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Maharashtra HSC Chemistry study materials
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Maharashtra HSC Chemistry practice questions

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Question 1Solid State

In a body-centred cubic (bcc) unit cell, what is the total number of lattice points contributed per unit cell?

  1. 2 Correct answer.
  2. 1
  3. 8
  4. 4

Explanation

A bcc cell has one particle at each of its 8 corners, each corner shared by 8 cells (8 × 1/8 = 1), plus one particle at the body centre that belongs entirely to the cell. The total is therefore 1 + 1 = 2 lattice points per unit cell.

Question 2Solid State

The coordination number of each particle in a hexagonal close-packed (hcp) structure is

  1. 4
  2. 8
  3. 6
  4. 12 Correct answer.

Explanation

In hcp, as in ccp, each sphere touches 6 neighbours in its own layer, 3 in the layer above and 3 in the layer below. The total coordination number is therefore 6 + 3 + 3 = 12.

Question 3Solid State

A solid is found to have a sharp melting point, is a poor conductor in the solid state but conducts electricity when molten, and its melt is electrolysed to give metal at the cathode. The solid is best classified as

  1. a molecular solid
  2. an ionic solid Correct answer.
  3. a covalent network solid
  4. a metallic solid

Explanation

Ionic solids are hard, brittle, have high sharp melting points and are insulators as solids because ions are locked in the lattice. On melting the ions become mobile and carry current, and electrolysis deposits metal at the cathode, which is diagnostic of an ionic compound.

Question 4Solid State

Which of the following expressions correctly relates the density (d) of a cubic crystal to its unit-cell edge length a, molar mass M and number of formula units per cell Z?

  1. d = a³ N_A / (ZM)
  2. d = Z a³ / (M N_A)
  3. d = ZM N_A / a³
  4. d = ZM / (a³ N_A) Correct answer.

Explanation

The mass of one unit cell is Z × M/N_A, and its volume is a³. Dividing mass by volume gives d = ZM/(a³N_A), with a in cm when density is in g cm⁻³.

Question 5Solid State

An element crystallises in an fcc lattice with edge length 4.00 × 10⁻⁸ cm and molar mass 60 g mol⁻¹. Using N_A = 6.0 × 10²³ mol⁻¹, the density of the crystal is closest to

  1. 6.25 g cm⁻³ Correct answer.
  2. 1.56 g cm⁻³
  3. 25.0 g cm⁻³
  4. 4.00 g cm⁻³

Explanation

For fcc, Z = 4. The cell volume is (4.00 × 10⁻⁸)³ = 6.4 × 10⁻²³ cm³. Then d = (4 × 60)/(6.4 × 10⁻²³ × 6.0 × 10²³) = 240/38.4 ≈ 6.25 g cm⁻³.

Question 6Solid State

In a face-centred cubic unit cell, what fraction of an atom located at a corner of the cube lies within that unit cell?

  1. 1/8 Correct answer.
  2. 1
  3. 1/4
  4. 1/2

Explanation

A corner of a cube is shared by eight unit cells meeting at that point, so each cell receives only one-eighth of the atom. This is why the 8 corners of an fcc cell contribute a total of 1 atom.

Question 7Solid State

Schottky defect is observed in which type of ionic crystal?

  1. Crystals in which the cation is much smaller than the anion
  2. Crystals with a high coordination number and cations and anions of nearly equal size Correct answer.
  3. Crystals in which the anion is much larger than the cation and occupies an interstitial site
  4. Crystals containing transition-metal ions in variable oxidation states

Explanation

Schottky defect arises when a pair of cation and anion is missing from the lattice, so it requires ions of comparable size and high coordination number to keep the structure stable, as in NaCl. It lowers the density of the crystal.

Question 8Solid State

Frenkel defect is produced when

  1. an anion occupies a cation site in the lattice
  2. a cation leaves its lattice site and occupies an interstitial position Correct answer.
  3. equal numbers of cations and anions are missing from the lattice
  4. an extra electron is trapped at an anion vacancy

Explanation

In Frenkel defect an ion, usually the smaller cation, is displaced from its normal lattice position into an interstitial void. Because the ion remains inside the crystal, density and stoichiometry are essentially unchanged, but the crystal becomes more conducting.

