Solutions MCQs

Solid solutions are homogeneous mixtures where the solute atoms occupy lattice positions of the solvent crystal.

In a metal alloy like brass, Zn atoms replace some Cu atoms in the copper lattice, forming a single solid phase.

Both components are solids and remain in the same crystalline structure throughout the mixture.

Hence the correct answer is Brass (Cu and Zn alloy).

Mass of urea = 10 g, mass of water = 90 g, total mass of solution = 10 g + 90 g = 100 g.

Mass percent = (mass of solute / total mass) × 100 = (10 g / 100 g) × 100 = 10 %.

Hence the correct answer is 10%.

Raoult’s law relates the vapor pressure of a component to its concentration in the liquid phase.

For an ideal solution the partial pressure \(p_i\) equals the mole fraction \(x_i\) times the pure component pressure \(p_i^\circ\): \(p_i = x_i p_i^\circ\).

Thus the proportionality is with the mole fraction of the component in the solution.

Hence the correct answer is mole fraction in solution.

In a solution, negative deviation from Raoult’s law occurs when intermolecular forces between unlike molecules are stronger than those in the pure liquids.

Chloroform (CHCl₃) can donate a hydrogen bond from its slightly acidic H, while acetone (CH₃COCH₃) has a carbonyl O that can accept a hydrogen bond.

The H‑bond between chloroform’s H and acetone’s O is stronger than the original dipole–dipole forces in each pure component, reducing vapor pressure.

Hence the mixture shows negative deviation because hydrogen bonding forms between chloroform H and acetone O.

Moles of glucose = 18 g ÷ 180 g mol⁻¹ = 0.10 mol.

Molality = 0.10 mol ÷ 1 kg = 0.10 m.

ΔTb = Kb × m = 0.52 K kg mol⁻¹ × 0.10 m = 0.052 K.

Hence the boiling‑point elevation is 0.052 K.

Mass of solute = 20 mg = 20 × 10⁻³ g.

Mass of solvent = 1 kg = 10³ g, so total mass ≈ 10³ g (since solute amount is negligible).

ppm = (mass of solute / mass of solution) × 10⁶ = (20 × 10⁻³ / 10³) × 10⁶ = 20.

Thus the concentration is 20 ppm.

Use Henry’s law in mole‑fraction form: p = K_H · x, where p is the gas partial pressure and K_H is the Henry constant.

Re‑arrange to find the mole fraction: x = p / K_H.

Substitute p = 2 bar and K_H = 50 bar: x = 2 / 50 = 0.04.

Hence the mole fraction of the gas in solution is 0.04.

Both components are non‑polar, have similar molecular sizes and intermolecular forces (London dispersion).

Their enthalpy of mixing is essentially zero and volume change on mixing is negligible.

Thus Raoult’s law is obeyed closely, giving behavior close to an ideal solution.

n‑hexane and n‑heptane.

ΔTb = Kb · m (boiling‑point elevation formula).

Substitute Kb = 0.52 K kg mol⁻¹ and m = 0.5 mol kg⁻¹.

ΔTb = 0.52 × 0.5 = 0.26 K.

Hence the elevation in boiling point is 0.26 K.

Osmotic pressure (π) of a solution depends on the number of solute particles (π = iMRT).

If two solutions exert the same π at the same temperature, they contain the same total concentration of solute particles.

Thus they have equal tendency to cause osmotic flow and are said to be isotonic.

Correct answer: Isotonic

Oxygen and nitrogen are both gases under the conditions given, so each component remains in the gaseous phase.

When one gas dissolves uniformly in another, the resulting homogeneous mixture is called a gaseous solution.

Since both solute (O₂) and solvent (N₂) are gases, the solution is a gas‑in‑gas system.

Moles needed = concentration × volume = 0.5 mol L⁻¹ × 2 L = 1 mol.

Mass = moles × molar mass = 1 mol × 180 g mol⁻¹ = 180 g.

Thus 180 g of glucose is required. 180 g.

Total vapour pressure of an ideal solution equals the sum of partial pressures of each component.

Partial pressure of A = pA⁰·xA (Raoult’s law) and of B = pB⁰·xB.

Add the two partial pressures to get the overall pressure.

Thus Ptotal = pA⁰ xA + pB⁰ xB.

Azeotropes occur when the total vapour pressure curve has a maximum (or minimum) at a certain composition.

For a maximum‑boiling azeotrope the liquid mixture has a lower vapour pressure than predicted by Raoult’s law, i.e., the vapour pressure curve bends downward.

This downward bend is characteristic of a negative deviation from Raoult’s law.

Hence the correct answer is Negative deviation from Raoult's law.

Molal elevation constant Kb relates the rise in boiling point (ΔTb) to the molality (m) of a solute: ΔTb = Kb·m.

Since it quantifies boiling‑point elevation, it is called the ebullioscopic constant.

Thus Kb is the ebullioscopic constant.

Sugar is a solid at room temperature and it disperses uniformly in water forming a homogeneous phase.

The solute (sugar) is a solid, the solvent (water) is a liquid, giving a solid‑in‑liquid solution.

Thus the correct choice is Sugar dissolved in water.

Molality is defined as the amount of solute (in moles) divided by the mass of the solvent (in kilograms).

It directly relates the solute quantity to the solvent mass, unlike molarity which uses solution volume.

Hence the expression “moles of solute per kilogram of solvent” corresponds to Molality.

Apply Raoult’s law for an ideal solution: total vapour pressure = Σ (mole fraction × vapour pressure of pure component).

For component A: 0.5 × 100 mm Hg = 50 mm Hg.

For component B: 0.5 × 60 mm Hg = 30 mm Hg.

Add the contributions: 50 mm Hg + 30 mm Hg = 80 mm Hg, which is the total vapour pressure.

Hence the correct answer is 80 mm Hg.

Depression of freezing point, osmotic pressure and elevation of boiling point all depend only on the number of solute particles, not their nature, so they are colligative properties.

Optical activity arises from the ability of chiral molecules to rotate plane‑polarized light, which depends on the molecular structure, not on particle concentration.

Therefore optical activity is not a colligative property. Optical activity.

Molal depression constant Kf relates the lowering of freezing point to the molality of a solute.

The formula ΔTf = Kf·m shows Kf is the proportionality factor for freezing‑point depression.

Since this constant is used in cryoscopy (freezing‑point measurements), it is called the cryoscopic constant.