Solutions MCQ Quiz - Objective Question with Answer for Solutions - Download Free PDF

CONCEPT:

Type of Solutions and Their Examples

• A solution is a homogeneous mixture composed of a solute and solvent, which can be in different phases (solid, liquid, or gas).
• Gaseous solutions consist of gases dissolved in other gases, liquid solutions consist of liquids dissolved in other liquids, and solid solutions are solid mixtures.
• These solutions can be classified into gas-liquid, solid-liquid, and solid-solid solutions depending on the phase of the solute and solvent.

EXPLANATION:

• 1. Mixture of oxygen and nitrogen gases → Gaseous Solutions
  • This is a gaseous solution where oxygen and nitrogen gases mix homogeneously in the gas phase.
• 2. Ethanol dissolved in water → Liquid Solutions
  • Ethanol (a liquid) dissolves in water (a liquid), forming a liquid solution.
• 3. Solution of hydrogen in palladium → Solid Solutions
  • This is an example of a solid solution, where hydrogen gas dissolves in solid palladium, forming a homogeneous mixture in the solid phase.
• 4. Oxygen dissolved in water → Gas-Liquid Solutions
  • This is a gas-liquid solution, where oxygen (gas) dissolves in water (liquid), forming a homogeneous solution.
• 5. Copper dissolved in gold → Solid-Solid Solutions
  • Copper and gold are both solids, and when they mix to form an alloy, it is a solid-solid solution.
• The correct matching is as follows  1 - A, 2 - D, 3 - E, 4 - B, 5 - C

Therefore, the correct answer is: b) 1 - A, 2 - D, 3 - E, 4 - B, 5 - C

 CONCEPT:

Henry’s Law Constant (K_H) and Temperature

• Henry’s Law relates the partial pressure of a gas to its mole fraction in a liquid:
  p = K_H × x
• As temperature increases, the solubility of gases in liquids decreases.
• Thus, K_H increases with temperature, since more pressure is needed to dissolve the same amount of gas.
• This explains why aquatic life thrives better in cold water than in warm water.

EXPLANATION:

• The correct graphical representation is the one where K_H increases with temperature.
• Graph I shows a linear increase of K_H with T, which matches theoretical expectations.

Therefore, the correct answer is I

CONCEPT:

Osmotic Pressure and Van't Hoff Factor

• The osmotic pressure (π) of a solution is given by the formula:

  π = iCRT

  where:
  • i is the Van't Hoff factor, which represents the number of particles into which a solute dissociates in solution.
  • C is the concentration of the solution in molarity (M).
  • R is the ideal gas constant.
  • T is the temperature in Kelvin.
• The osmotic pressure is directly proportional to the product of the Van't Hoff factor (i) and concentration (C).
• In dilute solutions where 100% ionization is assumed, we can calculate the effective number of particles produced in solution, which is the Van't Hoff factor, i.

EXPLANATION:

• A. 0.500 M C₂H₅OH (aq):
  • Ethanol does not ionize in solution,
  • so i = 1.
  • For 0.25 M KBr (aq):
  • KBr ionizes into K⁺ and Br⁻, so i = 2.
• B. 0.100 M K₄[Fe(CN)₆] (aq):
  • ​ K₄[Fe(CN)₆] dissociates into 5 ions (4 K⁺ and 1 Fe(CN)₆⁴⁻),
  • so i = 5.
  • For 0.100 M FeSO₄(NH₄)₂SO₄ (aq):
  • It dissociates into 5 ions , so i = 5.
• C. 0.05 M K₄[Fe(CN)₆] (aq): i = 5.
  • For 0.25 M NaCl (aq): NaCl dissociates into Na⁺ and Cl⁻, so i = 2.
• D. 0.15 M NaCl (aq): i = 2.
  • For 0.1 M BaCl₂ (aq): BaCl₂ dissociates into Ba²⁺ and 2 Cl⁻, so i = 3.
• E. 0.02 M KCl•MgCl₂•6H₂O (aq): This dissociates into 5 ions (K⁺, Mg²⁺, and 3 Cl⁻), so i = 5.
  For 0.05 M KCl (aq): i = 2.
• Now, we calculate the product of i and C for each pair:
  • A: i x C = 1 x 0.500 = 0.500 for C₂H₅OH and 2 x 0.25 = 0.500 for KBr.
  • B: i x C = 5 x 0.100 = 0.500 for K₄[Fe(CN)₆] and 5 x 0.100 = 0.500 for FeSO₄(NH₄)₂SO₄.
  • C: i x C = 5 x 0.050 = 0.250 for K₄[Fe(CN)₆] and 2 x 0.25 = 0.500 for NaCl.
  • D: i x C = 2 x 0.150 = 0.300 for NaCl and 3 x 0.100 = 0.300 for BaCl₂.
  • E: i x C = 5 x 0.020 = 0.100 for KCl•MgCl₂•6H₂O and 2 x 0.050 = 0.100 for KCl.

