Question

NCERT Solutions Class 12 Chemistry Chapter 2 Solutions have solutions to all the queries students might face. These NCERT Solutions are made by an expert mentor who has a long time of experience in the subjects. Our experts suggest that NCERT Solutions Class 12 is the best way to prepare for board exams.

• Solutions – a solution is a homogenous mixture of substances most probably two or more than two in various amounts which also vary many a times and also limit the solubility. The solution is commonly applied to the liquid state of the matter but rarely it is can be also used for the gaseous and some solids.
• Types of Solutions – there are different ways to form a solution. A solution requires both solvent and solute to form. In broad exams, the solution has three different types as
  1. Solid-liquid solutions
  2. Liquid-liquid solutions
  3. Gas-liquid solutions

S. No.	Solute	Solvent	Types of Solution	Example
1.	solid	solid	solid in solid	stainless steel
2.	Liquid	Solid	Liquid in Solid	washing soda
3.	Gas	Solid	Gas in Solid	H2 in palladium
4.	Solid	Liquid	Solid in liquid	salt and glucose solution
5.	Liquid	Liquid	Liquid in liquid	ethanol in water
6.	Gas	Liquid	Gas in liquid	aerated water, O2 in H2O
7.	Solid	Gas	Solid in gas	camphor in N2 gas
8.	Liquid	Gas	Liquid in gas	humidity of the air
9.	Gas	Gas	Gas in gas	air, all the gases mixed

• Solubility – the ability of any substance to form a solution when another kind of substance is mixed in it is called solubility.
• Vapour Pressure of Liquid Solutions – pressure applied by any gas on the liquid solvent when it is at the equilibrium at any certain temperature is called the vapor pressure.

NCERT Solutions Class 12 Chemistry is the best study material one can find online so our experts suggest going through our solutions for cracking their tough exams.

Answer 1

NCERT Solutions Class 12 Chemistry Chapter 2 Solutions

1. Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Answer:

A homogeneous mixture of two or more non-reacting substances whose composition varies within certain fixed limits, known as a solution. Generally, the component that is present in the largest quantity is known as solvent. Solvent determines the physical state in which solution exists. One or more components present in the solution other than solvent are called solutes. 

When the solutions contains two, three or four components, it is known as binary, ternary or quaternary solutions respectively. On the basis of physical state of components, the solutions are of the following types: 

1) When both solute and solvent are in solid state. Example: Alloys. 

2) When solute is in solid and solvent is in liquid state. Example: Sugar or salt solutions. 

3) When solute is in solid and solvent is in gaseous state. Example: iodine vapours in air. 

4) When both solute and solvent are in liquid state. Example: alcohol in water. 

5) When solute is in liquid and solvent is in solid state. Example: Zinc amalgam.

6) When solute is in liquid and solvent is in gaseous state. Example: water vapour in air. 

7) When both solute and solvent are in gaseous state. Example: air. 

8) When solute is in gaseous and solvent is in liquid state. Example: aerated drinks. 

9) When solute is in gaseous and solvent is in solid state. Example: dissolve gas in mineral.

2. Give an example of a solid solution in which the solute is a gas.

Answer:

Dissolved gases in minerals.

3. Define the following terms: 

(i) Mole fraction 

(ii) Molality 

(iii) Molarity 

(iv) Mass percentage.

Answer:

(i) Mole fraction: Commonly used symbol for mole fraction is x and subscript used on the right hand side of x denotes the component. It is defined as:

\(Mole \,fraction \,of\,a\,component=\frac{Number\,of\,moles\,of\,the\,component}{Number\,of\,moles\,of\,all\,the\,component}\) 

(ii) Molality: Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:

(iii) Molarity: Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.

(iv) Mass percentage: The mass percentage of a component of a solution is defined as:

4. Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL^–1?

Answer:

In 100 g of nitric acid (HNO₃) solution, 

Mass of nitric acid (HNO₃) = 68 g

Molar mass of HNO₃ = 63 g/mol

Moles on 68 g of HNO₃ = \(\frac{68}{63}\) mole = 1.079 mole

Density of solution = 1.504 g mL^–1 

Therefore, volume of solution = \(\frac{Mass\,of\,solution}{Density \,of\,solution}=\frac{100}{1.504}mL = 0.0665 \,L\)

Molarity of the solution =  \(\frac{Moles\,of\,solute}{Volume\,of\, solution\,in\,L}= \frac{1.079}{0.0665}M= 16.23\, M\)

5. A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL^–1, then what shall be the molarity of the solution?

Answer:

10% w/w means 10 g glucose is present in 100 g solution. 

Mass of water = 100 – 10 = 90 g = 0.90 kg 

10 g glucose = \(\frac{10}{180}\) mol = 0.0555 moles 

Number of moles in 90 g of H₂O = \(\frac{90}{18}\) mol = 5 moles

Molality of the solution = \(\frac{0.0555}{0.090}m = 0.617 \,m\)

Mole fraction of glucose = \(\frac{0.0555}{5 + 0.0555}= 0.01\)

Mole fraction of H₂O =1 – 0.01 = 0.99

Volume of 100 g solution = \(\frac{100}{1.2}mL\) = 83.33 mL = 0.08333 L

Molarity = \(\frac{0.0555\,mol}{0.08333\,L} \) = 0.67 M

6. How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?

Answer:

Step 1: To calculate the number of moles of the components in the mixture: 