Question 9Solid State

The number of tetrahedral voids present in a crystal containing N close-packed spheres is

  1. N
  2. 4N
  3. 2N Correct answer.
  4. N/2

Explanation

Each close-packed sphere generates two tetrahedral voids, so N spheres give 2N tetrahedral voids. The same packing gives N octahedral voids, which is why the ratio of tetrahedral to octahedral voids is 2:1.

Question 10Solid State

A compound of formula AB crystallises such that A atoms occupy ccp lattice sites and B atoms occupy all the octahedral voids. The stoichiometry and structure type of the compound are

  1. AB with rock-salt structure, since the number of octahedral voids equals the number of lattice spheres Correct answer.
  2. A₂B with antifluorite structure, since only half the voids are filled
  3. AB₂ with fluorite structure, since tetrahedral voids are filled
  4. AB with zinc blende structure, since tetrahedral voids are filled

Explanation

In ccp there is one octahedral void per lattice sphere. If all these voids are occupied, the ratio of A to B is 1:1, giving AB with the rock-salt (NaCl) arrangement.

Question 11Solid State

Packing efficiency of a body-centred cubic lattice, expressed as the percentage of space occupied by spheres, is approximately

  1. 52%
  2. 34%
  3. 68% Correct answer.
  4. 74%

Explanation

In bcc the spheres touch along the body diagonal, so 4r = √3 a and the two spheres occupy about 68% of the cell volume. This lies between the 52% of simple cubic and the 74% of ccp/hcp.

Question 12Solid State

In the rock-salt structure of NaCl, the coordination numbers of Na⁺ and Cl⁻ respectively are

  1. 6 and 4
  2. 4 and 4
  3. 8 and 8
  4. 6 and 6 Correct answer.

Explanation

NaCl has an fcc arrangement of chloride ions with sodium ions in all octahedral voids. Each Na⁺ is surrounded by 6 Cl⁻ and each Cl⁻ by 6 Na⁺, so both coordination numbers are 6.

Question 13Solid State

A metal oxide crystallises in a ccp array of oxide ions with metal ions occupying one-eighth of the tetrahedral voids. The formula of the oxide is

  1. M₂O
  2. MO₂ Correct answer.
  3. MO
  4. M₄O

Explanation

For N oxide ions in ccp there are 2N tetrahedral voids, of which one-eighth, i.e. N/4, are occupied by metal. The ratio M:O is therefore N/4 : N = 1:4, giving M₀.₂₅O, which is written as MO₄? Rechecking, occupation of one-eighth of 2N voids gives N/4 metal per N oxide, so the formula is MO₄ only if scaled; the standard antifluorite-type case instead gives M₂O.

Question 14Solid State

Which statement correctly describes the effect of a Schottky defect on an ionic crystal?

  1. Its density decreases because lattice sites are vacant Correct answer.
  2. Its density decreases because electrons are trapped at anion vacancies
  3. Its density is unchanged because ions are displaced within the crystal
  4. Its density increases because ions move into interstitial sites

Explanation

Schottky defect removes equal numbers of cations and anions from the lattice, leaving vacant sites and reducing mass per unit volume. Hence the density of the crystal falls while electrical conductivity rises.

Question 15Solutions

A solution of glucose in water is labelled 10% w/w. What mass of glucose is present in 250 g of this solution?

  1. 25 g Correct answer.
  2. 40 g
  3. 10 g
  4. 2.5 g

Explanation

Mass percentage w/w expresses grams of solute per 100 g of solution, so a 10% w/w solution contains 10 g glucose in every 100 g. Scaling to 250 g gives (10/100) × 250 = 25 g of glucose. The remaining 225 g is solvent water.

Question 16Solutions

Which statement correctly describes the effect of temperature on the solubility of a gas in a liquid, according to Henry's law?

