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Questions:

Case Study Questions

14 x 4 = 56

Observe the table in which azeotropic mixtures are given along with their boiling points of pure components and azeotropes and answer the questions that follow.

Some Azeotropic Mixtures
A	B	Minimum Boiling Azeotropes	Boiling Points
A	B	Minimum Boiling Azeotropes	A	B	Mixture Azeotropes
H₂O	C₂H₅OH	95.37%	373K	35 13K	351.15
H₂O	C₃H₇OH	71.69%	373K	370.19K	350.72
CH₃CQCH₃	CS₂	67%	329.25K	319.25K	312.30
A	B	Maximum Boiling Azeotropes	A	B	Mixture Azeotropes
H₂O	HCI	20.3%	373K	188K	383K
H₂O	HNO₃	68.0%	373K	359K	393.5K
H₂O	HClO₄	71.6%	373K	383K	476K

(a) What type of deviation is shown by minimum boiling azeotropes?
(b) Why does H₂O and HCI mixture form maximum boiling azeotropes?
(c) How can be separate azeotropic mixture?
(d) Give one example of ideal solution. What type of liquids form ideal solutions?
(e) What are azeotropes?
(f) What will be vapour pressure of maximum boiling azeotrops?
(g) At what mole fraction of A', vapour pressure of A (P^o_A= 450 mm) and vapour pressure of B (P^o_B = 200 mm) in solution will be equal if both A and B form ideal solution.

Observe the graph between mole fraction of HCI gas dissolved in cyclohexane Vs equilibrium pressure of HCI(g) and answer the questions based on graph.

(a) Which law is depicted by this graph?
(b) What is mathematical expression for Henry's law?
(c) What does slope represent?
(d) What is effect of temperature and pressure on solubility of gas in liquid?
(e) Name two factors which affect the value of K_H?

Solution playa very important role in our daily life. Alloys, homogeneous mixture of metal are solution of solid in solid. 1 ppm (parts per million) of fluoride ions prevent tooth decay. All intravenous injections must be isotonic with our body fluids, i.e. should have same concentration as blood plasma. Diabetic patients are more likely to have heart attack and high blood pressure due to higher glucose level in blood. Common salt increase blood pressure because Na⁺ mixes up with blood. Aquatic species are more comfortable in cold water than warm water.
(a) 0.1 M glucose is not isotonic with 0.1 M KCI solutions. Why?
(b) A solution contains 5.85 g of NaCI (molar mass 58.5 g mol^-1 per litre of the solution, has osmotic pressure 4.75 atm of 27°c Calculate the degree of dissociation of NaCI in this solution. [R = 0.0821 L atm K^-1 mol ^-1]
(c) What will happen if blood cells are placed in saline water (hypertonic solution)?
(d) Calculate the molality of ethanol solution in which mole fraction of water is 0.88.
(e) What will happen if pressure applied on solution side is more than osmotic pressure, when solvent and solution are separated by semipermeable membrane?

Solutions are homogeneous mixture of two or more substances. Ideal solution follow Raoult's law. The vapour pressure of each component is directly proportional to their mole fraction if both solute and solvent are volatile. The relative lowering of vapour pressure is equal to mole fraction of solute if only solvent is volatile.
Non-ideal solution form azeotropes which cannot be separated by 'tractional distillation. Henry's law is special case of Raoult's law applicable to gases dissolved in liquids.
Colligative properties depend upon number of particles of solute. Relative lowering of vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure are colligative properties which depend upon mole fraction of solute, molality and molarity of solutions. When solute undergoes either association or dissociation, molecular mass determined by colligative property will be abnormal.
van't Hofffactor is used in such cases which is ratio of normal molecular mass over observed molar mass.
(a) 50 ml of an aqueous solution of glucose (Molar mass 180 g/mol) contains 6.02 x 10²² molecules. What is molarity?
(b) Identify which liquid has lower vapour pressure at 90°C if boiling point of liquid 'A' and 'B' are 140°C and 180° respectively.
(c) What type of azeotropes are formed by nonideal solution showing negative deviation from Raoult's law?
(d) For a 5% solution of area (molar mass 60 g mol ^-1), calculate the osmotic pressure at 300 K (R = 0.0821 L atm k^-1 )
(e) Predict the van't Hoff factor
(i) CH3 COOH dissolved in water,
(ii) dissolved in benzene.
(f) Why meat is preserved for longer time by salting?
(g) Why 0.1 M KCI has higher boiling point than 0.1 M glucose solution?

