CBSE Class 12 Chemistry Solution Exam Notes

CBSE Class 12 Chemistry Solution Exam Notes. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

SOLUTIONS (EXAM NOTES)

SOLUTIONS

A solution may be defined as a homogeneous mixture of a single phase containing one or more chemical species dispersed on a molecular scale.

SOLUTE AND SOLVENT

1. The component whose physical state is the same as that of the resulting solution is called the solvent, and the other component is termed the solute. For example, in a homogeneous mixture of Na(s) and Hg(l) resulting in the formation of NaHg(s), Na(s) is considered the solvent and Hg(l) the solute, which is contrary to the common belief that liquids are solvents.
2. When both components have the same phase as the resulting solution, the component present in the greater amount is generally regarded as the solvent.

METHODS OF EXPRESSING CONCENTRATIONS OF SOLUTIONS

The concentration of a solution can be expressed in the following ways. Solute is designated as 'b' and solvent as 'a'.

1. Molarity (M): number of moles of solute per litre of solution.
   \[M=\frac{\text{number of moles of solute}}{\text{volume of solution in litres}}=\frac{n_b}{V_{(L)}}=\frac{W_b\times 1000}{M_b\times V_{(mL)}}\]where \(V_{(mL)}\) is the volume of the solution in millilitres.
2. Molality (m): number of moles of solute per kilogram of solvent.
   \[m=\frac{\text{number of moles of solute}}{\text{mass of solvent in kg}}=\frac{n_b}{W_a\,(\text{kg})}=\frac{W_b\times 1000}{M_b\times W_a}\]where \(W_a\) is the mass of the solvent in grams.
3. Normality (N): number of gram equivalents of solute per litre of solution.
   \[\text{Gram equivalents}=\frac{\text{weight of solute}}{\text{equivalent mass of solute}};\qquad \text{Equivalent mass}=\frac{\text{molar mass of solute}}{n\text{-factor}}\]\[N=M\times n_f\]where \(M\) is the molarity and \(n_f\) is the \(n\)-factor.
4. Mole fraction (X): ratio of the number of moles of one of the components to the total number of moles present in the solution.
   \[X_b=\frac{n_b}{n_a+n_b+n_c+\cdots};\qquad X_a=\frac{n_a}{n_a+n_b+n_c+\cdots}\]The total of the mole fractions of a solution is 1; for two components, \(X_a+X_b=1\).
5. Mass percentage (w/W %): \[\frac{\text{mass of solute (g)}}{\text{mass of solution (g)}}\times 100\]A 10% aqueous NaCl solution contains 10 g of NaCl per 100 g of solution. Ten grams of NaCl is mixed with 90 g of water to form 100 g of solution.
6. Volume percentage (v/V %): \[\frac{\text{volume of solute (mL)}}{\text{volume of solution (mL)}}\times 100\]For example, a 10% volume percentage of aqueous ethanol means 10 mL of ethanol is present in 100 mL of the solution.
7. Mass by volume percentage (w/V %): \[\frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}}\times 100\]
8. Parts per million (ppm): \[\frac{\text{number of parts of the component}}{\text{total number of parts of all the components of the solution}}\times 10^{6}\]Concentration in parts per million can also be expressed as mass to mass, volume to volume and mass to volume.

WORKED EXAMPLES ON CONCENTRATION

Calculate the molality and mole fraction of the solute in an aqueous solution containing 3.0 g of urea (molar mass = 60 g mol\(^{-1}\)) per 250 g of water.

Answer: Molality = 0.2 m; mole fraction of urea = 0.0036

Weight of solute = 3 g; molar mass of urea \(M_b\) = 60 g mol\(^{-1}\); weight of solvent = 250 g.

\[m=\frac{W_b\times1000}{M_b\times W_a}=\frac{3\times1000}{60\times250}=0.2\ \text{m}\]

\[n_b=\frac{3}{60}=0.05;\qquad n_a=\frac{250}{18}=13.89;\qquad X_b=\frac{0.05}{0.05+13.89}=0.0036\]

Calculate the molarity and normality of a solution containing 9.8 g of H₂SO₄ in 250 cm\(^3\) of the solution. [Molar mass of H₂SO₄ = 98 g mol\(^{-1}\)]

Answer: Molarity = 0.4 M; normality = 0.8 N

\[M=\frac{W_b\times1000}{M_b\times V_{(mL)}}=\frac{9.8\times1000}{98\times250}=0.4\ \text{M}\]

Normality = \(M\times n_f\) = 0.4 \(\times\) 2 = 0.8 N (the \(n\)-factor of H₂SO₄ is 2).

Find the molarity and molality of a 15% solution of H₂SO₄ (density = 1.020 g cm\(^{-3}\)).

Answer: Molarity = 1.56 M; molality = 1.8 m

Let the solution be 100 g. Weight of solute = 15 g; weight of solvent = 85 g.

\[V=\frac{100}{1.020}\ \text{mL};\qquad M=\frac{W_b\times1000}{M_b\times V_{(mL)}}=\frac{15\times1000}{98\times\frac{100}{1.020}}=1.56\ \text{M}\]

\[m=\frac{W_b\times1000}{M_b\times W_a}=\frac{15\times1000}{98\times85}=1.8\ \text{m}\]

A solution of ethanol in water is 1.54 molal. How much of it is there in 2500 g of the solvent?

Answer: 177.1 g of ethanol

Molality = 1.54 m; weight of solvent = 2500 g; molar mass of ethanol = 46 g mol\(^{-1}\).

\[m=\frac{W_b\times1000}{M_b\times W_a}\ \Rightarrow\ 1.54=\frac{W_b\times1000}{46\times2500};\qquad W_b=177.1\ \text{g}\]

What volume of 95% H₂SO₄ (density = 1.85 g cm\(^{-3}\)) and what mass of water must be taken to prepare 100 cm\(^3\) of a 15% solution of H₂SO₄ (density = 1.10 g cm\(^{-3}\))?

Answer: 9.33 mL of the 95% acid is taken and water is added to make 100 mL, i.e. 90.67 mL of water by volume (about 92.7 g of water by mass).

