//M0//QN1//CS//DL0//EQ

At a certain temperature every pure liquid has a certain vapour pressure. But when a solid or liquid solute is added to it, its vapour pressure changes. colligative properties for the vapour pressure of solutions were studied in 1986 by F. M. Raoult and the rule given by him is called Raoult's law. When both the solute and the solvent are liquids, their own vapour pressures are called their partial pressures. It depends on the mole fraction of the component present in the solution. If both the liquid components in the solution are not equally volatile, their mole fractions are diﬀerent in the vapour state than in the solution state. If the solute is a solid, the vapour pressure of the solution is lower than that of the pure solvent. This decrease is called relative decrease in vapour pressure. The relative decrease in vapour pressure is proportional to the mole fraction of solute in the solution. This rule is useful for calculating the molecular mass of a solute.(i) The vapour pressures of pure components A and B at a certain temperature are 108 and 36 torr respectively. What will be the vapour pressure of solution of equal moles of components A and B in solution? (ii) What will be mole fraction of component B in vapour phase?(iii) The vapour pressure of a solvent decreased by 10 mm Hg when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in vapour pressure is to be 20 mm Hg?(iv) The vapour pressure of pure liquid A is 10 torr and at the same temperature when 1 g solid

B is dissolved in 20 g of A , its vapour pressure is reduced to 9.0 torr. If the molecular mass of A is 200 amu, then the molecular mass of B is :

//X

(i) 72 torr
P T = P°A χ A + P°B χ B
P T = (108 × 0.5) + (36 × 0.5)
P T = 54 + 18
= 72 torr
(ii) 0.25
Y B = =
= =
= 0.25
(iii) 0.6
∆P = P° × χ solute
P° = = 50 mm HG
∆P = P° × χ New solute
χ New solute = = 0.4
∴ χ solute + χ solvent = 1 ⇒ χ sol vent = 1 – 0.4 = 0.6
(iv) 90 a.m.u.
= χ B
= χ B ⇒ χ B = 0.1
n A = = 0.1 mol
n B =
∴ χ B = ⇒ = 0.1
= 0.1
M B = = 90

//M0//QN2//CS//DL0

Ideal solutions follow Raoult's law while non-ideal solutions do not follow Raoult's law. Similarly, volume and enthalpy do not change when ideal solutions are mixed, whereas for non-ideal solutions these properties vary depending on their characteristics. Non-ideal solutions can be classiﬁed into two forms, positive deviation and negative deviation. At particular composition both solutions form an azeotropic mixture.(i) Which of the following is not correct for ideal solution?

(A) ∆S mix = 0 (B) ∆V mix = 0
(C) ∆H mix = 0 (D) It follows Raoult's law
(ii) In an azeotropic mixture option of HCl and H 2 O is:
(A) 48% HCl (B) 22.2% HCl
(C) 36% HCl (D) 20.2% HCl
(iii) Which of the following did not show a positive deviation from Raoult's law?
(A) Benzene-chloroform
(B) Benzene-acetone
(C) Benzene-ethanol
(D) Benzene-carbon tetrachloride

//X

(i) (A) ∆S mix = 0
For a process to occur spontaneously, the Entropy of mixing (ΔSmix) must always be positive ( > 0).
(ii) (D) 20.2% HCl
Hydrochloric acid (HCl) and water (H 2 O) form a maximum boiling azeotrope. This occurs when a solution shows a large negative deviation from Raoult's Law.
The specific composition that boils at a constant temperature (approx. 108.6°C) is 20.2% HCl and 79.8% water by mass.
At this point, the liquid phase and vapor phase have the exact same composition, making it impossible to separate them further by simple distillation.
(iii) (A) Benzene-chloroform
Benzene-Chloroform: In this specific mixture, there is a slight interaction between the π-electrons of the benzene ring and the acidic hydrogen of chloroform (CHCl 3). Because the A−B forces are stronger, the molecules are less likely to escape into the vapor phase, resulting in a lower vapor pressure (negative deviation).