Read the, passage given below and answer the following questions:
If some solute is added to a solvent, the boiling point of solution increases. This is known as elevation in boiling point.
\(\Delta T_{b}=K_{b} m\) ,where, K_b = Molal elevation constant
\(\Delta T_{b} \propto m\)
Hence, it is a colligative property.
Also, \(K_{b}=\frac{M R T_{b}^{2}}{\Delta_{\text {vap }} H \times 1000}\)
where, M= Molar mass of solvent
\(\Delta_{\text {vap }} H\) = Enthalpy of vaporisation
Molar mass can also be calculated using elevation in boiling point.
\(M_{B}=\frac{K_{b} \times W_{B} \times 1000}{\Delta T_{b} \times W_{A}}\)
In these questions (i-iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
(i) Assertion: In a pressure cooker, the water is brought to boil. The cooker is then removed from the stove. Now on removing the lid of pressure cooker, the water starts boiling again.
Reason: The impurities in water bring down its boiling point.
(ii) Assertion: On dissolving 3.24 g of sulphur in 40 g of benzene, boiling point of solution get higher than that of benzene by 0.081 K, then the formula of sulphur is S₈. (K_b for benzene = 2.53 K kg mol
Reason: Molecular mass of sulphur comes out to be 253.
(iii) Assertion: When sugar is added to water, boiling point of water increases.
Reason: When a non-volatile solute is added to a solvent, elevation in boiling point is observed.
(iv) Assertion: Cooking time in pressure cookers is reduced.
Reason: Boiling point inside the pressure cooker in raised.

Read the passage given below and answer the following questions:
The phenomenon of the flow of solvent through a semipermeable membrane from pure solvent to the solution is called osmosis.
Sometimes a pressure is applied to stop the process of osmosis, this is known as osmotic pressure. It is denoted by \(\pi \). Osmotic pressure is expressed as: \(\pi \) = CRT
Since, osmotic pressure depends upon the molar concentration of solution, therefore it is a colligative property.
In these questions (i-iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
(i) Assertion: If red blood cells were removed from the body and placed in pure water, pressure inside the cells increases.
Reason: The concentration of salt content in the cells increases.
(ii) Assertion: When a solution is separated from the pure solvent by a semipermeable membrane, the solvent molecules pass through it from pure solvent side to the solution side.
Reason: Diffusion of solvent occurs from a region of high concentration to a region of low concentration solution.
(iii) Assertion: Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.
Reason: Osmotic pressure is not a colligative property.
(iv) Assertion: The preservation of meat by salting and fruits by adding sugar protects against bacterial action.
Reason: A bacterium on salted meat or candid fruit loses water due to osmosis shrivels and ultimately dies.

