CBSE Class 12 Chemistry Solutions Case Study Questions Set 01

Case Study 1

The spontaneous flow of the solvent through a semipermeable membrane from a pure solvent to a solution or from a dilute solution to a concentrated solution is called osmosis. The phenomenon of osmosis can be demonstrated by taking two eggs of the same size. In an egg, the membrane below the shell and around the egg material is semi-permeable. The outer hard shell can be removed by putting the egg in dilute hydrochloric acid. After removing the hard shell, one egg is placed in distilled water and the other in a saturated salt solution. After some time, the egg placed in distilled water swells-up while the egg placed in salt solution shrinks.
The external pressure applied to stop the osmosis is termed as osmotic pressure (a Colligative property). Reverse osmosis takes place when the applied external pressure becomes larger than the osmotic pressure.

Question 1. What do you expect to happen when red blood corpuscles (RBC’s) are placed in 0.5% NaCl solution?
Answer:
RBC’s are isotonic with 0.9% NaCl solution, so they will swell and may even burst when placed in 0.5% NaCl solution.

Teacher's Note:
1. 0.5% NaCl is hypotonic to RBCs, so water enters the cells by osmosis.

Question 2. Which one of the following will have higher osmotic pressure in 1 M KCl or 1 M urea solution?
Answer:
1 M KCl will have higher osmotic pressure because it dissociates to give K⁺ and Cl⁻ ions while urea does not dissociate into ions in the solution.

Teacher's Note:
1. Use π = iCRT: i = 2 for KCl and 1 for urea, so π(KCl) is about twice π(urea).

Question 3. Name one SPM which can be used in the process of reverse osmosis.
Answer:
Cellulose acetate placed on a suitable support.

Teacher's Note:
1. The membrane for reverse osmosis must withstand high pressure and let only water pass, which is why cellulose acetate is used.

Question 4. What are isotonic solutions?
Answer:
Solutions having equal osmotic pressure are called isotonic solutions.

Teacher's Note:
1. Mention “at the same temperature” for a complete definition; no osmosis occurs between isotonic solutions.

Question 5. Write van’t Hoff equation for dilute solution.
Answer:
πV = nRT,
where, π = Osmotic pressure, n = Number of moles, V = Volume of solution in litre, R = Gas constant, T = Temperature

Teacher's Note:
1. Dividing by V gives π = CRT, the form most often used in numericals.

Case Study 2

A solution which obeys Raoult’s law strictly is called an ideal solution, while a solution which shows deviations from Raoult’s law is called a non-ideal solution or real solution. Suppose the molecules of the solvent and solute are represented by A and B respectively, and let γ_AB, γ_AA and γ_BB are the attractive forces between A–B, A–A and B–B respectively. An ideal solution of the components A and B is defined as the solution in which the intermolecular interactions between the components A–B are of the same magnitude as the intermolecular interactions found in the pure components A–A and B–B. Similarly, a non-ideal solution of the components A and B is defined as the solution in which the intermolecular interactions between the components A–B are of the different magnitude as the intermolecular interactions found in the pure components A–A and B–B.

Question 6. What type of liquids form the ideal solution?
Answer:
Liquids that have similar structures and polarities form ideal solutions.

Teacher's Note:
1. Similar structure and polarity make A–B interactions nearly equal to A–A and B–B interactions.

Question 7. Give one example of an ideal solution.
Answer:
Benzene + Toluene

Teacher's Note:
1. Other examples: n-hexane + n-heptane, bromoethane + chloroethane.

Question 8. Write two characteristics of non-ideal solution.
Answer:
ΔH_mix ≠ 0, ΔV_mix ≠ 0.

Teacher's Note:
1. A third characteristic you can add: it does not obey Raoult’s law over the entire range of concentration.

Question 9. On mixing liquid A and liquid B, volume of the resulting solution decreases, what type of deviation from Raoult’s law is shown by?
Answer:
The solution will show negative deviation from Raoult’s law.

Teacher's Note:
1. Volume decreases when A–B attractions are stronger and pull molecules closer: Δ_mixV < 0 goes with negative deviation.

Question 10. Which type of deviation will be shown by the solution, if γ_AB < γ_AA.
Answer:
Solution will show positive deviation.

Teacher's Note:
1. Weaker A–B attraction lets molecules escape more easily, so the vapour pressure is higher than predicted.