Therefore, the pairs with the same osmotic pressure are:

• The pairs with the same osmotic pressure (i x C) are:
  • A (C₂H₅OH and KBr)
  • B (K₄[Fe(CN)₆] and FeSO₄(NH₄)₂SO₄)
  • D (NaCl and BaCl₂)
  • E ( KCl•MgCl2•6H2O and  KCl.)

Therefore, the number of pairs with the same osmotic pressure is 4.

CONCEPT:

Osmotic Pressure and Molar Mass Determination

• The osmotic pressure (π) of a solution can be calculated using the formula:

  π = (w / m) × (R × T / V)

• Where:
  • π is the osmotic pressure,
  • w is the weight of the solute,
  • m is the molar mass of the macromolecule,
  • R is the universal gas constant (8.3 J·K^-1·mol^-1),
  • T is the temperature in Kelvin (300 K),
  • V is the volume of the solution.
• Osmotic pressure is also related to the height (h) of the liquid column and the density (ρ) of the solution using the formula:

  π = ρ × g × h

EXPLANATION:

• Given data:
  • Density (ρ) = 1.00 g/cm³ = 1000 kg/m³
  • Height (h) = 2.00 cm = 0.02 m
  • Acceleration due to gravity (g) = 10 m/s²
•  π = ρgh:
  • π = 1000 × 10 × 0.02 = 200 Pa
  • Convert to atm: π = 200 × 10^-5 atm
• Apply the osmotic pressure formula:

  π = (w / m) × (R × T / V)

  Rearrange to solve for m (molar mass):

  m = (w × R × T) / (π × V)

• w = 2.00 g
• R = 8.3 J·mol^-1·K^-1
• T = 300 K
• π = 200 × 10^-5 atm (converted to correct units for calculation)
• m =  2.49 × 10⁶ g/mol.

Therefore, the molar mass of the macromolecule is 2.49 × 10⁶ g/mol, and the value of X is 2.49.

CONCEPT:

Determination of Molar Mass using Osmotic Pressure

• Osmotic pressure (π) is a colligative property and is related to molar mass by the formula:

  π = CRT

  where:
  • π = Osmotic pressure (atm)
  • C = Concentration in mol/L
  • R = Gas constant (0.083 L·atm·K–1·mol–1)
  • T = Temperature in Kelvin
• For solutions of macromolecules like PVC, concentration is expressed in g/L, and:

  C = (W/V) × (1/M) ⇒ π = C × (RT/M)

• If we plot π/C vs C, the slope gives RT/M, so:

  M = RT / slope

EXPLANATION:

\(π=\mathrm{M}^{\prime} \mathrm{RT}=\left(\frac{\mathrm{W} / \mathrm{M}}{\mathrm{~V}}\right) \mathrm{RT}\)

⇒ \(π=\left(\frac{\mathrm{W}}{\mathrm{~V}}\right)\left(\frac{1}{\mathrm{M}}\right) \mathrm{RT}=\mathrm{C}\left(\frac{\mathrm{RT}}{\mathrm{M}}\right)\)