Read the passage given below and answer the following questions:
According to Raoult's law, the partial pressure of two components of the solution may be given as:
\(p_{A}=p_{A}^{\circ} x_{A} \text { and } p_{B}=p_{B}^{\circ} x_{B}\)
For an ideal solution (obeys Raoult's law always)
\(\Delta H_{\operatorname{mix}}=0, \Delta V_{\operatorname{mix}}=0\)
All solutions do not obey Raoult's law over entire range of concentration. These are known as non-ideal solutions.
For non-ideal solutions, \(p_{A} \neq p_{A}^{\circ} x_{A} \text { or } p_{B} \neq p_{B}^{\circ} x_{B}\)
Positive deviation \(\Rightarrow p_{A}>p_{A}^{\circ} x_{A} \text { and } p_{B}>p_{A}^{\circ} x_{B}\)
Negative deviation \(\Rightarrow p_{A}<p_{A}^{\circ} x_{A} \text { and } p_{B}<p_{B}^{\circ} x_{B}\)
In these questions (i-iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statemeht but reason is correct statement.
(i) Assertion: An ideal solution obeys Raoults law.
Regson : In an ideal solution, solute-solute as well as solvent-solvent interactions are similar to solutesolvent interactions.
(ii) Assertion: Acetone and aniline show negative deviations.
Reason: H-bonding between acetone and aniline is stronger than that between acetone-acetone and aniline-aniline.
(iii) Assertion: The solutions which show negative deviations from Raoult's law are called maximum boiling azeotropes.
Reason: 68% nitric acid and 32% water by mass form maximum boiling azeotrope.
(iv) Assertion: \(\Delta H_{\mathrm{mix}}\) and \(\Delta V_{\operatorname{mix}}\)are positive for an ideal solution.
Reason: The interactions between the particles of the components of an ideal solution are almost identical as between particles in the liquids.

Read the passage given below and answer the following questions:
At the freezing point of a solvent, the solid and the liquid are in equilibrium. Therefore, a solution will freeze when its vapour pressure becomes equal to the vapour pressure of the pure solid solvent.
It has been observed that when a non-volatile solute is added to a solvent, the freezing point of the solution is
always lower than that of the pure solvent. Depression in freezing point can be given as, \(\Delta T_{f}=K_{f} m\)
Where, K_f = Molal freezing point depression constant
or we can write, \(\Delta T_{f}=\frac{K_{f} \times W_{B} \times 1000}{W_{A}^{1} \times M_{B}}\)
In these questions (i-iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
(i) Assertion: 0.1 M solution of glucose has same depression in the freezing point as 0.1 M solution of urea.
Reason: K_ffor both has same value.
(ii) Assertion: Larger the value of cryoscopic constant of the solvent, lesser will be the freezing point of the solution.
Reason: Extent of depression in the freezing point depends on the nature of the solvent.
(iii) Assertion: The water pouch of instant cold pack for treating athletic injuries breaks when squeezed and NH₄NO₃ dissolves thus lowering the temperature.
Reason: Addition of non-volatile solute into solvent results into depression of freezing point of solvent.
(iv) Assertion: If a non-volatile solute is mixed in a solution then elevation in boiling point and depression in freezing point both will be same.
Reason: Elevation in boiling point and depression in freezing point both depend on number of particles of solute.

Read the passage given below and answer the following questions:
The solubility of gases increases with increase of pressure. William Henry made a systematic investigation of the solubility of a gas in a liquid. According to Henry's law "the mass of a gas dissolved per unit volume of the solvent at constant temperature is directly proportional to the pressure of the gas in equilibrium with the solution".
Dalton during the same period also concluded independently that the solubility of a gas in a liquid solution depends upon the partial pressure of the gas. If we use the mole fraction of gas in the solution as a measure of its solubility, then Henry's law can be modified as "the partial pressure of the gas in the vapour phase is directly proportional to the mole fraction of the gas in the solution":
(i) Henry's law constant for the solubility of methane in benzene at 298 K is 4.27 X 10⁵ mm Hg. The solubility of methane in benzene at 298 K under 760 mm Hg is

(a) 4.27 x 10^-5	(b) 1.78 x 10^-3
(c) 4.27 X 10^-3	(d) 1.78 x 10^-5

(ii) The partial pressure of ethane over a saturated solution containing 6.56 x 10^-2 g of ethane is 1 bar. If the solution contains 5.00 x 10^-2 g of ethane then what will be the partial pressure (in bar) of the gas?

(a) 0.762

(b) 1.312

(c) 3.81

(d) 5.0

(iii) K_H (K bar) values for Ar_(g), CO_2(g), HCHO_(g)and CH_4(g) are 40.39, l.67, 1.83 x 10^-5 and 0.413 respectively.
Arrange these gases in the order of their increasing solubility.