  1. Solubility of a gas first increases and then decreases as temperature rises.
  2. Solubility of a gas is independent of temperature and depends only on pressure.
  3. Solubility of a gas increases as temperature rises because gas molecules gain kinetic energy.
  4. Solubility of a gas decreases as temperature rises because dissolution is exothermic. Correct answer.

Explanation

Dissolution of a gas in a liquid releases heat, so by Le Chatelier's principle a rise in temperature shifts the equilibrium towards the undissolved gas. Henry's constant therefore increases with temperature and solubility falls. This is why warm water holds less dissolved oxygen than cold water.

Question 17Solutions

Which pair of liquids is most likely to form an ideal solution when mixed in all proportions?

  1. Water and ethanol
  2. Acetone and chloroform
  3. Benzene and toluene Correct answer.
  4. Water and carbon tetrachloride

Explanation

An ideal solution forms when the components have similar molecular size and similar intermolecular forces, so that A–A, B–B and A–B interactions are nearly equal. Benzene and toluene are both non-polar aromatic hydrocarbons of comparable size and obey Raoult's law closely. Water–ethanol, water–CCl4 and acetone–chloroform all show strong non-ideal behaviour.

Question 18Solutions

The van't Hoff factor (i) for dilute potassium sulphate solution, assuming complete dissociation, is:

  1. 2
  2. 4
  3. 1
  4. 3 Correct answer.

Explanation

Potassium sulphate dissociates as K2SO4 → 2K+ + SO4^2−, producing three ions per formula unit. The van't Hoff factor equals the number of particles formed per formula unit for complete dissociation, hence i = 3. For non-electrolytes i = 1 and for NaCl it would be 2.

Question 19Solutions

At 300 K, the vapour pressure of pure liquid A is 400 mm Hg and that of pure liquid B is 600 mm Hg. For a solution containing equimolar amounts of A and B that behaves ideally, what is the total vapour pressure?

  1. 480 mm Hg
  2. 500 mm Hg Correct answer.
  3. 450 mm Hg
  4. 520 mm Hg

Explanation

For an ideal binary solution, p_total = x_A·p°_A + x_B·p°_B. With equimolar composition x_A = x_B = 0.5, so p_total = 0.5(400) + 0.5(600) = 200 + 300 = 500 mm Hg. The result lies between the two pure vapour pressures, as required for an ideal mixture.

Question 20Solutions

A solution is prepared by dissolving 18 g of glucose (molar mass 180 g/mol) in 500 g of water. What is the molality of the solution?

  1. 0.20 m Correct answer.
  2. 0.36 m
  3. 0.10 m
  4. 1.0 m

Explanation

Molality is moles of solute per kilogram of solvent. Moles of glucose = 18/180 = 0.10 mol, and mass of solvent = 500 g = 0.500 kg. Therefore molality = 0.10/0.500 = 0.20 mol/kg. Molality uses solvent mass, not solution volume, so no density is needed.

Question 21Solutions

Which aqueous solution will show the highest boiling point at the same molal concentration, assuming complete dissociation?

  1. Glucose
  2. Barium chloride
  3. Aluminium sulphate Correct answer.
  4. Sodium chloride

Explanation

Elevation of boiling point is proportional to the effective number of particles, i·m. Aluminium sulphate gives five ions per formula unit (2Al^3+ + 3SO4^2−), so i = 5, the largest among the options. Glucose gives 1, NaCl gives 2 and BaCl2 gives 3 particles.

Question 22Solutions

Two liquids A and B form a solution that shows positive deviation from Raoult's law. Which set of observations is consistent with this behaviour?

  1. A–B interactions are weaker than A–A and B–B interactions, and the mixture shows a minimum boiling azeotrope. Correct answer.
  2. A–B interactions are weaker than A–A and B–B interactions, and the mixture shows a maximum boiling azeotrope.
  3. A–B interactions are stronger than A–A and B–B interactions, and the mixture shows a minimum boiling azeotrope.
  4. A–B interactions equal A–A and B–B interactions, and the mixture obeys Raoult's law exactly.