Given solution: take 100 g of solution; weight of solute = 95 g; density = 1.85 g cm\(^{-3}\).

\[V=\frac{100}{1.85}=54.04\ \text{mL};\qquad M=\frac{95\times1000}{98\times54.04}=18\ \text{M}\]

Required solution: take 100 g of solution; weight of solute = 15 g; density = 1.10 g cm\(^{-3}\).

\[V=\frac{100}{1.10}=90.91\ \text{mL};\qquad M=\frac{15\times1000}{98\times90.91}=1.68\ \text{M}\]

To prepare the required solution from the given one: \(M_1V_1=M_2V_2\).

\[18\times V_1=1.68\times100\ \Rightarrow\ V_1=9.33\ \text{mL}\]

The rest of the volume is due to the added water: 100 \(-\) 9.33 = 90.67 mL.

Added to complete the answer (mass of water): 100 cm\(^3\) of the 15% solution has a mass of 100 \(\times\) 1.10 = 110 g. The 9.33 mL of 95% acid has a mass of 9.33 \(\times\) 1.85 = 17.26 g. So the mass of water to be added = 110 \(-\) 17.26 \(\approx\) 92.7 g.

PRACTICE QUESTIONS

Calculate the mole fraction of ethylene glycol (C₂H₆O₂) in a solution containing 20% of C₂H₆O₂ by mass.

Answer: 0.068

Calculate the molarity of a solution of 5 g of NaOH in 450 mL of solution.

Answer: 0.278 mol dm\(^{-3}\)

Calculate the molality of 2.5 g of ethanoic acid (CH₃COOH) in 75 g of benzene.

Answer: 0.556 mol kg\(^{-1}\)

Calculate (a) molality, (b) molarity and (c) mole fraction of KI if the density of 20% (m/m) aqueous KI is 1.202 g mL\(^{-1}\).

Answer: (a) 1.5 mol kg\(^{-1}\); (b) 1.45 M; (c) 0.0264

Calculate the molarity of each of the following solutions: (a) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL; (b) 30 g of Co(NO₃)₂.6H₂O in 4.3 L of solution.

Answer: (a) 0.03 M; (b) 0.024 M

SOLUBILITY

Saturated solution: the solution in which no more solute can be dissolved at the same temperature and pressure.

Solubility: the maximum amount of a solute that can be dissolved in a given volume of solution at a constant temperature and pressure is called its solubility.

FACTORS AFFECTING THE SOLUBILITY OF A SOLID IN A LIQUID

1. Temperature: the solubility of a solid in a liquid is significantly affected by temperature changes. Being a dynamic equilibrium, it follows Le Chatelier's principle. In general, in a nearly saturated solution, if the dissolution process is endothermic (\(\Delta_{sol}H\gt 0\)), the solubility should increase with a rise in temperature, and if it is exothermic (\(\Delta_{sol}H\lt 0\)), the solubility should decrease. These trends are also observed experimentally.
2. Pressure: pressure does not have any significant effect on the solubility of solids in liquids. This is because solids and liquids are highly incompressible and practically remain unaffected by a change in pressure.

HENRY'S LAW

The law states that at a constant temperature the solubility of a gas in a liquid is directly proportional to the pressure of the gas. The most commonly used form of Henry's law states that "the partial pressure of the gas in the vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution" and is expressed as:

\[p=K_H\,x\]

Here \(K_H\) is Henry's law constant.

From the equation \(p=K_Hx\), we can say that the higher the value of \(K_H\) at a given pressure, the lower is the solubility of the gas in the liquid. The \(K_H\) values for both N₂ and O₂ increase with an increase of temperature, indicating that the solubility of gases increases with a decrease of temperature. It is due to this reason that aquatic species are more comfortable in cold waters than in warm waters.

LIMITATIONS OF HENRY'S LAW

1. It is applicable at moderate temperature and pressure, and fails when the pressure used is too high and the temperature too low.
2. The gas should not get dissociated or associated when it is dissolved in the solvent.
3. The solubility of the gas should not be high in the solvent (for example NH₃, CO₂ and HCl in water).
4. The gas should not form a new species by reacting with the solvent when it is dissolved. For example, NH₃ does not follow Henry's law for this reason.

APPLICATIONS OF HENRY'S LAW

1. In aerated soft drinks, carbon dioxide is sealed under high pressure to increase the solubility of CO₂ in the drink for the desired taste.
2. Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure underwater. Increased pressure increases the solubility of atmospheric gases in blood. When the divers come towards the surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries and creates a medical condition known as bends, which are painful and dangerous to life. To avoid bends, as well as the toxic effects of high concentrations of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen).
3. At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers. Low blood oxygen causes climbers to become weak and unable to think clearly, symptoms of a condition known as anoxia.

EFFECT OF TEMPERATURE ON THE SOLUBILITY OF A GAS

According to Le Chatelier's principle, solubility should increase with an increase of temperature for endothermic processes (\(\Delta_{sol}H\gt 0\)) and decrease with an increase of temperature for exothermic processes (\(\Delta_{sol}H\lt 0\)). When a gas dissolves, its molecules are present in the liquid phase and the process can be considered similar to condensation. Heat is evolved in this process, so the solubility of a gas decreases as the temperature rises.

H₂S, a toxic gas with a rotten-egg-like smell, is used for qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry's law constant.

Answer: \(K_H\) = 2.86 \(\times\) 10\(^2\) atm

At STP, \(p\) = 1 atm and the solubility is 0.195 m, i.e. 0.195 mol in 1000 g of water (1000/18 = 55.5 mol).

\[x=\frac{0.195}{0.195+55.5}=3.5\times10^{-3};\qquad K_H=\frac{p}{x}=\frac{1}{3.5\times10^{-3}}=2.86\times10^{2}\ \text{atm}\]

Henry's law constant for CO₂ in water is 1.67 \(\times\) 10\(^8\) Pa at 298 K. Calculate the quantity of CO₂ per mL of soda water when packed under 2.5 atm CO₂ pressure at 298 K.

Answer: About 3.6 \(\times\) 10\(^{-3}\) g of CO₂ per mL of soda water

1 atm = 101325 Pa, so \(K_H\) = 1.67 \(\times\) 10\(^8\) Pa = 1648 atm = 1.648 \(\times\) 10\(^3\) atm.

\[p=K_Hx\ \Rightarrow\ x=\frac{2.5}{1.648\times10^{3}}=1.5\times10^{-3}\]

1 mL of water = 1 g = 1/18 = 0.055 mol. Let \(n\) mol of CO₂ dissolve in it.

\[x=\frac{n}{n+0.055}=1.5\times10^{-3}\ \Rightarrow\ n\approx\frac{0.055}{665.67}=8.3\times10^{-5}\ \text{mol}\]

\[\text{Mass of CO}_2=8.3\times10^{-5}\times44=3.6\times10^{-3}\ \text{g}\]

PRACTICE QUESTIONS

If N₂ gas is bubbled through water at 293 K, how many millimoles of N₂ gas would dissolve in 1 litre of water? Assume that N₂ exerts a partial pressure of 0.987 bar. Given that Henry's law constant for N₂ at 293 K is 76.48 kbar.

