they dimerise and their no. becomes almost half and so molecular mass doubles as a result.

Question. When a pressure higher than the osmotic pressure is applied on the surface of the solution separated from a solvent by semi permeable membrane, what will happen?
Answer: Reverse osmosis will take place. We will observe the movement of solvent molecules from the solution to solvent phase and the level of solution will decrease.

Question. The freezing depression of 0.1M sodium chloride solution is nearly twice that of 0.1 M glucose solution. Explain?
Answer: Sodium chloride being ionic compound ionizes as \( \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \) in aqueous solution. The concentration of solute particles in this case becomes approximately 0.2 M which is twice the concentration of glucose solution. Consequently, freezing point depression of NaCl solution is also approximately twice that of glucose solution.

Question. The depression in freezing point is a colligative property. Explain.
Answer: The freezing point depression depends upon the molal concentration of the solute and does not depend upon the nature of the solute. It is therefore, a colligative property.

Question. Equimolar solution of glucose and Common salt are not isotonic. Why?
Answer: Glucose is a non electrolyte, when added to water it do not break up into ions whereas Common salt is an electrolyte when added to water it breaks up to give Sodium and chloride ions. The number of particles in solution of Common salt are nearly double the number of particles in the solution of glucose so the osmotic pressure of common salt solution is nearly twice that of Glucose solution.

Questions For 3 Marks

Question. A 5% solution of sucrose \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \) is isotonic with 3% solution of an unknown organic substance. Calculate the molecular mass of unknown substance.
Answer:
\( \pi_1 = \pi_2 \)
\( C_1 RT = C_2 RT \)
\( \frac{n_1}{V} = \frac{n_2}{V} \)
\( \Rightarrow \frac{W_1}{m_1} = \frac{W_2}{m_2} \)
\( \Rightarrow \frac{5}{342} = \frac{3}{m_2} \)
\( \dots m_2 = \frac{3 \times 342}{5} = 205.2 \)

Question. A solution of Barium Chloride is prepared by dissolving 3.100 g of it in 250 g of water. The solution boils at \( 100.083^\circ\text{C} \). Calculate the Van’t Hoff factor and Molality of this solution. (\( K_b \) for water = \( 0.52\text{ K kg mol}^{-1} \), Molar mass of \( \text{BaCl}_2 = 208.3\text{ g mol}^{-1} \))
Answer:
Observed \( \Delta T_b = 100.083 - 100 = 0.083^\circ\text{C} \)
Molality of the solution \( = \frac{W_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent}}} \)
\( \text{Molality} = \frac{3.1 \times 1000}{208.3 \times 250} = 0.0595\text{ m} \)
Calculated (Normal) \( \Delta T_b = K_b \times m = 0.52 \times 0.0595 = 0.03095\text{ K} \)
\( i = \frac{\text{Observed } \Delta T_b}{\text{Normal } \Delta T_b} = \frac{0.083}{0.03095} = 2.68 \)

Question. Why semi permeable membrane is so important in the phenomenon of osmosis? What are isotonic, hypo tonic and hyper tonic solutions? Does osmosis take place in all three types of solutions?
Answer: The semi permeable membrane is very important in the phenomenon of osmosis because they only permit the movement of solvent molecules through them.
Two solutions having similar osmotic pressure at a given temperature are called isotonic solutions. If the given solution has less osmotic pressure it is called hypo tonic and it is hyper tonic if its osmotic pressure is higher than the solution on the other side of semi permeable membrane. Osmosis takes place only in hypo tonic and hypertonic solutions.

Question. Which will have more osmotic pressure and why? Solution prepared by dissolving 6g/L of \( \text{CH}_3\text{COOH} \) or Solution prepared by dissolving 7.45g/L of \( \text{KCl} \)
Answer:
Moles of \( \text{CH}_3\text{COOH} = \frac{\text{Mass}}{\text{Mol. Wt.}} = \frac{6}{60} = 0.1 \)
Moles of \( \text{KCl} = \frac{\text{Mass}}{\text{Mol. Wt.}} = \frac{7.45}{74.5} = 0.1 \)
Molar concentration of both the solutions is same.
\( \text{KCl} \) ionizes into \( \text{K}^+ \) and \( \text{Cl}^- \) whereas \( \text{CH}_3\text{COOH} \) does not ionize.
Osmotic pressure is a colligative property.
Its value depends on number of particles.
Since, \( \text{KCl} \) produces more ions so, osmotic pressure of \( \text{KCl} \) will be more than that of \( \text{CH}_3\text{COOH} \).