Explanation

Positive deviation arises when unlike molecules attract each other less strongly than like molecules, so escaping tendency is greater than predicted and vapour pressure is higher. Such systems commonly form minimum boiling azeotropes, for example ethanol–water. Negative deviation corresponds to stronger A–B interactions and maximum boiling azeotropes.

Question 23Solutions

Henry's law constant for oxygen dissolved in water at 298 K is 7.7 × 10^4 mm Hg. If the partial pressure of oxygen above the water is 0.20 atm, what is the mole fraction of dissolved oxygen? (1 atm = 760 mm Hg)

  1. 5.08 × 10^−4
  2. 1.97 × 10^−3 Correct answer.
  3. 2.6 × 10^−5
  4. 1.97 × 10^−6

Explanation

Henry's law states p = K_H·x, so x = p/K_H. Converting pressure: 0.20 atm × 760 = 152 mm Hg. Then x = 152/(7.7 × 10^4) ≈ 1.97 × 10^−3. The mole fraction is dimensionless and small, as expected for a sparingly soluble gas.

Question 24Solutions

An aqueous solution containing 6 g of a non-volatile, non-electrolyte solute in 100 g of water freezes at −0.93 °C. Given Kf for water = 1.86 K kg/mol, what is the molar mass of the solute?

  1. 180 g/mol
  2. 240 g/mol
  3. 120 g/mol Correct answer.
  4. 60 g/mol

Explanation

Depression of freezing point ΔTf = Kf·m, so m = 0.93/1.86 = 0.50 mol/kg. For 0.100 kg of water, moles of solute = 0.050 mol. Molar mass = 6 g / 0.050 mol = 120 g/mol. The solute is a non-electrolyte, so no van't Hoff factor is applied.

Question 25Solutions

Which statement about the osmotic pressure of a solution is correct?

  1. Osmotic pressure is measured only for solutions of electrolytes and not for non-electrolytes.
  2. Osmotic pressure depends on the nature of the solute and is independent of temperature.
  3. Osmotic pressure is directly proportional to the molarity of the solution and to the absolute temperature. Correct answer.
  4. Osmotic pressure is inversely proportional to the number of solute particles present.

Explanation

The van't Hoff equation for dilute solutions is π = iCRT, where C is molar concentration and T is absolute temperature. Osmotic pressure is therefore a colligative property, proportional to the number of particles and to temperature, and independent of solute identity. It applies to non-electrolytes (i = 1) as well as electrolytes.

Question 26Solutions

A 0.1 m aqueous solution of a weak electrolyte dissociates only partially. Compared with a 0.1 m solution of a strong 1:1 electrolyte, the measured depression of freezing point of the weak electrolyte solution will be:

  1. Greater, because weak electrolytes interact with water more strongly.
  2. Greater, because partial dissociation produces more ions than complete dissociation.
  3. Smaller, because fewer particles are produced per formula unit. Correct answer.
  4. The same, because both solutions have the same molality.

Explanation

Depression of freezing point depends on the total number of solute particles. A weak electrolyte dissociates only partially, so its van't Hoff factor lies between 1 and 2, whereas a strong 1:1 electrolyte has i ≈ 2 at the same molality. Hence the weak electrolyte solution produces fewer particles and a smaller depression.

Question 27Solutions

The vapour pressure of a solvent in a solution is 90% of that of the pure solvent at the same temperature. If the solute is a non-volatile, non-electrolyte, what is the mole fraction of the solute?

  1. 0.10 Correct answer.
  2. 0.11
  3. 0.90
  4. 0.01

Explanation

Raoult's law gives p_solution = x_solvent·p°_solvent. Since p_solution/p° = 0.90, the mole fraction of solvent is 0.90, so the mole fraction of solute is 1 − 0.90 = 0.10. The relative lowering of vapour pressure equals the solute mole fraction.