//M0//QN3//CS//DL0//EQ

Pure Solvent when a solute is added to form a homogeneous solution, the properties of the pure solvent such as boiling point, freezing point and vapour pressure etc. change. All these properties are called Colligative properties. Colligative properties are very useful in everyday life. For example, a mixture of ethylene glycol and water in vehicle radiators is useful as an anti-freezing agent. Solution M is made by mixing the given ethanol with water. The ethanol mole-fraction in this mixture is 0.9. Molal depression constant of water (Kf water) = 1.86 K. kg mol–1, Molal depression constant of ethanol (Kf ethanol) = 2.0 K.kg.mol–1, Molal elevation constant of water (Kb water) 0.52

K.kg.mol–1, Molal elevation constant of ethanol

(Kb ethanol) = 1.2 K.kg.mole–1, Standard freezing point of water = 273K, Standard freezing point of ethanol = 155.7K Standard boiling point of water = 373K, Standard boiling point of ethanol = 351.5K, Vapour pressure of pure water = 32.8 mm Hg, Vapour pressure of pure ethanol = 40 mm Molecular mass of water = 18 g.mol–1, Molecular mass of ethanol = 46 g.mol–1. Assuming the solvent is non-volatile and immiscible and the solution is dilute and ideal, answer the following questions:(i) What will be the freezing point of solution M?(ii) What will be the vapour for pressure of

solution M?(iii) If water is added to solution M and mole-function of water in the solution is 0.9. What will be the boiling point of the solution?

//X

(i) 150.9K
n water = 1 – 0.9 = 0.1 mol
W ethenol = 0.9 × 46 = 41.4 gm = 0.0414 kg.
∆T g = Kg . m ⇒ 2 × = 4.83 K
T f = T°f – ∆T f = 155.7 – 4.83 = 150.9 K
(ii) 36.0 mm Hg
P° = 40 mm Hg, X = 0.9
P s = P°C 2 H 5 OH ⋅ X C 2 H 5 OH
= 40 × 0.9
= 36.0 mm Hg
(iii) 376.2K
W water = 0.9 × 18 = 16.2g = 0.0162 Kg
n Ethanol = 1 – 0.9 = 0.1 mol
m = ≈ 6.173 mol / kg
∆T b = K b × m ⇒ 0.52 × 6.173
= 3.2 K
T b = T°b × ∆T b = 373 + 3.2
= 376.2 K

//M0//QN4//CS//DL0//EQ

When any colligative property is calculated using any formula in electrolysis the experimental value is always greater than the theoretical value (the value obtained by calculation). Because electrolytes always dissociate in solution. Similarly, when electrolytes is associated, (e.g., acetic acid in benzene) the experimental value of a colligative property is less than the theoretical value. Thus, the experimentally observed molecular mass diﬀers from the theoretical value. It is called abnormal molar mass. For all these cases the correction factor '(i)' is introduced as the Van't Hoﬀ factor. If the value of the Van't Hoﬀ factor (i) is known, the degree of dissociation or degree of association for the solute can be calculated.(i) Which of the following aqueous solution will have the highest freezing point?

(A) 0.1 M urea (B) 0.1 M sucrose
(C) 0.1 M AlCl 3 (D) 0.1 M K 4 [Fe(CN)6]
(ii) The boiling point of 0.1 molal K 4[Fe(CN)6] solution will be (K b for water = 0.52 K.kg. mole–1)(iii) The Van't Hoff factor for 0.1 M, Ba(NO 3)2 solution is 2.74. The degree of dissociation is:

//X

(i) (B) 0.1 M sucrose
∆T f a The number of particles
NaCl → Na+ + Cl– ⇒ i = 2
Sucrose → Non-electrolyte ⇒ i = 1
AlCl 3 → Al 3+ + 3Cl– ⇒ i = 4
K u [Fe(CN)6] → 4K+ + [Fe(CN)6]4– ⇒ i = 5
Since, sucrose produce the fewest particles
⇒ Highest freezing point.
(ii) 100.26°C
K 4 [Fe(CN)6] → 4K+ + [Fe(CN)6]4–
Assuming 100% dissociation
i = 4 + 1 = 5
∆T b = i ⋅ K b ⋅ m ⇒ 5 × 0.52 × 0.1 = 0.26°C
T b = 100°C + 0.26°C ⇒ 100.26°C
(iii) 87%
Ba(NO 3)2 → Ba 2+ + 2NO–3
i = 1 + a (n – 1)
2.74 = 1 + a (3 – 1)
a = ⇒ 0.87
a = 87%