\(\text {(a) } \mathrm{HCHO}<\mathrm{CH}_{4}<\mathrm{CO}_{2}<\mathrm{Ar}\)	\(\text { (b) } \mathrm{HCHO}<\mathrm{CO}_{2}<\mathrm{CH}_{4}<\mathrm{Ar}\)
\(\text {(c) } \mathrm{Ar}<\mathrm{CO}_{2}<\mathrm{CH}_{4}<\mathrm{HCHO}\)	\(\text { (d) } \mathrm{Ar}<\mathrm{CH}_{4}<\mathrm{CO}_{2}<\mathrm{HCHO}\)

(iv) Which of the following statements is correct

(a) K_H increases with increase of temperature

(b) K_H decreases with increase of temperature

(c) K_H remains constant with increase oftemperature

(d) K_H first increases then decreases, with increase of temperature.

10.

Read the passage given below and answer the following questions:
Few colligative properties are:
(a) relative lowering of vapour pressure: depends only on molar concentration of solute (mole fraction) and independent of its nature.
(b) depression in freezing point: it is proportional to the molal concentration of solution.
(c) elevation of boiling point: it is proportional to the molal concentration of solute.
(d) osmotic pressure: it is proportional to the molar concentration of solute.
A solution of glucose is prepared with 0.052 g at glucose in 80.2 g of water. (K_f = 1.86 K kg mol^-1 and K_b = 5.2 K kg mol^-1)
The following questions are multiple choice questions. Choose the most appropriate answer:
(i) Molality of the given solution is

(a) 0.0052 m

(b) 0.0036 m

(c) 0.0006 m

(d) 1.29 m

(ii) Boiling point for the solution will be

(a) 373.05 K

(b) 373.15 K

(c) 373.02 K

(d) 372.98 K

(iii) The depression in freezing point of solution will be

(a) 0.0187 K

(b) 0.035 K

(c) 0.082 K

(d) 0.067 K

(iv) Mole fraction of glucose in the given solution is

(a) 6.28 x 10^-5

(b) 1.23 x 10^-4

(c) 0.00625

(b) 0.00028

11.

Read the passage given below and answer the following questions:
The properties of the solutions which depend only on the number of solute particles but not on the nature of the solute are called colligative properties. Relative lowering in vapour pressure is also an example of colligative properties.
For an experiment, sugar solution is prepared for which lowering in vapour pressure was found to be 0.061 mm of Hg. (Vapour pressure of water at 20°C is 17.5 mm of Hg.)
The following questions are multiple choice questions. Choose the most appropriate answer:
(i) Relative lowering of vapour pressure for the given solution is

(a) 0.00348

(b) 0.061

(c) 0.122

(d) 1.75

(ii) The vapour pressure (mm of Hg) of solution will be

(a) 17.5

(b) 0.61

(c) 17.439

(d) 0.00348

(iii) Mole fraction of sugar in the solution is

(a) 0.00348

(b) 0.9965

(c) 0.061

(d) 1.75

(iv) The vapour pressure (mm of Hg) of water at 293 K when 25 g of glucose is dissolved in 450 g of water is

(a) 17.2

(b) 17.4

(c) 17.120

(d) 17.02

12.

Read the passage given below and answer the following questions:
An ideal solution may be defined as the solution which obeys Raoult's law exactly over the entire range of concentration. The solutions for which vapour pressure is either higher or lower than that predicted by Raoult's law are called non-ideal solutions.
Non-ideal solutions can show either positive or negative deviations from Raoult's law depending on whether the A-B interactions in solution are stronger or weaker than A - A and B - B interactions.
The following questions are multiple choice questions. Choose the most appropriate answer:
(i) Which of the following solutions is/are ideal solution(s)?
(i) Bromoethane and iodoethane (ii) Acetone and chloroform
(iii) Benzene and acetone (iv)n-heptane and n-hexane

(a) only 1

(b) I and II

(c) II and III

(d) I and IV

(ii) Which of the following is not true for positive deviations?

(a) The A-B interactions in solution are weaker than the A -A and B -B interactions.