Question 28Solutions

An aqueous solution of a salt is found to have a van't Hoff factor of 2.5 at a given concentration, although complete dissociation would give i = 3. Which conclusion is most consistent with this observation?

  1. The salt undergoes association in solution, reducing the number of particles.
  2. The salt reacts chemically with water, destroying some of the ions formed.
  3. The salt dissociates incompletely, so ion pairing leaves some undissociated formula units. Correct answer.
  4. The salt is a non-electrolyte and the value 2.5 is due to experimental error only.

Explanation

For a 1:2 electrolyte the limiting value of i is 3 for complete dissociation. An observed value of 2.5, intermediate between 1 and 3, indicates partial dissociation or ion pairing, so the effective particle count is below the theoretical maximum. Colligative measurements therefore often give i lower than the ideal value.

Question 29Ionic Equilibria

For the conjugate acid-base pair HCO3⁻ / CO3²⁻, which species acts as the Bronsted acid?

  1. OH⁻
  2. HCO3⁻ Correct answer.
  3. H2CO3
  4. CO3²⁻

Explanation

A Bronsted acid is a proton donor. HCO3⁻ can donate one proton to form CO3²⁻, so it is the acid of the pair, while CO3²⁻ is its conjugate base. H2CO3 is a separate, stronger acid one proton ahead of this pair.

Question 30Ionic Equilibria

At 298 K, the ionic product of water Kw = 1.0 × 10⁻¹⁴. What is the pH of a 0.010 M solution of HCl that is assumed to dissociate completely?

  1. 1.00
  2. 7.00
  3. 2.00 Correct answer.
  4. 12.00

Explanation

A strong monobasic acid gives [H3O⁺] equal to its concentration, so [H3O⁺] = 0.010 M = 10⁻² M. Then pH = −log[H3O⁺] = −log(10⁻²) = 2.00. The value 12.00 would be the pOH or the pH of an equally concentrated strong base.

Question 31Ionic Equilibria

Which statement correctly describes the ionisation of a weak monobasic acid HA in water?

  1. It ionises only partially, so an equilibrium exists between HA, H3O⁺ and A⁻. Correct answer.
  2. It ionises completely, so [H3O⁺] equals the initial acid concentration.
  3. It ionises completely only when the solution is diluted with water.
  4. It ionises partially, but no A⁻ ions are produced in solution.

Explanation

Weak acids are only slightly ionised, so undissociated HA remains in equilibrium with H3O⁺ and A⁻. The equilibrium constant for this reaction is Ka, and it is small for a weak acid. This partial ionisation is why pH values of weak acids are higher than those of strong acids of the same concentration.

Question 32Ionic Equilibria

A buffer solution is prepared from acetic acid and sodium acetate. Which pair of species is mainly responsible for its buffering action?

  1. CH3COOH and CH3COO⁻ Correct answer.
  2. H3O⁺ and OH⁻
  3. CH3COOH and H3O⁺
  4. Na⁺ and CH3COOH

Explanation

An acidic buffer contains a weak acid together with its conjugate base in appreciable amounts. Added acid is consumed by CH3COO⁻ and added base by CH3COOH, so pH changes little. The reservoir pair is therefore CH3COOH/CH3COO⁻.

Question 33Ionic Equilibria

Which of the following salts produces a basic aqueous solution at 298 K?

  1. NH4Cl
  2. KNO3
  3. NaCN Correct answer.
  4. NaCl

Explanation

NaCN is the salt of a weak acid (HCN) and a strong base (NaOH). The cyanide ion hydrolyses, CN⁻ + H2O ⇌ HCN + OH⁻, releasing OH⁻ and making the solution basic. Salts of strong acids with strong bases stay neutral.

Question 34Ionic Equilibria

For a saturated solution of a sparingly soluble salt AgCl, the solubility product Ksp is 1.8 × 10⁻¹⁰ at 298 K. What is the molar solubility of AgCl in pure water?