Answer: 0.717 millimoles (7.17 \(\times\) 10\(^{-4}\) mol)

Calculate the concentration of CO₂ in a soft drink that is bottled with a partial pressure of CO₂ of 4 atm over the liquid at 25 \(^\circ\)C. The Henry's law constant for CO₂ in water at 25 \(^\circ\)C is 3.1 \(\times\) 10\(^{-2}\) mol per litre per atm.

Answer: 0.124 M

The Henry's law constant for oxygen dissolved in water is 4.34 \(\times\) 10\(^4\) atm at 25 \(^\circ\)C. If the partial pressure of oxygen in air is 0.2 atm under atmospheric conditions, calculate the concentration of dissolved oxygen in water in equilibrium with air at 25 \(^\circ\)C.

Answer: Mole fraction of O₂ = 4.6 \(\times\) 10\(^{-6}\); molar concentration = 2.56 \(\times\) 10\(^{-4}\) M

Calculate the solubility of H₂ in water at 25 \(^\circ\)C if the partial pressure above the solution is 1 bar. Given that Henry's constant for H₂ in water at 25 \(^\circ\)C is 71.18 kbar.

Answer: Mole fraction of H₂ = 1.40 \(\times\) 10\(^{-5}\)

Henry's law constant for the molarity of methane in benzene at 298 K is 4.27 \(\times\) 10\(^5\) mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Answer: Mole fraction of methane = 1.78 \(\times\) 10\(^{-3}\)

VAPOUR PRESSURE OF LIQUID SOLUTIONS

Vapour pressure is the pressure exerted by the vapours of a liquid on its liquid surface at equilibrium. The important points regarding vapour pressure are:

1. The more volatile the substance, the higher its vapour pressure.
2. The lesser the intermolecular force of attraction between liquid molecules, the higher the vapour pressure (as the liquid is more volatile).
3. Vapour pressure of a liquid is not considered a surface phenomenon. Although evaporation is a surface phenomenon, condensation also occurs at the surface.
4. Vapour pressure of a liquid does not depend on the size or shape of the vessel.
5. Vapour pressure of a liquid depends on temperature. As the temperature increases, the vapour pressure increases.

RAOULT'S LAW

Let the total vapour pressure at this stage be \(p_{total}\), and let \(p_1\) and \(p_2\) be the partial vapour pressures of the two components 1 and 2 respectively. These partial pressures are related to the mole fractions \(x_1\) and \(x_2\) of the two components. The quantitative relationship between them is known as Raoult's law, which states: "For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction."

Thus, for component 1: \(p_1\propto x_1\) and \(p_1=p_1^0x_1\) (where \(p_1^0\) = saturated vapour pressure of pure component 1).

Similarly, for component 2: \(p_2=p_2^0x_2\) (where \(p_2^0\) = saturated vapour pressure of pure component 2).

According to Dalton's law of partial pressures, \(p_{total}=p_1+p_2\). Substituting the values of \(p_1\) and \(p_2\), we get

\[p_{total}=p_1+p_2=(1-x_2)\,p_1^0+x_2\,p_2^0=p_1^0+(p_2^0-p_1^0)\,x_2\]

The composition of the vapour phase in equilibrium with the solution is determined by the partial pressures of the components. If \(y_1\) and \(y_2\) are the mole fractions of components 1 and 2 respectively in the vapour phase, then, using Dalton's law of partial pressures, \(p_1=y_1p_{total}\) and \(p_2=y_2p_{total}\); in general, \(p_i=y_ip_{total}\).

RAOULT'S LAW AS A SPECIAL CASE OF HENRY'S LAW

If we compare the equations for Raoult's law (\(p_1=x_1p_1^0\)) and Henry's law (\(p=K_Hx\)), we see that the partial pressure of the volatile component of a gas is directly proportional to its mole fraction in solution. Only the proportionality constant \(K_H\) differs from \(p_1^0\). Thus, Raoult's law becomes a special case of Henry's law in which \(K_H\) becomes equal to \(p_1^0\) only if the solution is ideal. That is why Raoult's law is applicable only to ideal solutions, whereas Henry's law is applicable to non-ideal solutions as well. A plot between the vapour pressure and the mole fraction of the solvent is linear.

Benzene (C₆H₆) and toluene (C₇H₈) form an ideal solution at 313 K. The vapour pressures of pure benzene and toluene are 160 mm Hg and 60 mm Hg respectively. Calculate the partial pressures of benzene and toluene and the total pressure over the following solutions: (i) containing equal weights of benzene and toluene; (ii) containing 1 mole of benzene and 4 moles of toluene; (iii) containing equal numbers of molecules of benzene and toluene.

Answer: (i) 86.6 mm Hg (benzene), 27.5 mm Hg (toluene), 114.1 mm Hg (total); (ii) 32 mm Hg, 48 mm Hg, 80 mm Hg; (iii) 80 mm Hg, 30 mm Hg, 110 mm Hg

(i) Take 1 g of each (molar masses 78 and 92 g mol\(^{-1}\)): \(n_B\) = 1/78 = 0.01282 mol and \(n_T\) = 1/92 = 0.01087 mol, so \(x_B\) = 0.541 and \(x_T\) = 0.459. Then \(p_B\) = 160 \(\times\) 0.541 = 86.6 mm Hg, \(p_T\) = 60 \(\times\) 0.459 = 27.5 mm Hg, and \(p_{total}\) = 114.1 mm Hg.

(ii) \(x_B\) = 1/5 = 0.2 and \(x_T\) = 0.8. Then \(p_B\) = 160 \(\times\) 0.2 = 32 mm Hg, \(p_T\) = 60 \(\times\) 0.8 = 48 mm Hg, and \(p_{total}\) = 80 mm Hg.