(b) \(P_{A}<P_{A}^{\circ} x_{A} \text { and } P_{B}<P_{B}^{\circ} x_{B}\)

(c) Carbon tetrachloride and chloroform mixture is an example of positive deviations.

(d) All of these

(iii) For water and nitric acid mixture which of the given graph is correct?


(C) Both of these	(d) None of these

(iv) Water- HCl mixture
I. shows positive deviations II. forms minimum boiling azeotrope
III. shows negative deviations IV. forms maximum boiling azeotrope

(a) I and II	(b) II and III
(c) I and IV	(d) III and IV

13.

Read the passage given below and answer the following questions:
At 298 K, the vapour pressure of pure benzene, C₆H₆ is 0.256 bar and the vapour pressure of pure toluene
C₆H₅CH₃ is 0.0925 bar. Two mixtures were prepared as follows:
(i) 7.8 g of C₆H₆ + 9.2 g of toluene
(ii) 3.9 g of C₆H₆ + 13.8 g of toluene
The following questions are multiple choice questions. Choose the most appropriate answer:
(i) The total vapour pressure (bar) of solution 1 is

(a) 0.128

(b) 0.174

(c) 0.198

(d) 0.258

(ii) Which of the given solutions have higher vapour pressure?

(a) I	(b) II
(c) Both have equal vapour pressure	(d) Cannot be predicted

(iii) Mole fraction of benzene in vapour phase in solution 1 is

(a) 0.128

(b) 0.174

(c) 0.734

(d) 0.266

(iv) Solution I is an example of a/an

(a) ideal solution	(b) non-ideal solution with positive deviation
(c) non-ideal solution with negative deviation	(d) can't be predicted

14.

Read the passage given below and answer the following questions:
The concentration of a solute is very important in studying chemical reactions because it determines how often molecules collide in solution and thus indirectly determine the rate of reactions and the conditions at equilibrium.
There are several ways to express the amount of solute present in a solution. The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. Concentration can be expressed in terms of molarity, molality, parts per million, mass percentage, volume percentage, etc.
The following questions are multiple choice questions. Choose the most appropriate answer:
(i) The molarity (in mol L^-1) of the given solution will be

(a) 1.56

(b) 1.89

(c) 0.263

(d) 1.44

(ii) Which of the following is correct relationship between mole fraction and molality?

\(\text { (a) } x_{2}=\frac{m M_{1}}{1+m M_{1}}\)	\(\text { (b) } x_{2}=\frac{m M_{1}}{1-m M_{1}}\)
\(\text { (c) } x_{2}=\frac{1+m M_{1}}{m M_{1}}\)	\(\text { (d) } x_{2}=\frac{1-m M_{1}}{m M_{1}}\)

(iii) Which of the following is temperature dependent?

(a) Molarity	(b) Molality
(c) Mole fraction	(d) Mass percentage

(iv) Which of the following is true for an aqueous solution of the solute in terms of concentration?

(a) 1 M = 1 m	(b) 1M > 1m
(c) 1M < 1 m	(d) Cannot be predicted

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Answers Key:

Case Study Questions

14 x 4 = 56

(a) Positive deviation from Raoult's law.
(b) It is because force of attraction between H₂O and HCI is more than H₂O-H₂O and HCI-HCI.
(c) Azeotropic distillation. Add benzene to water and ethanol, all three will get separated.
(d) Hexane and heptane form ideal solution. Those compounds of same family having similar forces of attraction form ideal solution.
(e) These are constant boiling mixtures which distill out unchanged in their composition.
(f) The vapour pressure of azeotropic mixture will lower than vapour pressure of each component due to stronger force of attraction, therefore, boiling point will be higher.
(g) \(P_{A}^{\circ} x_{A}=P_{B}^{\circ} x_{B}\)
450 x x_A= 200 (1 -x_A ) [\(\therefore\) xB = 1 -x_A ]
450 x _A = 200 - 200 x _A
650 x _A = 200
\(x_{A}=\frac{200}{650}=\frac{4}{13}=0.30\)