Case Study 3

Liquid solutions are the solutions in which the solvent is liquid. Solution of gases in liquid is governed by Henry’s law. Considering the cases when the solutes are liquids or solids, and further considering simple cases when there is only one solute (i.e. studying binary solutions only), if the solute is liquid, it is volatile and if the solute is a solid, it is non-volatile. In either case, vapours will be present above the solution taken in a closed vessel. The pressure exerted by the vapours in equilibrium with the solution at a particular temperature is called ‘vapour pressure’. It increases with increase of temperature. The effect was quantitatively studied by Clausius and Clapeyron. For binary solutions, as a result of detailed studies, a French chemist, F.M. Raoult in 1886, put forward a law known after him as Raoult’s law. For liquid in liquid, it states that “the vapour pressure of any component is equal to the mole fraction of that component in the solution multiplied by vapour of that component in the pure state.” In case of solid in liquid, it states that “relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution.” These definitions help us in the calculation of different parameters related to solutions of liquid-in-liquid and solid-in-liquid.

Question 11. Two gases A and B have values for Henry’ law constant as 35 and 77 at 293 K. Which will have greater solubility in water at the same pressure and why ?
Answer:
Gas A will have greater solubility because according to Henry’s law expression, p_A = K_Hx_A, greater the value of K_H, lower is the value of x_A at the same value of p_A.

Teacher's Note:
1. Solubility is inversely related to K_H: x = p/K_H.

Question 12. Which equation is used to study quantitatively the effect of temperature on the vapour pressure of a liquid ? Write the equation. What do different symbols represent ?
Answer:
The equation is called Clausius-Clapeyron equation. It is written as
\( \log \left(\frac{P_2}{P_1}\right) = \frac{\Delta H_{vap}}{2.303\,R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \)
where P₁ is the vapour pressure at temperature T₁, P₂ is the vapour pressure at temperature T₂, ΔH_vap = enthalpy of vaporisation of the liquid, R = gas constant

Teacher's Note:
1. Temperatures must be in kelvin, and R = 8.314 J K⁻¹ mol⁻¹ if ΔH_vap is in J mol⁻¹.

Question 13. At a given temperature, the vapour pressure in mm of Hg of a solution of two volatile liquids A and B is given by the equation
p = 120 − 80 x_B (x_B = mole fraction of B)
Calculate the vapour pressure of pure A and B at the same temperature.
Answer:
By Raoult’s law, p = p_A°x_A + p_B°x_B = p_A°(1 − x_B) + p_B°x_B = p_A° + (p_B° − p_A°)x_B
Comparing with p = 120 − 80 x_B, p_A° = 120 mm and p_B° − p_A° = −80 or p_B° = 40 mm
Hence, p_A° = 120 mm Hg, p_B° = 40 mm Hg

Teacher's Note:
1. Quick check: put x_B = 0 to get pure A (120 mm) and x_B = 1 to get pure B (40 mm).

Question 14. The vapour pressures of benzene and toluene at 293 K are 75 mm and 22 mm of Hg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed. If the two form an ideal solution, calculate the mole fraction of benzene in the vapour phase assuming that the vapours are in equilibrium with the liquid mixture at this temperature.
Answer:
In the liquid phase, n_B = 23.4/78 = 0.3, n_T = 64.4/92 = 0.7. Hence, x_B = 0.3, x_T = 0.7
p_B = 0.3 × 75 = 22.5 mm, p_T = 0.7 × 22 = 15.4 mm ;
Total vapour pressure = 22.5 + 15.4 = 37.9 mm
Mole fraction of benzene in the vapour phase = 22.5/37.9 = 0.59.

Teacher's Note:
1. Benzene is more volatile, so its mole fraction rises from 0.3 in the liquid to 0.59 in the vapour.

Case Study 4

There are many properties of the ideal solutions which do not depend upon the nature of the solute dissolved but depend only on the number of particles of the solute present in the solution. Such properties of the solution are called “colligative properties.” For example, when a non-volatile (solid) solute is dissolved in a liquid solvent, vapour pressure decreases, boiling point is raised while freezing point is depressed. The relative lowering of vapour that takes place or elevation in boiling point of the solvent or depression in the freezing point that take place are all colligative properties. These properties help us to calculate the molecular masses of the solutes. These properties are inter-related to one another. Thus, knowing the elevation in boiling (or the boiling point of the solution), depression in freezing point of that solution (or the freezing point of the solution) can be calculated. The study of depression in freezing point has a number of practical applications such as in making antifreeze solution or melting of ice on roads etc. As melting point of a crystalline solid has the same value as the freezing point of the melt, Rast method is an application of depression in melting point using a solid solvent to calculate the molecular mass of a solid solute.

Question 15. Solutions of different molalities of a solute were prepared and their boiling points determined. Then boiling points were plotted against molalities. What type of plot will be obtained ? How does it help to calculate the value of T_b° and K_b ?
Answer:
ΔT_b = K_b × m or T_b − T_b° = K_bm
or T_b = T_b° + K_bm
Hence, plot of T_b vs m will be linear with slope = K_b and intercept on T_b axis = T_b°. Hence, T_b° and K_b can be calculated.