(iii) Equal numbers of molecules means equal moles, so \(x_B\) = \(x_T\) = 0.5. Then \(p_B\) = 80 mm Hg, \(p_T\) = 30 mm Hg, and \(p_{total}\) = 110 mm Hg.

The vapour pressures of benzene and toluene at 293 K are 75 mm Hg and 22 mm 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 vapour is in equilibrium with the liquid mixture at this temperature.

Answer: Mole fraction of benzene in the vapour phase = 0.59

\(p_B^0\) = 75 mm Hg; \(p_T^0\) = 22 mm Hg; molar masses: benzene 78 g mol\(^{-1}\), toluene 92 g mol\(^{-1}\).

\[n_B=\frac{23.4}{78}=0.3;\quad n_T=\frac{64.4}{92}=0.7;\quad x_B=\frac{0.3}{0.3+0.7}=0.3;\quad x_T=0.7\]

\[p_B=p_B^0x_B=75\times0.3=22.5\ \text{mm Hg};\qquad p_T=p_T^0x_T=22\times0.7=15.4\ \text{mm Hg}\]

\[p_{total}=22.5+15.4=37.9\ \text{mm Hg}\]

\[y_B=\frac{p_B}{p_{total}}=\frac{22.5}{37.9}=0.59;\qquad y_T=\frac{15.4}{37.9}=0.41\]

IDEAL AND NON-IDEAL SOLUTIONS

Ideal solutions: solutions which obey Raoult's law over the entire range of concentration are known as ideal solutions. Ideal solutions have two other important properties. The heat of mixing of the pure components to form the solution is zero and the volume of mixing is also zero, i.e., \(\Delta_{mix}H=0\) and \(\Delta_{mix}V=0\).

Non-ideal solutions: when a solution does not obey Raoult's law over the entire range of concentration, it is called a non-ideal solution. The vapour pressure of such a solution is either higher or lower than that predicted by Raoult's law. If it is higher, the solution exhibits positive deviation, and if it is lower, it exhibits negative deviation from Raoult's law. If \(p_{solution}\gt p_{solvent}\) (the value predicted by Raoult's law), the solution shows positive deviation, and if \(p_{solution}\lt p_{solvent}\), the solution shows negative deviation from Raoult's law.

In case of positive deviation, the solute-solvent (A-B) intermolecular forces are weaker than the solvent-solvent (A-A) and solute-solute (B-B) intermolecular forces. So it is easier for solute and solvent molecules (A-B) to escape than in the pure state. This increases the vapour pressure and results in positive deviation. Examples: mixtures of ethanol and acetone, carbon disulphide and acetone, acetone and benzene, methyl alcohol and water, ethyl alcohol and water, carbon tetrachloride and chloroform, carbon tetrachloride and benzene, carbon tetrachloride and toluene show positive deviation.

In case of negative deviation from Raoult's law, the intermolecular attractive forces between solute-solute (B-B) and solvent-solvent (A-A) are weaker than those between solute-solvent (A-B), and this leads to negative deviation. Examples: mixtures of phenol and aniline, chloroform and diethyl ether, acetone and aniline, HCl and water, HNO₃ and water, acetic acid and pyridine, etc.

AZEOTROPES

Some liquid mixtures, having a definite composition, boil like a pure liquid and are called azeotropic mixtures or azeotropes, or constant boiling mixtures. The components of these mixtures cannot be separated by fractional distillation.

1. Minimum boiling azeotropes: solutions that show a large positive deviation from Raoult's law form minimum boiling azeotropes at a specific composition, e.g. an ethanol-water mixture containing 95.4% ethanol boils at 351.5 K; a chloroform-ethanol mixture with 6.8% ethanol.
2. Maximum boiling azeotropes: solutions that show a large negative deviation from Raoult's law form maximum boiling azeotropes at a specific composition, e.g. 68% nitric acid and 32% water by mass, with a boiling point of 393.5 K; a water-HCl mixture with 20.3% HCl.

COLLIGATIVE PROPERTIES

The properties of dilute solutions which are induced on addition of a non-volatile solute to a volatile solvent are called colligative properties. These properties depend on the number of solute particles irrespective of their nature, relative to the total number of particles present in the solution.

1. Lowering of vapour pressure
2. Depression in freezing point
3. Elevation in boiling point
4. Osmotic pressure

RELATIVE LOWERING OF VAPOUR PRESSURE

The vapour pressure of the solvent in a solution is less than that of the pure solvent. The relation between the vapour pressure of the solvent and its mole fraction is \(p_1=x_1p_1^0\), where \(p_1\) = vapour pressure of the solution, \(x_1\) = mole fraction of the solvent and \(p_1^0\) = vapour pressure of the pure solvent.

\[\frac{\Delta p_1}{p_1^0}=\frac{p_1^0-p_1}{p_1^0}=x_2=\frac{n_2}{n_1+n_2}\]

\(\dfrac{\Delta p_1}{p_1^0}\) is called the relative lowering of vapour pressure and is equal to the mole fraction of the solute. For dilute solutions, \(n_2\ll n_1\), hence neglecting \(n_2\) in the denominator we have

\[\frac{p_1^0-p_1}{p_1^0}=\frac{n_2}{n_1}\quad\text{or}\quad\frac{p_1^0-p_1}{p_1^0}=\frac{w_2\times M_1}{M_2\times w_1}\]

ELEVATION OF BOILING POINT

For dilute solutions, the elevation of boiling point (\(\Delta T_b\)) is directly proportional to the molal concentration of the solute in a solution. Thus, \(\Delta T_b\propto m\), i.e. \(\Delta T_b=K_bm\). \(K_b\) is called the boiling point elevation constant or molal elevation constant (ebullioscopic constant). The unit of \(K_b\) is K kg mol\(^{-1}\).

\[m=\frac{1000\times w_2}{M_2\times w_1};\qquad \Delta T_b=\frac{K_b\times1000\times w_2}{M_2\times w_1}\]

DEPRESSION OF FREEZING POINT

A solution freezes when its vapour pressure equals the vapour pressure of the pure solid solvent. According to Raoult's law, when a non-volatile solute is added to the solvent, its vapour pressure decreases and would become equal to that of the solid solvent at a lower temperature. Thus, the freezing point of the solvent decreases. The depression of freezing point (\(\Delta T_f\)) for a dilute (ideal) solution is directly proportional to the molality \(m\) of the solution.