(a) Henry's law.
(b) P_gas = KH X _gas
Where P_gas = Partial pressure of gas
X_gas= Mole fraction of gas
K_H = Henry's law constant.
(c) Slope = K_H, Henry's law constant.
(d) Solubility of gas in liquid increase with increase in pressure. Solubility of gas in liquid increase with decrease in temperature.
(e) (i) Nature of gas
(ii) Temperature

(a) This is because, they do not have same osmotic pressure as number of particles are different.
(b) Given:
Osmotic pressure (\(\pi\)) = 4.75 atm
W_B= 5.85 g mol^-1
M_B = 58.5 g mol^-1
T = 27°C = 273 + 27 = 300K
R = 0.0821 Latm K^-1 mol^-1
\(\alpha\) = ?
\(\pi\)V = inRT
NaCI \(\rightarrow\)Na⁺ + Cl^-1
\(4.75 \times 1=i \times \frac{W_{B}}{M_{B}} \times R \times T\)
n = 2
\(4.75=i \times \frac{5.85}{58.5} \times 0.0821 \times 300\)
\(\Rightarrow \ i=\frac{4.75}{2.463}=1.928\)
\(\alpha=\frac{i-1}{n-1}=\frac{1928-1}{2-1}\)
= 0.928
\(\Rightarrow \ \alpha=92.8 \%\)
(c) These will shrink
(d) X_H2O = 0.88
xC₂H₅OH = 1- 0.88 = 0.12
\(x \mathrm{~B}=\frac{\mathrm{m}}{\mathrm{m}+\frac{1000}{\mathrm{M}_{\mathrm{A}}}}\)
\(\Rightarrow \ 0.12=\frac{\mathrm{m}}{\mathrm{m}+\frac{1000}{18}}\)
\(\Rightarrow \ 0.12 \mathrm{~m}+\frac{120}{18}=\mathrm{m}\)
\(\Rightarrow \ 0.88 \mathrm{~m}=\frac{120}{18}\)
\(\mathrm{m}=\frac{500}{66}=7.57 \mathrm{~mol} / \mathrm{kg}\)
(e) The process of reverse osmosis will take place. It is used for desalination of water.

(a) \(\mathrm{M}=\frac{\text { No. of moles }}{\text { Litres of solution }}=\frac{6.02 \times 10^{22}}{6.02 \times 10^{23}} \times \frac{1000}{50}\)
= 2M
(b) 'B' will have lower vapour pressure because its boiling point is higher.
(c) Maximum boiling azeotropes.
(d) \(\pi V=n R T \Rightarrow \pi \times 0.1 \mathrm{~L}=\frac{5}{60} \times 0.0821 \times 300\)
\(\Rightarrow \pi=\frac{24.63}{1.2}=20.52 \mathrm{~atm}[\because 100 \mathrm{~mL}=0.1 \mathrm{~L}]\)
(e) (i) i > 1, because dissociation takes place.
(ii) i < 1, because association takes place.
(f) Salt inhibits the growth of microorganisms by drawing out water from microbial cells through osmosis 20% salt is needed to kill most species of unwanted bacteria.
(g) It is because KCI dissociates into ions, it has double particles as compared to glucose.
Therefore, elevation in boiling point is double.

(i) (c): In pressure cooker, water boils above 100°C. When the lid of cooker is opened, pressure is lowered so that boiling point decreases and water boils again.
(ii) (a) : \(M_{B}=\frac{K_{b} \times 1000 \times W_{B}}{\Delta T_{b} \times W_{A}}\)
\(K_{b}=2.53 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}, W_{B}=3.24 \mathrm{~g}\)
\(\Delta T_{b}=0.81 \mathrm{~K}, W_{A}=40 \mathrm{~g}\)
\(M_{B}=\frac{2.53 \times 1000 \times 3.24}{0.81 \times 40}=253\)
Let molecular formula of sulphur = S_x
\(x \times 32=253 \text { or } x=7.91 \approx 8\)
(iii) (a)
(iv) (a)