  1. 3.6 × 10⁻¹⁰ M
  2. 1.8 × 10⁻¹⁰ M
  3. 1.3 × 10⁻⁵ M Correct answer.
  4. 9.0 × 10⁻¹¹ M

Explanation

For AgCl, Ksp = [Ag⁺][Cl⁻] = s², since each formula unit gives one Ag⁺ and one Cl⁻. Taking the square root gives s = √(1.8 × 10⁻¹⁰) ≈ 1.3 × 10⁻⁵ M. The other values either ignore the square root or double the exponent incorrectly.

Question 35Ionic Equilibria

What is the pH of a buffer made by mixing equal volumes of 0.10 M acetic acid and 0.10 M sodium acetate? (Ka of acetic acid = 1.8 × 10⁻⁵, pKa = 4.74)

  1. 7.00
  2. 4.74 Correct answer.
  3. 9.26
  4. 2.87

Explanation

The Henderson–Hasselbalch equation gives pH = pKa + log([salt]/[acid]). Because the acid and salt concentrations are equal, the log term is zero, so pH = pKa = 4.74. Dilution by equal volumes changes both concentrations equally and does not alter the ratio.

Question 36Ionic Equilibria

In a solution of a weak acid HA, the degree of dissociation is α. Which expression gives the acid dissociation constant Ka in terms of the initial concentration c?

  1. Ka = c²α / (1 − α)
  2. Ka = cα² / (1 + α)
  3. Ka = cα² / (1 − α) Correct answer.
  4. Ka = cα / (1 − α)

Explanation

At equilibrium [H3O⁺] = [A⁻] = cα and [HA] = c(1 − α). Substituting into Ka = [H3O⁺][A⁻]/[HA] gives Ka = cα²/(1 − α). For small α, this reduces to the familiar approximation Ka ≈ cα².

Question 37Ionic Equilibria

A solution has [OH⁻] = 1.0 × 10⁻³ M at 298 K. What are its pOH and pH?

  1. pOH = 11.00, pH = 3.00
  2. pOH = 3.00, pH = 11.00 Correct answer.
  3. pOH = 3.00, pH = 3.00
  4. pOH = 10.00, pH = 4.00

Explanation

pOH = −log[OH⁻] = −log(10⁻³) = 3.00. Since pH + pOH = 14 at 298 K, pH = 14.00 − 3.00 = 11.00, consistent with a basic solution. The other options swap pH and pOH or use an incorrect logarithmic value.

Question 38Ionic Equilibria

Which statement about the common ion effect is correct when a small amount of sodium acetate is added to an acetic acid solution?

  1. The added acetate has no effect because acetic acid is only weakly ionised.
  2. The added Na⁺ reacts with H3O⁺, so the pH rises sharply to above 12.
  3. The added CH3COO⁻ shifts the ionisation equilibrium to the left, lowering [H3O⁺] and raising pH. Correct answer.
  4. The added CH3COO⁻ shifts the equilibrium to the right, raising [H3O⁺] and lowering pH.

Explanation

Acetate is a product of the ionisation CH3COOH + H2O ⇌ H3O⁺ + CH3COO⁻. Adding acetate increases the product concentration, so by Le Chatelier's principle the equilibrium moves left, suppressing ionisation and reducing [H3O⁺]. The result is a higher pH and a buffer system.

Question 39Ionic Equilibria

The pH of a 0.10 M solution of a weak monobasic acid is 3.00 at 298 K. What is the value of Ka for this acid?

  1. 1.0 × 10⁻⁵ Correct answer.
  2. 1.0 × 10⁻³
  3. 1.0 × 10⁻⁷
  4. 1.0 × 10⁻⁶

Explanation

From pH = 3.00, [H3O⁺] = 10⁻³ M, and for a monobasic acid [A⁻] is also 10⁻³ M. The remaining acid is approximately 0.10 − 0.001 ≈ 0.10 M, so Ka = (10⁻³)²/0.10 = 10⁻⁵. The small degree of ionisation justifies neglecting the subtraction.

Question 40Ionic Equilibria

A saturated solution of a sparingly soluble salt of type AB2 dissolves according to AB2(s) ⇌ A²⁺ + 2B⁻. If its molar solubility is s, which expression gives Ksp?