\[\Delta T_f\propto m\ \text{or}\ \Delta T_f=K_fm;\qquad \Delta T_f=T_f^0-T_f\]

The proportionality constant \(K_f\), which depends on the nature of the solvent, is known as the freezing point depression constant, molal depression constant or cryoscopic constant. The unit of \(K_f\) is K kg mol\(^{-1}\).

\[\Delta T_f=\frac{K_f\times w_2\times1000}{M_2\times w_1};\qquad K_f=\frac{R\times M_1\times T_f^2}{1000\times\Delta_{fus}H};\qquad K_b=\frac{R\times M_1\times T_b^2}{1000\times\Delta_{vap}H}\]

Here \(R\) and \(M_1\) are the gas constant and the molar mass of the solvent respectively, and \(T_f\) and \(T_b\) denote the freezing point and the boiling point of the pure solvent respectively in kelvin. \(\Delta_{fus}H\) and \(\Delta_{vap}H\) are the enthalpies of fusion and vapourisation of the solvent respectively.

OSMOSIS AND OSMOTIC PRESSURE

When a solution is separated from the pure solvent by a semi-permeable membrane, it is observed that the solvent flows from the region of pure solvent into the solution. The passage of solvent from pure solvent, or from a solution of lower concentration, into a solution of higher concentration through a semi-permeable membrane is called osmosis. The osmotic pressure of the solution at a given temperature may be defined as the excess hydrostatic pressure that builds up when the solution is separated from the solvent by a semi-permeable membrane. It is denoted by \(\pi\).

DETERMINATION OF MOLAR MASS FROM OSMOTIC PRESSURE

To determine the molar mass of an unknown non-volatile compound, a known mass (say \(w\) g) of the compound is dissolved to prepare a known volume (say \(V\) litre) of solution, and the osmotic pressure of the solution is determined. From this, the molar mass of the compound is calculated as follows. According to the van't Hoff equation,

\[M=\frac{wRT}{\pi V}\]

CONDITIONS FOR GETTING AN ACCURATE VALUE OF MOLAR MASS

1. The solute must be non-volatile.
2. The solution must be dilute. The solute should not undergo dissociation or association in the solution.

REVERSE OSMOSIS

If the pressure applied on the solution is greater than the osmotic pressure, then the solvent starts passing from the solution into the pure solvent. This is called reverse osmosis. It is generally used for the purification of sea water or hard water. (In the set-up, the salt water is on one side of a semi-permeable membrane under a piston with pressure greater than the osmotic pressure, and fresh water collects on the other side.)

SUMMARY OF COLLIGATIVE PROPERTIES

\(w_1\) = weight of solvent, \(M_1\) = molar mass of solvent, \(w_2\) = weight of solute, \(M_2\) = molar mass of solute, \(R\) = gas constant.

Left: the vapour pressure curve for the solution lies below the curve for the pure solvent; \(\Delta T_b\) is the elevation of boiling point (\(T^0\) is the boiling point of the pure solvent and \(T\) that of the solution). Right: \(\Delta T_f\) shows the depression of the freezing point of a solvent in a solution on the vapour pressure-temperature diagram.

ABNORMAL MOLAR MASSES

• Abnormal molar masses are observed when the solution is non-ideal, i.e. the solution is not dilute.
• Abnormal molar masses are also observed when the solute undergoes dissociation or association in the solution.

Van't Hoff factor i, which accounts for the extent of association or dissociation, is the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

\[i=\frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}=\frac{\text{Observed colligative property}}{\text{Calculated colligative property}}=\frac{\text{No. of moles of particles after association/dissociation}}{\text{No. of moles of particles before association/dissociation}}\]

PRACTICE PROBLEMS

An aqueous solution freezes at \(-0.186\ ^\circ\)C (\(K_f\) = 1.86 K kg mol\(^{-1}\); \(K_b\) = 0.512 K kg mol\(^{-1}\)). What is the elevation in boiling point?

Answer: 0.051 \(^\circ\)C

\(m=\dfrac{\Delta T_f}{K_f}=\dfrac{0.186}{1.86}=0.1\) mol kg\(^{-1}\), so \(\Delta T_b=K_bm=0.512\times0.1=0.051\) K.

Calculate the amount of NaCl that must be added to one kilogram of water so that the freezing point is depressed by 3 K. (Given \(K_f\) for water = 1.86 K kg mol\(^{-1}\).)

Answer: About 47.2 g (taking NaCl to dissociate completely, \(i\) = 2)

NaCl gives two particles (Na⁺ and Cl^-) in water, so \(i\) = 2 and \(\Delta T_f=i\,K_fm\).

\[m=\frac{3}{2\times1.86}=0.806\ \text{mol kg}^{-1};\qquad \text{mass of NaCl}=0.806\times58.5=47.2\ \text{g}\]

(If \(i\) were wrongly taken as 1, one would get 94.35 g, which is the value printed in the source.)

Ethylene glycol (molar mass = 62 g mol\(^{-1}\)) is a common automobile antifreeze. Calculate the freezing point of a solution containing 12.4 g of this substance in 100 g of water. Would it be advisable to keep this substance in the car radiator during summer? Given: \(K_f\) for water = 1.86 K kg mol\(^{-1}\), \(K_b\) for water = 0.512 K kg mol\(^{-1}\).

Answer: Freezing point = \(-3.72\ ^\circ\)C. Yes, it is advisable in summer too, because the boiling point is raised to about 101.02 \(^\circ\)C.

\(m=\dfrac{12.4/62}{0.100}=2\) mol kg\(^{-1}\); \(\Delta T_f=1.86\times2=3.72\) K, so the freezing point is \(-3.72\ ^\circ\)C.

Added to complete the answer (summer use): \(\Delta T_b=0.512\times2=1.024\) K, so the solution boils at about 101.02 \(^\circ\)C. The coolant therefore does not boil away easily in summer, and it can be kept in the radiator all year.

The boiling point of water at 750 mm Hg is 96.63 \(^\circ\)C. How much sucrose is to be added to 500 g of water such that it boils at 100 \(^\circ\)C? (Molal elevation constant for water is 0.52 K kg mol\(^{-1}\).)

Answer: 1.108 kg (1108 g)

A solution of a non-volatile solute in water freezes at \(-0.30\ ^\circ\)C. The vapour pressure of pure water at 298 K is 23.51 mm Hg and \(K_f\) for water is 1.86 K kg mol\(^{-1}\). Calculate the vapour pressure of this solution at 298 K.