Teacher's Note:
1. Compare with y = c + mx: the intercept is the boiling point of the pure solvent and the slope is K_b.

Question 16. The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g is added to 39.0 g of benzene (molar mass 78 g mol⁻¹). The vapour pressure of the solution then is 0.845 bar. What is the molar mass of the solid substance ?
Answer:
p° = 0.850 bar, p_s = 0.845 bar, w₂ = 0.5 g, w₁ = 39.0 g, M₁ = 78 g mol⁻¹
For a dilute solution, \( \frac{p^\circ - p_s}{p^\circ} = \frac{w_2 \times M_1}{M_2 \times w_1} \)
\( \frac{0.850 - 0.845}{0.850} = \frac{0.5 \times 78}{M_2 \times 39} \) or \( M_2 = \frac{0.5 \times 78 \times 0.850}{0.005 \times 39} \) = 170 g mol⁻¹

Teacher's Note:
1. A tiny pressure drop (0.005 bar) must be kept to three significant figures; rounding it changes the answer a lot.

Question 17. The vapour pressure of an aqueous solution of cane sugar (mol. mass = 342) is 732 mm at 100°C. Calculate the boiling point of the solution (K_b for water = 0.52°C).
Answer:
At 100°C, p° for water = 760 mm, p_s = 732 mm
For a dilute solution, \( \frac{p^\circ - p_s}{p^\circ} = \frac{n_2}{n_1} \) ∴ \( \frac{n_2}{n_1} = \frac{760 - 732}{760} = \frac{28}{760} \) = 0.03684
Molality = moles of solute per 1000 g water = \( \frac{n_2}{n_1} \times \frac{1000}{18} \) = 0.03684 × 55.56 = 2.047 m
ΔT_b = K_b × m = 0.52 × 2.047 = 1.064°
Boiling point of solution = 100 + 1.064 = 101.064°C

Teacher's Note:
1. At the normal boiling point of water its vapour pressure is 760 mm; this is the hidden data in the question.

Question 18. Calculate the amount of ice that will separate out when a solution containing 50 g of ethylene glycol in 200 g of water is cooled to − 9.3°C. (K_f for H₂O = 1.86 K kg mol⁻¹)
Answer:
ΔT_f = 9.3°, molar mass of ethylene glycol (C₂H₆O₂) = 62 g mol⁻¹
Let w₁ g be the mass of water left as liquid at −9.3°C. Then \( \Delta T_f = \frac{1000\, K_f\, w_2}{M_2\, w_1} \)
\( 9.3 = \frac{1000 \times 1.86 \times 50}{62 \times w_1} \) or w₁ = 161.3 g
∴ Ice separated = 200 − 161.3 = 38.7 g

Teacher's Note:
1. Only water freezes out; the glycol stays in the remaining liquid, so the solution becomes more concentrated until its freezing point equals −9.3°C.

Case Study 5

Colligative properties depend upon the number of particles of the solute present in a definite amount of the solvent. The calculation of molecular masses of solutes which do not undergo any dissociation or association in the solution is simple. However if a solute undergoes dissociation or association in the solution, number of particles increases in the former case and decreases in the latter case. For example, if NaCl is the solute and it dissociates completely, 1 mole of NaCl will produce 2 moles particles viz. Na⁺ and Cl⁻ ions (NaCl → Na⁺ + Cl⁻, in water). Similarly, substances like acetic acid, benzoic acid dimerise in solvents like benzene. If dimerisation is complete, 1 mole of benzoic acid will produce 0.5 mole of particles viz. dimers (2 C₆H₅COOH → (C₆H₅COOH)₂, in C₆H₆). In such cases, the observed value of colligative property is different from the theoretically calculated value. Consequently, molecular masses calculated will be different. In other words, ‘abnormal molecular masses’ will be observed. Further, dissociation or association may not be 100%. To account for all such complications, van’t Hoff introduced a factor i known as van’t Hoff factor and all the formulae for calculation of colligative properties have been modified.

Question 19. Arrange the following solutions in order of decreasing freezing points :
0.1 M K₃[Fe(CN)₆], 0.1 M BaCl₂, 0.1 M NaNO₃, 0.1 M glucose
Answer:
Glucose → no dissociation,
NaNO₃ → Na⁺ + NO₃⁻ (2 particles)
BaCl₂ → Ba²⁺ + 2 Cl⁻ (3 particles),
K₃[Fe(CN)₆] → 3 K⁺ + [Fe(CN)₆]³⁻ (4 particles).
Greater the number of particles, greater is the depression in freezing point or lower is the actual freezing point. Hence, correct order of freezing point is
0.1 M glucose > 0.1 M NaNO₃ > 0.1 M BaCl₂ > 0.1 M K₃[Fe(CN)₆].