⇒ \(\frac{π}{\mathrm{C}}=\frac{\mathrm{RT}}{\mathrm{M}} \neq \mathrm{f}(\mathrm{c})\)

If we assume graph between \(\rm \frac{π}{C}\) and C

Assuming π vs C graph

Slope = \(\frac{\mathrm{RT}}{\mathrm{M}}=\frac{0.083 \times 300}{\mathrm{M}}=6 \times 10^{-4}\)

∴ \(\rm M=\frac{0.083 \times 300}{6 \times 10^{-4}}=\frac{830 \times 300}{6}=41,500\) gm/mole

Therefore, the molar mass of PVC is 41,500 g/mol (nearest integer).

The correct answer is Sugar in water.

Key Points

• The physical properties help in separating the homogenous mixtures.
• Those mixtures in which the substances are completely mixed together and are indistinguishable from one another are called homogeneous mixtures.
• A homogeneous mixture is a mixture in which the composition is uniform throughout the mixture.
• Many homogeneous mixtures are commonly referred to as solutions.
• Some of the examples of homogeneous mixtures (or solutions) are Sugar solution, Salt solution, Copper sulphate solution, Seawater, Alcohol and water mixture, Petrol and oil mixture, Soda water etc.

• Heterogeneous mixture:
  • A heterogeneous mixture is a mixture with a non-uniform composition that contains components in different phases.
  • The composition varies from one region to another with at least two phases that remain separate from each other, with clearly identifiable properties.
  • Heterogeneous mixtures contain particles that retain their chemical properties when they are mixed and can be distinguished after they are mixed.
  • The components of heterogeneous mixtures can be separated by the filtration of chemical procedures.
  • The two types of heterogeneous mixtures are suspensions and colloids.
  • Sugar and sand form a heterogeneous mixture. If you look closely, you can identify tiny sugar crystals and particles of sand.
  • Ice cubes in cola form a heterogeneous mixture.

The correct answer is Increase.

Concept:

• Molarity:
  • It is defined as the moles of a solute per litres of a solution.
  • It is also known as the molar concentration of a solution.
• Molality:
  • It is defined as the number of moles of solute per kilogram of solvent.
• Mole fraction:
  • It is the ratio of moles of a component with the total moles of solute and solvent.
• Mass%:
  • It is the percentage of the mass of solute or solvent w.r.t total mass of solution.
• Formulas:

Explanation:

• Molarity depends on the volume of the solution.
• And volume is directly proportional to temperature.
• And when we increase the temperature the volume will increase.
• So the increase in volume leads to a decrease in Molarity as Molarity is inversely proportional to the volume of solution.

​Additional Information

Notes:

• Normality:
  • It is defined as the number of gram equivalent per litre of solution.
  • Also known as equivalent concentration.
  • Normality = Number of gram equivalents / [volume of solution in litres]
• Normality is inversely proportional to temperature.
• ​Volume is directly proportional to temperature.
• Molarity is inversely proportional to volume.
• Molarity is inversely proportional to temperature.
• Molality is not dependent on temperature.

The correct answer is isotonic solutions.

Concept:

• Colligative properties are the properties that depend upon the number of solute particles present in the solution. 
  They are : 
• Lowering of Vapour pressure:
  • The vapour pressure exerted by solute molecules on the surface of the solution decreases as solute particles are added to the solution.
  • The relative lowering vapour pressure as given by Raoult's law is:

​\(\Delta p = p_0 × x_2\)

• Elevation of the boiling point:
• The boiling point of a solution increases as we add solute particles to the pure solvent.
• The elevation in boiling point is directly proportional to the molality of the solution.

​ΔT_b = k_b × m; where m = molality of the solution and kb = molal elevation constant.

• Depression of freezing point:
  • The freezing point of a solution decreases as we add solute particles to the pure solvent.
  • The depression of the freezing point is also proportional to the molality of the solution.