Question. What is Bends? If a diver had the "bends", describe how this can be treated.
Answer: Scuba divers cylinder has a mixture of helium, nitrogen and oxygen, as they go down at high pressure which increases the solubility of these gases in the blood when they come up pressure decreases and nitrogen is released as the solubility decreases and the bubbles of nitrogen gas can block capillaries causing condition called bends which is both painful and dangerous. In order to avoid formation of bends in blood the divers are subjected to decompression chambers where pressure is lowered down gradually releasing the gas from blood slowly.

Question. At 300 K, 18g of glucose present per litre of its solution has an osmotic pressure of 4.98 bars. If the osmotic pressure of solution is 1.52 bars at the same temperature, what would be its concentration?
Answer:
For solution A:
\( 4.98 \times 1\text{ L} = \frac{18}{180} \times R \times T \)
For solution B:
\( 1.52 \times 1\text{ L} = n \times R \times T \)
\( \frac{1.52 \times 1}{n} = \frac{4.98 \times 1}{0.1} \)
\( \frac{1.52}{n} = \frac{4.98}{0.1} \)
\( n = \frac{1.52 \times 0.1}{4.98} = 0.035\text{ moles} \)
\( c = 0.035\text{ mol L}^{-1} \)

Question. The freezing depression of 0.1M sodium chloride solution is nearly twice that of 0.1 M glucose solution. Explain.
Answer: Sodium chloride being ionic compound ionizes as \( (\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-) \) in aqueous solution. The concentration of solute particles in this case becomes approximately 0.2 M which is twice the concentration of glucose solution. Consequently, freezing point depression of NaCl solution is also approximately twice that of glucose solution.

Question. Calculate the amount of NaCl must be added to 1000 ml of water so as to reduce its freezing point by two Kelvin. For water \( K_f = 1.86\text{ K kg mol}^{-1} \), given that the density of water is \( 1.0\text{ g ml}^{-1} \) and NaCl is completely dissociated.
Answer:
Mass of water = density \(\times\) volume = \( 1 \times 1000 = 1000\text{ g} = 1\text{ kg} \).
\( \Delta T_f = i K_f m \)
\( 2 = \left( 2 \times 1.86 \times \frac{z}{58.5 \times 1000} \right) \times 100 \)
\( z = \frac{58.5}{1.86} = 31.45\text{ g} \)

Question. Predict the Boiling point of solution prepared by dissolving 25.0g of urea and 25.0 g of thiourea in 100 gram of water. Given for water \( K_b = 0.52\text{ K kg mol}^{-1} \) and Boiling point of pure water is 373.15 K.
Answer:
No of moles of urea = \( \frac{\text{mass of urea}}{\text{molar mass of urea}} = \frac{25}{60} = 0.42 \)
No of moles of thiourea = \( \frac{\text{mass of thiourea}}{\text{molar mass of thiourea}} = \frac{25}{76} = 0.33 \)
molality of solution = \( \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)
molality of solution = \( \left( \frac{\text{moles of solute}}{\text{mass of solvent in g}} \right) \times 1000 \)
molality of solution = \( \frac{0.33 + 0.42}{100} \times 1000 = 1.50\text{ m} \)
\( \Delta T_b = 0.52 \times 1.50 = 5.44\text{ K} \)
\( T_b \text{(solution)} - T_b \text{(solvent)} = 5.44\text{ K} \)
\( T_b \text{(solution)} = 5.44\text{ K} + 373.15 = 378.59\text{ K} \)