Teacher's Note:
1. The complex ion [Fe(CN)₆]³⁻ does not break up further, so count it as one particle.

Question 20. What observed molecular mass do you expect for NaCl using colligative properties and why ?
Answer:
Expected observed molecular mass of NaCl will be half of its normal value, i.e., \( \frac{58.5}{2} \) = 29.25.
This is because NaCl dissociates completely in the solution to form two particles (Na⁺ and Cl⁻). Hence, observed colligative property value will be double and molecular mass will be half.

Teacher's Note:
1. Observed molar mass = normal molar mass / i; here i = 2.

Question 21. A decimolar solution of potassium ferrocyanide is 50% dissociated at 300 K. The osmotic pressure of the solution is nearly
(Given solution constant (R) = 8.314 JK⁻¹ mol⁻¹)
Answer:
K₄[Fe(CN)₆] ⇄ 4 K⁺ + [Fe(CN)₆]⁴⁻; n = 5 particles, α = 0.5
i = 1 + (n − 1)α = 1 + (5 − 1) × 0.5 = 3
C = 0.1 mol L⁻¹ = 0.1 × 10³ mol m⁻³ = 100 mol m⁻³
π = iCRT = 3 × 100 × 8.314 × 300 = 7.48 × 10⁵ N m⁻²
Osmotic pressure is nearly 7.48 × 10⁵ Pa (about 7.4 atm).

Teacher's Note:
1. With R in J K⁻¹ mol⁻¹, concentration must be in mol m⁻³ to get the answer in pascals.

Question 22. Molality of sodium chloride solution whose elevation in boiling point is numerically equal to the depression in freezing point of 0.2 m aluminium sulphate solution in water is close to (assume complete dissociation and take K_b and K_f for water as 0.52 and 1.86 K kg mol⁻¹ respectively)
Answer:
Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻, i = 5
ΔT_f = i K_f m = 5 × 1.86 × 0.2 = 1.86 K
For NaCl, i = 2. ΔT_b = i K_b m = 2 × 0.52 × m = 1.86
∴ m = \( \frac{1.86}{1.04} \) = 1.79 m

Teacher's Note:
1. Set the two expressions equal including i on both sides; forgetting i for either salt gives a wrong answer.

Case Study 6

There are many phenomena which we observe in nature or at home. For example, raw mangoes shrivel when placed in brine (salt solution), wilted flowers revive when placed in fresh water, blood cells collapse when suspended in saline water etc. If we look into these processes, we find one thing common in all, i.e., all these substances are bound by membranes. These membranes contain microscopic holes or pores. Small solvent molecules can pass through them, bigger molecules of the solute cannot. This process of flow of the solvent through semipermeable membranes is called ‘osmosis’. If a solvent and a solution are connected through semipermeable membrane, the pressure that has to be applied on the solution to prevent the entry of the solvent into the solution is called ‘osmotic pressure’. Osmotic pressure increases with increase in concentration and increase in temperature (i.e. π ∝ C × T or π = CRT, where R is constant, similar to gas constant). This expression is applicable only to non-electrolytes. For electrolytes, the formula is modified. Like other colligative properties, osmotic pressure is also a colligative property and can be used for calculation of molar masses of the solutes (especially for proteins, polymers etc.). In osmosis, there is a net flow of solvent from solvent to solution. However, this can be reversed by applying pressure greater than osmotic pressure on the solution. This process is called ‘reverse osmosis’ and is used in the purification of water (e.g. desalination of sea water). Also, it may be remembered that solutions having same molar concentration will have same osmotic pressure. Such solutions are called ‘isotonic solutions’.

Case Study 7

In normal life, we rarely come across pure substances. Most of these are mixtures containing two or more pure substances in varying composition. If a mixture of two or more chemically non-reacting substances is homogeneous in which composition of the components can be varied within certain limits, it is called a ‘solution’. The simplest solutions are made up of two components called solute and solvent. Depending upon the physical state of the solvent (solid, liquid or gas), solutions are classified as solid solutions, liquid solutions and gaseous solutions, the most significant being liquid solutions. Depending upon the amount of solute dissolved in a definite amount of the solvent, there are different ways in which the concentration of the solution can be expressed such as percentage, strength, molarity, molality, normality, mole fraction, mass fraction, parts per million parts (ppm) etc. If a solution is diluted, the new normality or molarity of the solution can be calculated by applying normality equation or molarity equation. Similarly, if two non-reacting solutions are mixed, the normality or molarity of the final solution formed can be calculated by applying similar equations.