​​ΔTf = k_f × m; where m = molality of the solution and k_f = molal depression constant

• Osmotic pressure:
  • ​When a solution and a pure solvent are separated by a semi-permeable membrane, due to differences in concentration, the solvent particles start moving towards the solution via the membrane. This phenomenon is called osmosis. However, the diffusion can be stopped by applying pressure over the membrane in the solution.
  • The osmotic pressure of a solution is the pressure required to stop osmosis when the solution is separated from pure solvent by a semi-permeable membrane.

​Explanation:

Isotonic solutions:

• Diffusion of solvent molecules takes place when there is a difference of chemical potential (or simply concentration) between two solutions connected by a semi-permeable membrane.
• The diffusion can be stopped by applying a pressure π over the solution. This is the osmotic pressure.
• The osmotic pressure is independent of the nature of the membrane but depends on the following factors:
• The temperature remains constant, the osmotic pressure of a solution is directly proportional to its concentration.

​π = kC ; where C = concentration and k = proportionality constant

• Concentration remaining constant, the osmotic pressure is directly proportional to absolute temperature.

π = kT; where T = temperature

• Combining the two laws, we get

π = CRT; where R = universal gas constant

• The osmosis takes place until and unless the chemical potential of both the solutions becomes the same. This is the state of equilibrium. at this point, both solutions have the same osmotic pressure.
• When two solutions of equimolar concentration having the same osmolarity, are separated by a semi-permeable membrane, no net osmosis will take place, then the solution is called isotonic solutions. This is because the osmotic pressure on both sides is the same.

​Hypotonic solutions:

• When the concentration of solutes in a solution is less as compared to the other solution, it is called a hypotonic solution.
• In a hypotonic solution, diffusion takes place in an inward direction.

​Hypertonic solutions:​

• When the concentration of solutes in a solution is more as compared to the other solution, it is called a hypertonic solution.
• In a hypotonic solution, diffusion takes place in an outward direction.

Hence, two solutions with equal osmotic pressure are called isotonic.

The correct answer is option 1, i.e. Boiling point of water will increase above 373K.

Concept:

Vapour Pressure-

• The pressure at which liquid and vapour can co-exist at a given temperature is called the vapour pressure of the liquid.
• When a liquid is kept in a closed vessel with some free space, it starts to vaporise.
• The vapourisation continues until a state of equilibrium is reached between vapourisation and condensation.
• At equilibrium, the state gets saturated and the pressure exerted by the vapour molecules is called vapour pressure.

Explanation:

Vapour Pressure of Solutions of Solids in Liquids - 

• Liquids vapourise at a given temperature and under equilibrium conditions, the pressure exerted by the vapours of the liquid over the liquid is called vapour pressure.
• If a non-volatile solute is added to a pure liquid, the vapour pressure of the solution at a given temperature is found to be lower than the pure solvent.
• Colligative properties of solutions connected with this decrease of vapour pressure are - 
  • Relative lowering of the vapour pressure of the solvent.
  • Depression of the freezing point of the solvent.
  • Elevation of the boiling point of the solvent, ex: Boiling point of water will increase above 373K. Hence Option 1 is correct.
  • The osmotic pressure of the solution.

Concept:

Mass by mass percentage

Mass by mass percentage of a solute in a solution is the mass of the solute present in 100 grams of the solution.
Formula:

Mass by mass percentage = (Mass of solute / Mass of solution) x 100

Explanation:

• Mass of solute (salt) = 40 g
• Mass of solvent (water) = 320 g

We know,

Mass of solution = Mass of solute + Mass of solvent

= 40 g + 320 g

= 360 g

\(mass\;percentage\;of\;solution = \frac{{mass\;of\;solute}}{{mass\;of\;solution}} \times 100\)

The equivalent weight of oxalic acid in C₂H₂O₄⋅2H₂O is 63.