Answer: 23.44 mm Hg

SOLVED EXAMPLES

Calculate the mole fraction of benzene in a solution containing 30% by mass of it in carbon tetrachloride.

Answer: 0.458

30% of benzene in carbon tetrachloride by mass means: mass of benzene in the solution = 30 g, mass of solution = 100 g, mass of CCl₄ = 100 \(-\) 30 = 70 g. Molar mass of benzene (C₆H₆) = 78 g mol\(^{-1}\); molar mass of CCl₄ = 12 + 4 \(\times\) 35.5 = 154 g mol\(^{-1}\).

\[n_{\text{benzene}}=\frac{30}{78}=0.385;\quad n_{\text{CCl}_4}=\frac{70}{154}=0.455;\quad x_{\text{benzene}}=\frac{0.385}{0.385+0.455}=0.458\]

H₂S, a toxic gas with a rotten-egg-like smell, is used for qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry's law constant.

Answer: \(K_H\) = 282 bar

Solubility of H₂S = 0.195 m, i.e. 0.195 mole in 1 kg of the solvent (water). 1 kg of water = 1000/18 = 55.55 moles.

\[x_{\text{H}_2\text{S}}=\frac{0.195}{0.195+55.55}=0.0035\]

Pressure at STP = 0.987 bar. Applying Henry's law, \(p_{\text{H}_2\text{S}}=K_H\,x_{\text{H}_2\text{S}}\):

\[K_H=\frac{0.987\ \text{bar}}{0.0035}=282\ \text{bar}\]

Henry's law constant for CO₂ in water is 1.67 \(\times\) 10\(^8\) Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at 298 K.

Answer: 1.854 g of CO₂

\[K_H=1.67\times10^8\ \text{Pa};\quad p_{CO_2}=2.5\ \text{atm}=2.5\times101325\ \text{Pa};\quad x_{CO_2}=\frac{p_{CO_2}}{K_H}=\frac{2.5\times101325}{1.67\times10^8}=1.517\times10^{-3}\]

\(x_{CO_2}=\dfrac{n_{CO_2}}{n_{H_2O}+n_{CO_2}}\simeq\dfrac{n_{CO_2}}{n_{H_2O}}\). For 500 mL of soda water, the water present is about 500 mL = 500 g = 500/18 = 27.78 mol.

\[n_{CO_2}=1.517\times10^{-3}\times27.78=42.14\times10^{-3}\ \text{mol};\qquad \text{mass}=42.14\times10^{-3}\times44=1.854\ \text{g}\]

The vapour pressures of pure liquids A and B are 450 and 700 mm Hg respectively at 350 K. Find out the composition of the liquid mixture if the total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Answer: Liquid: \(x_A\) = 0.40, \(x_B\) = 0.60. Vapour: \(y_A\) = 0.30, \(y_B\) = 0.70

\[p_A=x_Ap_A^0;\quad p_B=x_Bp_B^0=(1-x_A)p_B^0;\quad p_{total}=p_A+p_B=p_B^0+(p_A^0-p_B^0)x_A\]

Substituting the given values:

\[600=700+(450-700)x_A\ \Rightarrow\ 250x_A=100\ \Rightarrow\ x_A=\frac{100}{250}=0.40\]

Thus the composition of the liquid mixture is \(x_A\) = 0.40 and \(x_B\) = 1 \(-\) 0.40 = 0.60.

\[p_A=x_Ap_A^0=0.40\times450=180\ \text{mm};\quad p_B=x_Bp_B^0=0.60\times700=420\ \text{mm}\]

\[\text{Mole fraction of A in the vapour phase}=\frac{p_A}{p_A+p_B}=\frac{180}{180+420}=0.30;\quad \text{that of B}=1-0.30=0.70\]

The boiling point of water at 750 mm Hg is 96.63 \(^\circ\)C. How much sucrose is to be added to 500 g of water such that it boils at 100 \(^\circ\)C? Molal elevation constant for water is 0.52 K kg mol\(^{-1}\).

Answer: 1108.2 g (about 1.108 kg) of sucrose

Elevation of boiling point required: \(\Delta T_b\) = 100 \(-\) 96.63 = 3.37 \(^\circ\)C. Molar mass of sucrose = 342 g mol\(^{-1}\). Applying \(\Delta T_b=\dfrac{K_b\times w_2\times1000}{M_2\times w_1}\):

\[w_2=\frac{\Delta T_b\times M_2\times w_1}{K_b\times1000}=\frac{3.37\times342\times500}{0.52\times1000}=1108.2\ \text{g}\]

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of a polymer of molar mass 185,000 in 450 mL of water at 37 \(^\circ\)C.

Answer: 30.96 Pa

\[\pi=CRT=\frac{n}{V}RT=\frac{1}{185000}\ \text{mol}\times\frac{1}{0.45\ \text{L}}\times8.314\times10^{3}\ \text{Pa L K}^{-1}\text{mol}^{-1}\times310\ \text{K}=30.96\ \text{Pa}\]

The partial pressure of ethane over a saturated solution containing 6.56 \(\times\) 10\(^{-2}\) g of ethane is 1 bar. If the solution contains 5.00 \(\times\) 10\(^{-2}\) g of ethane, then what shall be the partial pressure of the gas?

Answer: 0.762 bar

Applying the relationship \(m=K_H\times p\), where \(m\) is the mass of the dissolved gas. In the first case, 6.56 \(\times\) 10\(^{-2}\) g = \(K_H\) \(\times\) 1 bar, so \(K_H\) = 6.56 \(\times\) 10\(^{-2}\) g bar\(^{-1}\).

\[\text{In the second case, }5.00\times10^{-2}\ \text{g}=(6.56\times10^{-2}\ \text{g bar}^{-1})\times p\ \Rightarrow\ p=\frac{5.00\times10^{-2}}{6.56\times10^{-2}}=0.762\ \text{bar}\]

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molecular mass of the solute?