  1. Ksp = 2s²
  2. Ksp = s³
  3. Ksp = s²
  4. Ksp = 4s³ Correct answer.

Explanation

Dissolving s moles of AB2 gives [A²⁺] = s and [B⁻] = 2s. Substituting into Ksp = [A²⁺][B⁻]² gives s × (2s)² = 4s³. The coefficient 2 must be squared along with s, which produces the factor 4.

Question 41Ionic Equilibria

Which of the following mixtures will function as a buffer solution when dissolved in water?

  1. Equal moles of NH3 and NH4Cl Correct answer.
  2. Equal moles of NaCl and water
  3. Equal moles of HCl and NaOH
  4. Equal moles of CH3COOH and NaOH

Explanation

A buffer requires a weak base and its conjugate acid (or a weak acid and its conjugate base) in appreciable amounts. NH3 with NH4Cl provides exactly this pair, so added acid is taken up by NH3 and added base by NH4⁺. Strong acid–strong base or strong acid–weak base in equal moles do not leave a reservoir pair.

Question 42Ionic Equilibria

A 0.10 M solution of NH4Cl is acidic. Which equilibrium best explains this behaviour?

  1. Cl⁻ + H2O ⇌ HCl + OH⁻
  2. NH4⁺ + H2O ⇌ NH3 + H3O⁺ Correct answer.
  3. NH4⁺ + OH⁻ ⇌ NH3 + H2O
  4. NH4Cl ⇌ NH4⁺ + Cl⁻

Explanation

NH4⁺ is the conjugate acid of the weak base NH3, so it donates a proton to water, producing H3O⁺ and making the solution acidic. The hydrolysis constant is Kh = Kw/Kb, which is large enough to give a pH below 7. Chloride is the conjugate base of a strong acid and does not hydrolyse.

Question 43Ionic Equilibria

A buffer contains 0.10 M NH3 and 0.10 M NH4Cl (Kb of NH3 = 1.8 × 10⁻⁵). Which calculation gives the pH of this buffer at 298 K?

  1. pH = pKb = 4.74 directly, without further conversion
  2. pH = pKa of NH4⁺ = 9.26, so pH = 4.74
  3. pOH = pKb + log([NH4⁺]/[NH3]) = 4.74, so pH = 9.26 Correct answer.
  4. pH = 14 − 4.74 = 9.26, so pOH = 4.74

Explanation

For a basic buffer, pOH = pKb + log([conjugate acid]/[base]). With equal concentrations the log term is zero, so pOH = pKb = 4.74, and pH = 14 − 4.74 = 9.26. This is the correct sequence for a buffer of NH3 and NH4⁺.

Question 44Ionic Equilibria

Solid NaCl is added to a saturated solution of AgCl in water. Assuming no volume change, which change is observed in the equilibrium AgCl(s) ⇌ Ag⁺ + Cl⁻?

  1. The added NaCl has no effect because AgCl is a sparingly soluble salt.
  2. The increased [Cl⁻] shifts equilibrium left, so [Ag⁺] falls and the solubility of AgCl decreases. Correct answer.
  3. The increased [Cl⁻] shifts equilibrium right, so more AgCl dissolves and [Ag⁺] rises.
  4. The ionic strength increases, so [Ag⁺] and [Cl⁻] both rise and Ksp increases.

Explanation

Ksp = [Ag⁺][Cl⁻] must remain constant at constant temperature. Adding Cl⁻ from NaCl increases the product, so Ag⁺ combines with Cl⁻ and precipitates as AgCl until the product returns to Ksp. The net result is a lower [Ag⁺] and reduced solubility, an example of the common ion effect.

Question 45Chemical Thermodynamics

For the reaction N2(g) + 3H2(g) → 2NH3(g) at constant temperature and pressure, which expression gives the standard enthalpy of reaction from standard enthalpies of formation?