//M0//QN5//CS//DL0//EQ

A pharmaceutical company is for mutating an 'Oral Rehydration Solution (ORS)' for dehydration treatment. ORS must be isotonic with human blood (osmotic pressure = 7.6 atm at 37°C) so that no harmful flow of water occurs across cell membranes. For one test batch, chemist dissolve the following in 500 ml of water:

18 g Glucose (molar mass = 180 g / mol)
0.90 g NaCl (molar mass = 58.5 g / mol)
0.75 g KCl (molar mass = 74.5 g / mol)
Because NaCl and KCl do not fully dissociate in real solution, their Van't Hoff factor (i) is taken as 1.9. The company want to determine whether this ORS sample is sufficiently isotonic to be safe for consumption.(i) Calculate the molarity of glucose in the solution.(ii) Using the Van't Hoff factor calculate the effective molarity of NaCl and KCl.(iii) Calculate the total osmotic pressure of the solution at 37°C. (R = 0.0821 atm K–1 mol–1)(iv) Is this ORS sample isotonic with blood? Justify your answer with calculated values.

//X

(i) Moles of Glucose = = 0.10 mol
∴ Molar ity = = 0.20 M
(ii) For NaCl For KCl
NaCl = KCl =
= 0.01538 mol = 0.01007 mol
Effective molarity Effective molarity
= × 1.9 = × 1.9
= 0.058 M = 0.038 M
(iii) Total effective molarity
= 0.20 + 0.058 + 0.038
= 0.296 M
Osmotic pressure (π) = MRT
= 0.296 × 0.0821 × 310
= 7.53 atm
(iv) Blood osmotic pressure = 7.6 atm
ORS = 7.53 atm → very close, hence nearly isotonic
∴ Safe for hydration.

//M0//QN6//CS//DL0//EQ

A chemical engineer is studying how a sparingly soluble salt, Ag 2 SO 4 behave when discharged in industrial wastewater. The solubility of the salt is temperature - dependent. At 25°C, the solubility of Ag 2 SO 4 is 1.42 g per litre, while at 40°C it rises to 2.10 g per litre. The dissociation of the salt is represented as: Ag 2 2Ag+(aq) + The engineer must determine:

The solubility of Ag 2 SO 4 in mol/L.
The solubility product (K sp).
Whether precipitation will occur in waste water.
How temperature influences solubility.
Molar mass of Ag 2 SO 4 = 312 g/mol The factory wastewater at one outlet contain: [Ag+] = 0.015 M [SO 4 2–] = 0.004 M(i) Calculate the solubility (s) of Ag 2 SO 4 in mol / L at 25°C.(ii) Using the dissociation formula, calculate the K sp of Ag 2 SO 4 at 25°C.(iii) Calculate the ion product (Q) for waste water and determine whether precipitation will occur.(iv) Explain why the solubility increases when temperature increases to 40°C.

//X

(i) Solubility (s) = =
(s) = 4.55 × 10–3 mol/L
(ii) Ag 2 2Ag+ + SO 4 2–
(2s) (s)
↓ ↓
9.10 × 10–3 4.55 × 10–3
mol/L mol/L
K sp = (2s)2 × (s)
= 4s 3
= 4 × (4.55 × 10–3)3
K sp = 3.78 × 10–7
(iii) Ionic product Q = [Ag+]2 × [SO 4 2–]
= (0.015)2 × (0.004)
= 9 × 10–7
Since Q > K sp precipitation will occur.
(iv) Dissolution of Ag 2 SO 4 is endothermic, so it absorbs heat.