Answer: 41.35 g mol\(^{-1}\)

Vapour pressure of pure water at the boiling point (\(p^0\)) = 1 atm = 1.013 bar; vapour pressure of the solution (\(p_s\)) = 1.004 bar.

\[\frac{p^0-p_s}{p^0}=\frac{n_2}{n_1+n_2}\simeq\frac{n_2}{n_1}=\frac{w_2/M_2}{w_1/M_1}=\frac{w_2}{w_1}\times\frac{M_1}{M_2}\]

\[\frac{1.013-1.004}{1.013}=\frac{2\ \text{g}}{98\ \text{g}}\times\frac{18\ \text{g mol}^{-1}}{M_2}\ \Rightarrow\ M_2=41.35\ \text{g mol}^{-1}\]

Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35.0 g of octane?

Answer: About 73.6 kPa

Molar masses: heptane (C₇H₁₆) = 100 g mol\(^{-1}\); octane (C₈H₁₈) = 114 g mol\(^{-1}\).

\[n_{\text{heptane}}=\frac{26.0}{100}=0.260;\quad n_{\text{octane}}=\frac{35.0}{114}=0.307;\quad x_{\text{heptane}}=\frac{0.260}{0.567}=0.459;\quad x_{\text{octane}}=0.541\]

\[p_{total}=0.459\times105.2+0.541\times46.8=48.3+25.3=73.6\ \text{kPa}\]

The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of a 1 molal solution of a non-volatile solute in it.

Answer: 12.08 kPa

Mole fraction of solute = \(\dfrac{1}{1+55.5}=0.0177\). Now \(\dfrac{p^0-p_s}{p^0}=x_2\), i.e. \(\dfrac{12.3-p_s}{12.3}=0.0177\), so \(p_s\) = 12.08 kPa.

Calculate the mass of a non-volatile solute (molar mass 40 g mol\(^{-1}\)) which should be dissolved in 114 g of octane to reduce its vapour pressure to 80%.

Answer: 10 g

Let the vapour pressure of pure octane be \(p^0\); the vapour pressure of the solution is then 0.8 \(p^0\), so the relative lowering is 0.2. Moles of octane = 114/114 = 1 mol. Let \(n_2\) be the moles of solute.

\[\frac{p^0-p_s}{p^0}=\frac{n_2}{n_1+n_2}\ \Rightarrow\ 0.2=\frac{n_2}{1+n_2}\ \Rightarrow\ n_2=0.25\ \text{mol}\]

\[\text{Mass of solute}=0.25\times40=10\ \text{g}\]

A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K. Calculate the freezing point of a 5% glucose solution in water if the freezing point of pure water is 273.15 K.

Answer: 269.06 K

Cane sugar (C₁₂H₂₂O₁₁, M = 342 g mol\(^{-1}\)): 5 g in 95 g of water. \(\Delta T_f\) = 273.15 \(-\) 271 = 2.15 K.

\[m=\frac{5/342}{0.095}=0.1539\ \text{mol kg}^{-1};\qquad K_f=\frac{\Delta T_f}{m}=\frac{2.15}{0.1539}=13.97\ \text{K kg mol}^{-1}\]

Glucose (C₆H₁₂O₆, M = 180 g mol\(^{-1}\)): 5 g in 95 g of water.

\[m=\frac{5/180}{0.095}=0.2924\ \text{mol kg}^{-1};\qquad \Delta T_f=K_fm=13.97\times0.2924=4.09\ \text{K}\]

Freezing point of the glucose solution = 273.15 \(-\) 4.09 = 269.06 K.

Two elements A and B form compounds having the molecular formulae AB₂ and AB₄. When dissolved in 20 g of benzene (C₆H₆), 1 g of AB₂ lowers the freezing point by 2.3 K, whereas 1.0 g of AB₄ lowers it by 1.3 K. The molal depression constant for benzene is 5.1 K kg mol\(^{-1}\). Calculate the atomic masses of A and B.

Answer: Atomic mass of A = 25.59 u; atomic mass of B = 42.64 u

\[M_2=\frac{1000\,K_f\,w_2}{w_1\,\Delta T_f};\quad M_{AB_2}=\frac{1000\times5.1\times1}{20\times2.3}=110.87\ \text{g mol}^{-1};\quad M_{AB_4}=\frac{1000\times5.1\times1}{20\times1.3}=196.15\ \text{g mol}^{-1}\]

Suppose the atomic masses of A and B are \(a\) and \(b\) respectively. Then \(a+2b\) = 110.87 and \(a+4b\) = 196.15. Subtracting, 2\(b\) = 85.28, so \(b\) = 42.64 u, and then \(a\) = 110.87 \(-\) 85.28 = 25.59 u.

Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain: cyclohexane, KCl, CH₃OH, CH₃CN.

Answer: KCl < CH₃OH < CH₃CN < cyclohexane

Cyclohexane and n-octane are both non-polar, hence they mix completely in all proportions. KCl is an ionic compound, while n-octane is non-polar, so KCl will not dissolve at all in n-octane. CH₃OH and CH₃CN are both polar, but CH₃CN is less polar than CH₃OH (which also has hydrogen bonding). As the solvent is non-polar, CH₃CN will dissolve more than CH₃OH in n-octane. Thus the order of solubility is KCl < CH₃OH < CH₃CN < cyclohexane.

If the solubility product of CuS is 6 \(\times\) 10\(^{-16}\), calculate the maximum molarity of CuS in aqueous solution.

Answer: 2.45 \(\times\) 10\(^{-8}\) mol L\(^{-1}\)

Maximum molarity of CuS in aqueous solution = solubility of CuS in mol L\(^{-1}\). If \(S\) is the solubility of CuS, then CuS(s) \(\rightleftharpoons\) Cu²⁺(aq) + S^2-(aq).

\[K_{sp}=[\text{Cu}^{2+}][\text{S}^{2-}]=S\times S=S^2;\qquad S=\sqrt{6\times10^{-16}}=2.45\times10^{-8}\ \text{mol L}^{-1}\]

The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.

Answer: The depression in freezing point is in the order: acetic acid < trichloroacetic acid < trifluoroacetic acid.

Fluorine, being more electronegative, has the highest electron-withdrawing inductive effect. Consequently, trifluoroacetic acid is the strongest acid while acetic acid is the weakest acid. Hence trifluoroacetic acid ionises to the largest extent while acetic acid ionises to the minimum extent to give ions in their solutions in water. The greater the ions produced, the greater is the depression in freezing point. Hence the depression in freezing point is maximum for trifluoroacetic acid and minimum for acetic acid.

Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water. \(K_a\) = 1.4 \(\times\) 10\(^{-3}\); \(K_f\) = 1.86 K kg mol\(^{-1}\).