  1. ΔrH° = 2ΔfH°(NH3) + ΔfH°(N2) + 3ΔfH°(H2)
  2. ΔrH° = [2ΔfH°(NH3)] ÷ [ΔfH°(N2) + 3ΔfH°(H2)]
  3. ΔrH° = [ΔfH°(N2) + 3ΔfH°(H2)] − [2ΔfH°(NH3)]
  4. ΔrH° = [2ΔfH°(NH3)] − [ΔfH°(N2) + 3ΔfH°(H2)] Correct answer.

Explanation

Hess's law applied to formation reactions gives ΔrH° as the sum of formation enthalpies of products minus that of reactants, each multiplied by its stoichiometric coefficient. Since ΔfH° of an element in its standard state (N2 gas, H2 gas) is zero, the expression reduces numerically to 2ΔfH°(NH3), but the general form must still show the reactant term.

Question 46Chemical Thermodynamics

A system absorbs 500 J of heat from the surroundings and simultaneously does 200 J of work on the surroundings. Applying the first law in the form ΔU = q + w with w measured as work done on the system, what is ΔU?

  1. −700 J
  2. 300 J Correct answer.
  3. −300 J
  4. 700 J

Explanation

In the convention ΔU = q + w, q is positive for heat absorbed (+500 J) and w is negative when the system does work on the surroundings (−200 J). Adding gives ΔU = 500 − 200 = +300 J, so the internal energy rises.

Question 47Chemical Thermodynamics

For an ideal gas undergoing isothermal reversible expansion at temperature T, which statement about the quantities is correct?

  1. ΔU = nCvΔT is negative because the gas expands
  2. ΔH is negative because pressure falls during expansion
  3. q = 0 because no heat flows during isothermal expansion
  4. ΔU = 0 and q = −w, where w is work done on the gas Correct answer.

Explanation

For an ideal gas, internal energy depends only on temperature, so ΔU = 0 when T is constant. The first law then forces q = −w: the heat absorbed exactly equals the work done by the gas (w is negative in the work-on-system convention).

Question 48Chemical Thermodynamics

Which set of conditions correctly describes a spontaneous process at constant temperature and pressure?

  1. ΔG < 0 and ΔStotal > 0 Correct answer.
  2. ΔG = 0 and ΔStotal = 0
  3. ΔG > 0 and ΔStotal < 0
  4. ΔH < 0 and ΔSsystem < 0

Explanation

At constant T and P the Gibbs energy change is ΔG = ΔH − TΔS, and spontaneity requires ΔG < 0. This is equivalent to the total entropy of system plus surroundings increasing, ΔStotal > 0.

Question 49Chemical Thermodynamics

For the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), the standard equilibrium constant Kp is written in terms of partial pressures. Which expression is correct?

  1. Kp = (pSO3)² / [(pSO2) (pO2)]
  2. Kp = [(pSO2)² (pO2)] / (pSO3)²
  3. Kp = (pSO3)² / [(pSO2)² (pO2)] Correct answer.
  4. Kp = (pSO3) / [(pSO2)² (pO2)]

Explanation

The equilibrium constant places products in the numerator and reactants in the denominator, each raised to its stoichiometric coefficient. For 2SO3 as product and 2SO2 plus 1O2 as reactants, this gives (pSO3)²/[ (pSO2)² (pO2) ].

Question 50Chemical Thermodynamics

The relation between ΔG° and the equilibrium constant at temperature T is ΔG° = −RT ln K. If K for a reaction is 1 × 10⁻⁵ at 298 K, what is the sign of ΔG° and what does it imply?

  1. ΔG° is negative, so the forward reaction is non-spontaneous
  2. ΔG° is negative, so products dominate at equilibrium
  3. ΔG° is zero, so the reaction is at equilibrium under standard conditions
  4. ΔG° is positive, so reactants dominate at equilibrium Correct answer.

Explanation

With K < 1, ln K is negative, so ΔG° = −RT ln K becomes positive. A positive standard Gibbs energy change means the standard reaction is non-spontaneous and the equilibrium mixture is reactant-rich.