• The molecular weight of oxalic acid (C₂H₂0₄) is 90.
• But since Oxalic Acid exists with 2 molecules of water, hence molecular weight of Oxalic acid (C₂H₂O₄⋅2H₂O) = 126
• Now, Equivalent weight = Molecular weight/Basicity
• Therefore, Equivalent Weight = 126/2 = 63 (As 2 is the basicity)
• Basicity here means Oxalic Acid release 2 H⁺ ions

Explanation:-

• Mole fraction is defined as the major concentration of a solute and solvent in terms of the number of moles.
• Mole fraction of a solute(X_A) = n _A  /n _A + n _B 
• Mole fraction of a solute(X_B) = n _B  /n A + n B 

Calculation:- 

⇒ No. of moles = Given mass / Molar mass

⇒ Mole fraction of NaCl = No. of moles in NaCl / Total no. of moles in the solution

⇒ No. of moles of NaCl =  1

⇒ No. of moles of H₂O = Given the mass of H₂O / Molar mass of H₂O 

⇒ No. of moles of H2O = 1000 / 18 = 55.55

Mole fraction of NaCl = No. of moles in NaCl / No. of moles in NaCl + No. of moles of H2O 

⇒ Mole fraction of NaCl =  1 / 1 + 55.5 

⇒ Mole fraction of NaCl = 0.0177 

The Correct Answer is Reverse Osmosis.

Key Points

•  Reverse osmosis is a water purification process that uses a partially permeable membrane to separate ions, unwanted molecules, and larger particles from drinking water. 
•  Applying an external pressure to reverse the natural flow of pure solvent, thus, is reverse osmosis.
• This application is mainly applied in the production of potable water in water plants and in industries. 
• Reverse osmosis works by reversing the principle of osmosis. 
• The salt solution is subjected to pressure and pressed against the semi-permeable membrane.
• Here, the applied pressure is greater than the osmotic pressure.
• Thus, the molecules move from a highly concentrated solution to a less concentrated solution.

Additional Information

• Osmosis: This is the process by which the molecules of a solvent pass through the semi-permeable membrane from a region of lower concentration to a higher concentration.
• It is a natural process.
• Occurs along the potential gradient.
• This is observed during the opening of stomata and absorption of water from the soil by the roots.

Concept:

Molarity (M) is defined as the number of moles of solute (n) divided by the volume (V) of the solution in litres. It is important to note that molarity is defined as moles of solute per litre of solution, not moles of solute per litre of solvent.

Molarity indicates the number of moles of solute present in 1 litre of solution. Formula is:

\(M=\frac{n}{V}=\frac{weight\ of\ the\ substance}{gram\ molecular\ weight}\times \frac{1000}{volume\ in\ ml}\)

Calculation:

Given that, 

Amount of NaOH solution = 10 % 

Molar mass of NaOH = 40 g/mol

Thus Number of moles in 1 gram of NaOH, n = 1/40

Whereas molar mass of 10% of NaOH = 10/40=1/4 moles

Volume of 10% of NaOH solution = 100 ml

Thus the Molar mass will be 

\(\begin{align} & M=\frac{(\frac{1}{4})mol}{100ml}=2.5\times {{10}^{-3}}mol/ml \\ & \therefore M=2.5mol/L \\ \end{align}\)

Hence, the correct option is (3).

Concept:

• Liquids vapourise at a given temperature and under equilibrium conditions, the pressure exerted by the vapours of the liquid over the liquid is called vapour pressure.
• If a non-volatile solute is added to a pure liquid, the vapour pressure of the solution at a given temperature is found to be lower than the pure solvent.
• Colligative properties of solutions connected with this decrease of vapour pressure are - 
  • Relative lowering of the vapour pressure of the solvent.
  • Depression of the freezing point of the solvent.
  • Elevation of the boiling point of the solvent, ex: Boiling point of water will increase above 373K.

Explanation:

From the above explanation, we can see that depression in freezing point of a solvent is considered as colligative property. 

Whereas freezing point, boiling point and melting point are the point at which substance feezes, boiled and melted at certain specific temperature and pressure