Answer: 0.65 \(^\circ\)C

\[\text{Molality of the solution }(m)=\frac{8.16\times10^{-2}\ \text{mol}}{250\ \text{g}}\times1000\ \text{g}=0.3264\ \text{mol kg}^{-1}\]

If \(\alpha\) is the degree of dissociation of the acid (HA), then at equilibrium, with initial concentration \(C\): [HA] = \(C(1-\alpha)\), [A^-] = \(C\alpha\), [H⁺] = \(C\alpha\).

\[K_a=\frac{C\alpha\cdot C\alpha}{C(1-\alpha)}\simeq C\alpha^2\ \Rightarrow\ \alpha=\sqrt{\frac{K_a}{C}}=\sqrt{\frac{1.4\times10^{-3}}{0.3264}}=0.065\]

For 1 mole of the acid taken initially: moles at equilibrium: HA = \(1-\alpha\), A^- = \(\alpha\), H⁺ = \(\alpha\); total = \(1+\alpha\).

\[i=\frac{1+\alpha}{1}=1+0.065=1.065;\qquad \Delta T_f=iK_fm=1.065\times1.86\times0.3264=0.65\ ^\circ\text{C}\]

19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point observed is 1.0 \(^\circ\)C. Calculate the van't Hoff factor and the dissociation constant of fluoroacetic acid. \(K_f\) for water is 1.86 K kg mol\(^{-1}\).

Answer: van't Hoff factor \(i\) = 1.0753; dissociation constant \(K_a\) = 3.07 \(\times\) 10\(^{-3}\) (worked out here; the source gives no answer)

Molar mass of CH₂FCOOH = 14 + 19 + 12 + 32 + 1 = 78 g mol\(^{-1}\).

\[m=\frac{19.5/78}{0.5}=0.5\ \text{mol kg}^{-1};\qquad \Delta T_f=iK_fm\ \Rightarrow\ i=\frac{1.0}{1.86\times0.5}=1.0753\]

From \(i=1+\alpha\): \(\alpha\) = 0.0753. Taking the molar concentration \(C\) as approximately equal to the molality (0.5):

\[K_a=\frac{C\alpha^2}{1-\alpha}=\frac{0.5\times(0.0753)^2}{1-0.0753}=3.07\times10^{-3}\]

Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Answer: 17.44 mm Hg

\[n_{\text{glucose}}=\frac{25}{180}=0.1389;\quad n_{\text{water}}=\frac{450}{18}=25;\quad x_{\text{glucose}}=\frac{0.1389}{25.1389}=0.00553\]

\[p_s=p^0(1-x_{\text{glucose}})=17.535\times(1-0.00553)=17.44\ \text{mm Hg}\]

Henry's law constant for the molarity of methane in benzene at 298 K is 4.27 \(\times\) 10\(^5\) mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Answer: Mole fraction of methane = 1.78 \(\times\) 10\(^{-3}\)

\[x=\frac{p}{K_H}=\frac{760}{4.27\times10^{5}}=1.78\times10^{-3}\]

100 g of liquid A (molar mass 140 g mol\(^{-1}\)) was dissolved in 1000 g of liquid B (molar mass 180 g mol\(^{-1}\)). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

Answer: Vapour pressure of pure A = about 280 torr; vapour pressure of A in the solution = 32 torr

\[n_A=\frac{100}{140}=0.714;\quad n_B=\frac{1000}{180}=5.556;\quad x_A=0.114;\quad x_B=0.886\]

\[p_B=x_Bp_B^0=0.886\times500=443\ \text{torr};\qquad p_A=475-443=32\ \text{torr}\]

\[p_A^0=\frac{p_A}{x_A}=\frac{31.9}{0.1139}\approx280\ \text{torr}\]

Benzene and toluene form an ideal solution over the entire range of composition. The vapour pressures of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in the vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Answer: 0.60

\[n_B=\frac{80}{78}=1.026;\quad n_T=\frac{100}{92}=1.087;\quad x_B=0.4856;\quad x_T=0.5144\]

\[p_B=50.71\times0.4856=24.63;\quad p_T=32.06\times0.5144=16.49;\quad y_B=\frac{24.63}{24.63+16.49}=0.60\]

Air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportions of 20% and 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K, if the Henry's law constants for oxygen and nitrogen are 3.30 \(\times\) 10\(^7\) mm and 6.51 \(\times\) 10\(^7\) mm respectively, calculate the composition of these gases in water.

Answer: Mole fraction of O₂ = 4.61 \(\times\) 10\(^{-5}\); mole fraction of N₂ = 9.22 \(\times\) 10\(^{-5}\)

Total pressure of air in equilibrium with water = 10 atm. As air contains 20% oxygen and 79% nitrogen by volume:

\[p_{O_2}=\frac{20}{100}\times10\ \text{atm}=2\ \text{atm}=2\times760=1520\ \text{mm};\qquad p_{N_2}=\frac{79}{100}\times10=7.9\ \text{atm}=7.9\times760=6004\ \text{mm}\]

\[p_{O_2}=K_Hx_{O_2}\ \Rightarrow\ x_{O_2}=\frac{1520}{3.30\times10^7}=4.61\times10^{-5};\qquad x_{N_2}=\frac{6004}{6.51\times10^7}=9.22\times10^{-5}\]

Determine the amount of CaCl₂ (\(i\) = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27 \(^\circ\)C.

Answer: 3.42 g

\[\pi=iCRT\ \Rightarrow\ C=\frac{0.75}{2.47\times0.0821\times300}=0.01233\ \text{mol L}^{-1}\]

\[\text{Moles in 2.5 L}=0.01233\times2.5=0.0308;\qquad \text{mass}=0.0308\times111=3.42\ \text{g}\]

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25 \(^\circ\)C, assuming that it is completely dissociated.

Answer: 5.27 \(\times\) 10\(^{-3}\) atm

K₂SO₄ dissociates completely as K₂SO₄ \(\rightarrow\) 2K⁺ + SO₄^2-; ions produced = 3, so \(i\) = 3.

\[\pi=iCRT=i\times\frac{n}{V}RT=3\times\frac{0.025\ \text{g}}{174\ \text{g mol}^{-1}}\times\frac{1}{2\ \text{L}}\times0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}\times298\ \text{K}=5.27\times10^{-3}\ \text{atm}\]