(i) (c) :If the red blood cells are placed in pure water, pressure inside the cells increases as the water is drawn in and the cell swells.
(ii) (b)
(iii) (c) : Osmotic pressure is a colligative property
(iv) (a)

(i) (a)
(ii) (a)
(iii) (b)
(iv) (d): For ideal solution \(\Delta H_{\operatorname{mix}}=0, \Delta V_{\operatorname{mix}}=0\)

(i) (b): Depression in freezing point is a colligative property which depends on the number of particles present in the solution. As both 0.1 M solution of glucose and 0.1 M solution of urea contain same number of moles (number of particles) therefore, both will have same depression in freezing point.
(ii) (a)
(iii) (a): Freezing point of a substance is defined as the temperature at which the vapour pressure of its liquid is equal to the vapour pressure of the corresponding solid. Since the addition of a non-volatile solute always lowers the vapour pressure of solvent, therefore it will be in equilibrium with solid phase at a lower pressure and hence at a lower temperature.
(iv) (d): Elevation in boiling point \(\left(\Delta T_{b}\right)=K_{b} \times m\)
Depression in freezing point \(\left(\Delta T_{f}\right)=K_{f} \times m\)
Elevation in boiling point and depression in freezing point are colligative properties i.e., they depend only on the number of particles of the solute. Value of K_band K_f are different, so \(\Delta T_{b}\) and \(\Delta T_{f}\) are also different.

(i) (b): K_H = 4.27 x 10⁵ mm Hg
P = 760mm Hg
According to Henry's law, \(p=K_{\mathrm{H}} \times x_{\mathrm{CH}_{4}}\)
\(x_{\mathrm{CH}_{4}}=\frac{p}{K_{\mathrm{H}}}=\frac{760}{4.27 \times 10^{5}}=1.78\ \times 10^{-3}\)
(ii) (a): According to Henry's law, m = K_H x P
6.56 X 10^-2 = K_H X 1
K_H = 6.56 X 10^-2
For another case,5x 10^-2 = 6.56 X 10^-2 x P
\(p=\frac{5 \times 10^{-2}}{6.56 \times 10^{-2}}=0.762 \mathrm{bar}\)
(iii) (c) : Higher the value of K_H at a given pressure, the lower is the solubility of the gas.
(iv) (a)

10.

(i) (b): m \(=\frac{0.052}{180} \times \frac{1000}{80.2}=0.0036\)
(ii) (c): \(\Delta T_{b}=K_{b} \times m=5.2 \times 0.0036=0.0187 \mathrm{~K}\)
\(T_{b}=373+0.0187=373.0187 \mathrm{~K} \approx 373.02 \mathrm{~K}\)
(iii) (d): \(\Delta T_{f}=K_{f} \times m=1.86 \times 0.0036=0.067 \mathrm{~K}\)
(iv) (a): Moles of glucose \(=\frac{0.052}{180}=0.00028\)
Moles 0f water = \(\frac{80.2}{18}=4.455\)
Mole fraction of glucose = \(\frac{0.00028}{4.45+0.00028}=6.28 \times 10^{-5}\)

11.

(i) (a) : Vapour pressure of water \(\left(p_{A}^{\circ}\right)\) = 17.5 mm of Hg
Lowering of vapour pressure \(\left(p_{A}^{\circ}-p_{A}\right)\)= 0.061
Relative lowering of vapour pressure
\(=\frac{p_{A}^{\circ}-p_{A}}{p_{A}^{\circ}}=\frac{0.061}{17.5}=0.00348\)
(ii) (c): P = Vapour pressure of solvent - lowering in vapour pressure = 17.5 - 0.061 = 17.439 mm of Hg
(iii) (a): \(\frac{p_{A}^{\circ}-p_{A}}{p_{A}^{\circ}}=x_{B}=0.00348\)
Hence, mole fraction of sugar = 0.00348
(iv) (b): \(\frac{p_{A}^{\circ}-p_{A}}{p_{A}^{\circ}}=x_{B}=\frac{w_{B} \times M_{A}}{M_{B} \times w_{A}}\)
\(\frac{17.5-p_{A}}{17.5}=\frac{25 \times 18}{450 \times 180}=5.56 \times 10^{-3}\)
\(17.5-p_{A}=17.5 \times 5.56 \times 10^{-3}\)
\(17.5-p_{A}=0.0973\)
P = 17.40 mm Hg

12.