Question. Predict the Boiling point of solution prepared by dissolving 3.42g of sugarcane in 100 gram of water. Given for water \( K_b = 0.52\text{ K kg mol}^{-1} \) and Boiling point of pure water is 373.15 K.
Answer:
\( \Delta T_b = K_b \times m \)
\( \Delta T_b = \left( 0.52 \times \frac{3.42\text{ g}}{342\text{ g mol}^{-1} \times 100} \right) \times 1000 \)
\( \Delta T_b = \frac{0.052 \times 100}{1000} = 0.0052\text{ K} \)
\( T_b \text{(solution)} - T_b \text{(solvent)} = 0.0052\text{ K} \)
\( T_b \text{(solution)} = 0.0052\text{ K} + 373.15 = 373.1552\text{ K} \)

Questions For 5 Marks

Question. (a) Difference between molarity and molality for a solution. How does a change in temperature influence their values?
(b) Calculate the freezing point of an aqueous solution containing 10.50 g of \( \text{MgBr}_2 \) in 200 g of water. (Molar mass of \( \text{MgBr}_2 = 184\text{ g} \)) (\( K_f \) for water = \( 1.86\text{ K kg mol}^{-1} \))
Answer:
(a) Molarity is defined as the number of moles of solute dissolved per litre of solution.
Mathematically, \( M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in litres (dm}^3)} \)
Molality of a solution is defined as the number of moles of solute dissolved in 1000 grams of solvent.
Mathematically, \( m = \frac{\text{Number of moles of the solute}}{\text{Mass of solvent in kg}} \)
While molarity decreases with an increase in temperature, molality is independent of temperature. This happens because molality involves mass, which does not change with a change in temperature, while molarity involves volume, which is temperature dependent.
(b) Given:
\( w_2 = 10.50\text{ g} \)
\( w_1 = 200\text{ g} \)
Molar mass of \( \text{MgBr}_2 \) (\( M_2 \)) = \( 184\text{ g mol}^{-1} \)
Using the formula:
\( \Delta T_f = \frac{1000 \times K_f \times w_2}{w_1 \times M_2} \)
\( \Delta T_f = \frac{1000 \times 1.86 \times 10.50}{200 \times 184} = 0.53\text{ K} \)
Now, \( T_f = T_o - \Delta T_f \)
\( T_f = 273 - 0.53 = 272.47\text{ K} \)

Question. (a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain.
(b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water. (\( K_b \) for water = \( 0.512\text{ K kg mol}^{-1} \)), Molar mass of NaCl = \( 58.44\text{ g} \).
Answer:
(a) Osmosis: The process of flow of solvent molecules from pure solvent to solution or from solution of lower concentration to solution of higher concentration through a semi-permeable membrane is called osmosis.
Osmotic pressure: The pressure required to just stop the flow of solvent due to osmosis is called osmotic pressure (\( \pi \)) of the solution.
Yes, the osmotic pressure of a solution is a colligative property. The osmotic pressure is expressed as:
\( \pi = \frac{n}{V} R T \)
Where, \( \pi \) = osmotic pressure, \( n \) = number of moles of solute, \( V \) = volume of solution, \( T \) = temperature.
From the equation, it is clear that osmotic pressure depends upon the number of moles of solute 'n' irrespective of the nature of the solute. Hence, osmotic pressure is a colligative property.
(b) Given:
\( K_b = 0.512\text{ K kg mol}^{-1} \)
\( w_2 = 15.00\text{ g} \)
\( w_1 = 250.0\text{ g} \)
\( M_2 = 58.44\text{ g mol}^{-1} \)
Using the formula:
\( \Delta T_b = \frac{1000 \times K_b \times w_2}{w_1 \times M_2} \)
\( \Delta T_b = \frac{1000 \times 0.512 \times 15.00}{250.0 \times 58.44} = 0.52\text{ K} \)
Now, \( T_b = T_o + \Delta T_b \)
\( T_b = 373 + 0.53 = 373.53\text{ K} \)

