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

CONCEPT:

Vapour Pressure and Deviations from Raoult's Law

• Raoult's Law states that the partial vapour pressure of a component in a solution is proportional to its mole fraction and the vapour pressure of the pure component.
• The total vapour pressure of an ideal solution can be calculated using:

  P_total = x_XP^o_X + x_YP^o_Y

  • Where x_X and x_Y are the mole fractions of components X and Y.
  • P^o_X and P^o_Y are the vapour pressures of pure components X and Y respectively.
• If the actual total vapour pressure deviates from the ideal total vapour pressure:
  • Positive deviation occurs when the actual vapour pressure is higher than expected (weaker intermolecular forces).
  • Negative deviation occurs when the actual vapour pressure is lower than expected (stronger intermolecular forces).

EXPLANATION:

• Mole fractions:
  • Total moles = 5 + 10 = 15
  • x_X = 5/15 = 1/3, x_Y = 10/15 = 2/3
• Ideal total vapour pressure (using Raoult's Law):
  • P_ideal = x_XP^o_X + x_YP^o_Y
  • = (1/3)(63) + (2/3)(78)
  • = 21 + 52
  • = 73 torr
• Actual total vapour pressure = 70 torr
• Ideal total vapour pressure = 73 torr

Since the actual pressure is lower than the ideal pressure, the solution exhibits a negative deviation from Raoult's Law.

Therefore, the solution shows negative deviation.

CONCEPT:

Elevation in Boiling Point

• The boiling point of a solution increases when a non-volatile solute is added to it. This phenomenon is known as elevation in boiling point.
• The elevation in boiling point (ΔTb) is directly proportional to the molality of the solution and the van 't Hoff factor (i), given by:

  ΔTb = i × Kb × m

• Here:
  • i = Van 't Hoff factor (number of particles the solute dissociates into).
  • Kb = Boiling point elevation constant.
  • m = Molality of the solution.

EXPLANATION:

• Van 't Hoff factors for the given solutes:
  • Urea (0.01 M): Does not dissociate, i = 1.
  • KNO₃ (0.01 M): Dissociates into K⁺ and NO₃⁻, i = 2.
  • Na₂SO₄ (0.01 M): Dissociates into 2 Na⁺ and SO₄²⁻, i = 3.
  • C₆H₁₂O₆ (0.015 M): Does not dissociate, i = 1.
• Based on the formula ΔTb = i × Kb × m:
  • Higher i leads to a greater elevation in boiling point.
  • Na₂SO₄ (i = 3) will exhibit the highest boiling point among the solutions.

Therefore, the solution with 0.01 M Na₂SO₄ will exhibit the highest boiling point.

CONCEPT:

Types of Solutions

• Solutions can exist in different phases, and they are classified based on the solute and solvent phases.
• Examples include:
  • Solid in solid (e.g., alloys)
  • Liquid in gas (e.g., humidity)
  • Solid in gas (e.g., smoke)
  • Liquid in solid (e.g., amalgams)

EXPLANATION:

• A. Humidity: Humidity refers to water vapor (liquid) dispersed in air (gas). This is an example of a liquid in gas solution. (A-II)
• B. Alloys: Alloys are homogeneous mixtures of metals or a metal and a non-metal in solid form. This is an example of a solid in solid solution. (B-I)
• C. Amalgams: Amalgams are alloys that consist of a metal (solid) dissolved in mercury (liquid). This is an example of a liquid in solid solution. (C-IV)
• D. Smoke: Smoke consists of tiny solid particles dispersed in a gas. This is an example of a solid in gas solution. (D-III)

Therefore, the correct option is 2) A-II, B-I, C-IV, D-III.

CONCEPT:

Osmotic Equilibrium via Semipermeable Partition

• Beaker A: 0.1 mol NaCl in 1 kg H₂O (NaCl dissociates into 2 ions)
• Beaker B: 0.1 mol sugar (non-electrolyte) in 1 kg H₂O
• Both solutions are placed in a closed container separated by a semipermeable membrane.
• At equilibrium, **vapour pressure of both solutions becomes equal**, which is governed by Raoult's law.
• For dilute solutions:

  \(\frac{\Delta p}{p^*} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}\)

  must be same for both beakers.

EXPLANATION:

• Let water transfer cause a change of mass by 'a' kg, then:
  • Mass of solvent in Beaker B becomes (1 - a) kg
  • Mass of solvent in Beaker A becomes (1 + a) kg
• Moles of H₂O = mass / 18
• Equating relative lowering of vapour pressure:

  \(\frac{0.1}{0.1 + \frac{(1 - a) \times 1000}{18}} = \frac{2 \times 0.1}{2 \times 0.1 + \frac{(1 + a) \times 1000}{18}}\)

• Solving the above equation gives:

  \(a = \frac{1}{3}\)

• New mass of water in Beaker A = (1 + 1/3) = 4/3 kg
• Molality of NaCl =

  \(\frac{0.1 \times 2}{\frac{4}{3}} = \frac{0.2 \times 3}{4} = \frac{3}{20} = \frac{x}{40} \Rightarrow x = 6\)

Therefore, the value of x = 6

Explanation:

• According to Henry's law, the solubility of a gas (S) in a liquid is given by.

S=K_H P

• Given Data,
  •  Henry's law constant\( (K_H) = 0.16 mol dm^{-3} atm^{-1}\)
  • Partial pressure of the gas (P) = 0.18 atm

Substitute the given values into the formula:
\(S = 0.16 \, \text{mol dm}^{-3} \, \text{atm}^{-1} \cdot 0.18 \, \text{atm}\ = 0.0288 \, \text{mol dm}^{-3} \)

Conclusion:

The closest option to \(0.0288 \ mol\ dm^{-3}\) is  \(0.029 \ mol\ dm^{-3}\)

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 

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).

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:

• 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