(i) (d) : II represents negative deviations and III represents positive deviations.
(ii) (b): For positive deviations \(p_{A}>p_{A}^{\circ} x_{A} \text { and } p_{B}>p_{B}^{\circ} x_{B}\)
(iii) (b): Water and nitric acid mixture shows negative deviations from Raoult's law, hence \(p_{A}<p_{A}^{\circ} x_{A} \text { and } p_{B}<p_{B}^{\circ} x_{B}\)
(iv) (d): Water-HCl mixture shows negative deviations from Raoult's law and solutions showing negative deviations from ideal behaviour form maximum boiling azeotrope.

13.

(i) (b) : Moles of C₆H₆ = \(\frac{7.8}{78}=0.1\)
Mole C₆H₅CH₃ = \(\frac{9.2}{92}=0.1\)
Mole fraction of C₆H₆ = \(\frac{0.1}{0.1+0.1}=0.5\)
=> Mole fraction of C₆H₅CH₃ = 0.5
Vapour pressure of toluene = Vapour pressure of pure toluene x mole fraction of toluene
= 0.0925 x 0.5 = 0.04625
Vapour pressure of benzene = 0.256 x 0.5 = 0.128
Total vapour pressure of solution = 0.17425
(ii) (a) : Moles of benzene in solution-II = \(\frac{3.9}{78}=0.05\)
Moles of toluene in solution-II = \(\frac{13.8}{92}=0.15\)
Vapour pressure of solution
= 0.256 x 0.05 + 0.0925 x 0.15
= 0.0128 + 0.013875 = 0.026675
(iii) (c) : Mole fraction of benzene in vapour phase
\(y_{\text {benzene }}=\frac{p_{\text {benzene }}}{P_{\text {total }}}=\frac{0.128}{0.17425}=0.734\)
(iv) (a) : Benzene and toluene form an ideal solution.

14.

(i) (d) : Density of solution = 1.202 g/mL
Volume of solution = \(\frac{100 \mathrm{~g}}{1.202 \mathrm{~g} / \mathrm{mL}}=83.2 \mathrm{~mL}\)
Molarity = \(\frac{n_{\mathrm{KI}}}{\text { Volume of solution in } \mathrm{L}}\)
\(=\frac{0.120 \mathrm{~mol}}{0.0832 \mathrm{~L}}=1.4423 \mathrm{~mol} \mathrm{~L}^{-1}\)
(ii) (a): \(x_{2}=\frac{n_{2}}{n_{1}+n_{2}} ; x_{1}=\frac{n_{1}}{n_{1}+n_{2}} ; \frac{x_{2}}{x_{1}}=\frac{n_{2}}{n_{1}}\)
\(\frac{x_{2}}{x_{1}}=\frac{m_{2} / M_{2}}{m_{1} / M_{1}}=\frac{m_{2}}{m_{1}} \times \frac{M_{1}}{M_{2}}\) ...(i)
Molality = \(\frac{n_{2}}{m_{1}}=\frac{m_{2}}{M_{2} \times m_{1}}\) ...(ii)
From(i) and (ii), m = \(\frac{x_{2}}{x_{1}} \times \frac{1}{M_{1}} ; x_{1}=1-x_{2}\)
Hence. x₂= \(\frac{m M_{1}}{1+m M_{1}}\)
(iii) (a) : Mass does not depend on temperature while volume does. Hence, molarity depends on temperature.
(iv) (b): 1M solution contains 1 mole of solute in less than 1000 g of the solvent whereas 1 m solution has 1 mole of the solute in 1000 g of the solvent.