Question. A 1.00 molal aqueous solution of trichloroacetic acid (\( \text{CCl}_3\text{COOH} \)) is heated to its boiling point. The solution has the boiling point of \( 100.18^\circ\text{C} \). Determine the van't Hoff factor for trichloroacetic acid. (\( K_b \) for water = \( 0.512\text{ K kg mol}^{-1} \))
Answer:
\( \Delta T = 100.18 - 100 = 0.18^\circ\text{C} = 0.18\text{ K} \)
Now \( i = \frac{\Delta T}{K_b \times m} \)
\( i = \frac{0.18\text{ K}}{(0.512\text{ K kg mol}^{-1})(1.00\text{ mol kg}^{-1})} = 0.35 \)

Question. Define the following terms:
(i) Mole fraction
(ii) Isotonic solutions
(iii) Van't Hoff factor
(iv) Ideal solution
Answer:
(i) Mole fraction: It is defined as the ratio of moles of a constituent to the total number of moles of the solution.
(ii) Isotonic solutions: Solutions which have the same osmotic pressure are called isotonic solution.
(iii) Van't Hoff factor: It is the ratio of experimental values of a colligative property to the calculated value of the property when the solution behaves ideally.
(iv) Ideal solution: A solution that obeys Raoult’s law over all ranges of temperature and concentration and shows no internal energy change on mixing and no attractive force between components.

Question. A solution is prepared by dissolving 30g of non-volatile non-electrolyte solute in 90g water. The vapour pressure of solution was 2.8 kPa at 298K. When 18g of water was further added to it, the vapour pressure became 2.9 kPa at 298K. Calculate molar mass of solute.
Answer:
According to Raoult's law:
\( P_A = P_A^\circ X_A \)
Here, A = water, and let \( n \) be the moles of solute.
Initially, mass of water = 90 g.
Moles of water = \( \frac{90}{18} = 5 \)
\( 2.8 = P_A^\circ \frac{5}{5 + n} \) ......(i)
When 18 g of water is further added, total mass of water = 90 + 18 = 108 g.
New moles of water = \( \frac{108}{18} = 6 \)
\( 2.9 = P_A^\circ \frac{6}{6 + n} \) ......(ii)
On dividing equation (i) by (ii), we get:
\( \frac{2.8}{2.9} = \frac{5(6 + n)}{6(5 + n)} \)
\( \frac{28}{29} = \frac{30 + 5n}{30 + 6n} \)
\( 28(30 + 6n) = 29(30 + 5n) \)
\( 840 + 168n = 870 + 145n \)
\( 23n = 30 \)
\( n = \frac{30}{23} \)
Since, \( n = \frac{\text{Mass of solute}}{\text{Molar mass of solute}} \)
\( \frac{30}{23} = \frac{30}{M} \)
\( M = 23\text{ g mol}^{-1} \)
Therefore, molar mass of solute is \( 23\text{ g mol}^{-1} \).

Question. a. Define the following terms:
i. Mole fraction
ii. Van't Hoff factor
b. 100 mg of a protein is dissolved in enough water to make 10.0 mL of a solution. If this solution has an osmotic pressure of 13.3 mmHg at \( 25^\circ\text{C} \), what is the molar mass of protein? (\( R = 0.0821\text{ L atm mol}^{-1}\text{ K}^{-1} \) and \( 760\text{ mmHg} = 1\text{ atm} \))
Answer:
a. i. Mole fraction: Mole fraction of a component is the ratio of number of moles of the component to the total number of moles of all the components.
ii. Van't Hoff factor: Van't Hoff factor is the ratio of normal molar mass to the abnormal molar mass (or the ratio of observed value of colligative property to calculated value of colligative property assuming no association or dissociation).
b. Calculation:
Given:
Mass of protein (\( w_2 \)) = \( 100\text{ mg} = 0.1\text{ g} \)
Volume of solution (\( V \)) = \( 10.0\text{ mL} = 0.01\text{ L} \)
Osmotic pressure (\( \pi \)) = \( 13.3\text{ mmHg} = \frac{13.3}{760}\text{ atm} \)
Temperature (\( T \)) = \( 25^\circ\text{C} = 298\text{ K} \)
Using formula:
\( \pi V = n R T \)
\( \pi V = \frac{w_2}{M} R T \)
\( \frac{13.3}{760} \times 0.01 = \frac{0.1}{M} \times 0.0821 \times 298 \)
\( M = \frac{0.1 \times 0.0821 \times 298 \times 760}{13.3 \times 0.01} = 13980\text{ g mol}^{-1} \) (or \( 13.98\text{ kg mol}^{-1} \))

Question. a. What is meant by:
i. Colligative properties
ii. Molality of a solution
b. What concentration of nitrogen should be present in a glass of water at room temperature? Assume a temperature of \( 25^\circ\text{C} \), a total pressure of 1 atmosphere and mole fraction of nitrogen in air of 0.78. (\( K_H \text{ for nitrogen} = 8.42 \times 10^{-7}\text{ M/mmHg} \))
Answer:
a. i. Colligative properties: All the properties which depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution are known as colligative properties.
ii. Molality of a solution: Molality of solution is the number of moles of solute present in 1 kilogram of solvent.
b. Calculation:
Partial pressure of Nitrogen, \( P_{\text{N}_2} = X_{\text{N}_2} \times P_{\text{total}} = 0.78 \times 760\text{ mmHg} = 592.8\text{ mmHg} \)
Given \( K_H = 8.42 \times 10^{-7}\text{ M/mmHg} \)
Since \( K_H \) is given in M/mmHg, concentration of nitrogen is:
\( C_{\text{N}_2} = K_H \times P_{\text{N}_2} \)
\( C_{\text{N}_2} = 8.42 \times 10^{-7}\text{ M/mmHg} \times 592.8\text{ mmHg} = 4.99 \times 10^{-4}\text{ M} \)
Alternatively, as per the calculation on the worksheet:
\( n_{\text{H}_2\text{O}} = \frac{1000}{18} = 55.5\text{ mol} \)
Since \( n_{\text{N}_2} \ll n_{\text{H}_2\text{O}} \), \( X_{\text{N}_2} = \frac{n_{\text{N}_2}}{n_{\text{H}_2\text{O}}} \)
\( n_{\text{N}_2} = 4.99 \times 10^{-4} \times 55.5 = 276.9 \times 10^{-4}\text{ M} \)

Question. (a) A weak electrolyte AB is 5% dissociated in aqueous solution. What is the freezing point of a 0.100 molal aqueous solution of AB? For water \( K_f = 1.86\text{ K kg mol}^{-1} \).
(b) 0.02 molal solution of acetic acid is 3% dissociated at \( 25^\circ\text{C} \). Calculate the osmotic pressure of the solution.
Answer:
(a)
For dissociation of AB:
\( \text{AB} \rightarrow \text{B}^- + \text{A}^+ \)
Initially: \( 1 \quad 0 \quad 0 \)
At equilibrium: \( 1-\alpha \quad \alpha \quad \alpha \)
Here, \( \alpha = \frac{5}{100} = 0.05 \)
Total particles at equilibrium (\( i \)) = \( 1 + \alpha = 1.05 \)
Effective molality = \( m \times (1 + \alpha) = 0.100 \times 1.05 = 0.105\text{ m} \)
\( \Delta T_f = K_f \times \text{Effective molality} = 1.86 \times 0.105 = 0.1953\text{ K} \)
\( T_f \text{(solution)} = 273.15\text{ K} - 0.1953\text{ K} = 272.95\text{ K} \)

(b)
For dissociation of \( \text{CH}_3\text{COOH} \):
\( \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}^+ \)
Initially: \( 1 \quad 0 \quad 0 \)
At equilibrium: \( 1-\alpha \quad \alpha \quad \alpha \)
\( i = 1 + \alpha = 1 + 0.03 = 1.03 \) (Note: worksheet calculates \( i = 1.003 \))
Using formula:
\( \pi = i C R T \)
\( \pi = 1.003 \times 0.0821 \times 300 \times 0.02 = 0.494\text{ atm} \)