Suppose mass of Na₂CO₃ present in the mixture = x g 

Mass of NaHCO₃ present in the mixture = (1 − x) g 

Molar mass of Na₂CO3 = 2 × 23 + 12 +3 × 16 = 106 g mol⁻¹

Molar mass of NaHCO₃ = 23+ 1 + 12+ 3 × 16 = 84 g mol⁻¹

Number of moles of Na₂CO₃ in x g = \(\frac x{106}\)

Number of moles of NaHCO₃ in (1 − x) g = \(\frac{1-x}{84}\)

As mixture contains equimolar amount of the two, therefore,

Now, Number of moles of Na₂CO₃ = \(\frac{0.558}{106}= 0.00526\)

Number of moles of NaHCO₃ = \(\frac{1-0.558}{84}=\frac{0.442}{84}=0.00526\)

Step 2: To calculate the number of moles of HCl required:

1 mole of Na₂CO₃ requires = 2 mole of HCl 

Therefore, 0.00526 mole of Na₂CO₃ requires = 0.00526 × 2 moles = 0.01052 mole 

1 mole of NaHCO₃ requires = 1 mole of HCl 

Therefore, 0.00526 mole of NaHCO₃ requires = 0.00526 mole 

Total HCl required = 0.01052 + 0.00526 = 0.01578 moles

Step 3: To calculate the volume of 0.1 M HCl:

0.1 mole of 0.1 M HCl are present in 1000 ml of HCl, therefore, 

0.01578 mole of 0.1 M HCl will be present in = \(\frac{1000}{0.1}\) × 0.01578 = 157.8 mL of HCl.

7. A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

Answer:

Total amount of solute present in the mixture is given by, 300 × \(\frac{25}{100}\) + 400 × \(\frac{40}{100}\) = 75 + 160 = 235 g

Total amount of solution = 300 + 400 = 700 g 

Therefore, mass percentage (w/w) of the solute in the resulting solution,

\(=\frac{235}{700}\times100\% = 33.57 \%\)

And, mass percentage (w/w) of the solvent in the resulting solution = (100 − 33.57) % = 66.43%

Answer 2

8. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^–1, then what shall be the molarity of the solution?

Answer:

Mass of the solute, C₂H₆O₂ = 222.6 g 

Molar mass of C₂H₆O₂ = 62 g mol^–1

Number of moles of solute = \(\frac{222.6g}{62\,g\,mol^{-1}}= 3.59\)

Mass of solvent = 200 g = 0.200 kg

Molality of the solution = \(\frac{3.59 \,moles}{0.200\,kg}=17.95\)

Total mass of the solution = 222.6 +200 = 422.6 g

Volume of the solution = \(\frac{Mass\,of\,solution}{Density\,of\,solution}=\frac{422.6g}{1.072 \,g \,mL^{-1}}=394.2 \,mL= 0.3942\,L\)

9. A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): 

(i) express this in percent by mass 

(ii) determine the molality of chloroform in the water sample.

Answer:

15 ppm means 15 part in one million (10⁶) parts by mass in the solution. Therefore, 

Percentage by mass = \(\frac{18}{10^6}\times100=15 \times10^{-4}\)

Taking 15 g chloroform in 10⁶ g of the solution, mass of the solvent = 10⁶ g [∵ Neglecting 15 g in comparison with 10⁶ g]

Molar mass of CHCl₃ = 12 + 1 + 3 × 35.5 = 119.5 g mol^–1

Therefore, molality = \(\cfrac{\frac{15}{119.5}}{10^6}\times100=1.25\times10^{-4}\,m\)

10. What role does the molecular interaction play in a solution of alcohol and water?

Answer:

There is strong intramolecular hydrogen bonding in alcohol molecules as well as water molecules separately. On mixing, intermolecular hydrogen bonding between alcohol and water molecules is comparatively week. Hence, they show positive deviations from ideal behaviour. As a result, the solution will have higher vapour pressure and lower boiling point than that of water and alcohol.

11. Why do gases always tend to be less soluble in liquids as the temperature is raised?

Answer:

Dissolution of gas in liquid is an exothermic process. 

Gas + Solvent ⇌ Solution + Heat

As the temperature is increased, equilibrium shifts backward, according to Le Chatelier’s Principle.

12. State Henry’s law and mention some important applications?

Answer:

According to Henry’s law, at a particular temperature, the solubility of a gas in given volume of a liquid is directly proportional to the pressure of the gas above the liquid. 

\(m\propto p\)

\(m = kp\)

Where, k = Henry’s constant. 

Here, m and p represent the mass of the dissolved gas and pressure respectively.

Some important applications of Henry’s law: 

♦ To increase the solubility of CO² in soft drinks, the bottle is sealed under high pressure. 

♦ Scuba divers face the problems of high concentration of dissolved gasses while breathing air at high pressure underwater. When the divers reach the surface, the pressure gradually decreases. This releases the dissolved gasses and leads to the formation of bubbles of nitrogen in blood. This blocks capillaries and creates a medical condition known bends which are painful and dangerous to life. To avoid bends, tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen). 

♦ The partial pressure of oxygen at high altitudes is smaller than at the sea level. This results in low concentration of oxygen in the blood and tissues. Low blood oxygen causes climbers to become weak and unable to think. Such symptoms are known as anoxia.

13. The partial pressure of ethane over a solution containing 6.56 × 10⁻³ g of ethane is 1 bar. If the solution contains 5.00 × 10⁻² g of ethane, then what shall be the partial pressure of the gas?

Answer:

Molar mass of ethane (C₂H₆) = 2 × 12 + 6 × 1 = 30 g mol⁻¹

\(\therefore\) Number of moles present in 6.56 × 10⁻² g of ethane = \(\frac{6.56\times10^{-2}}{30}=2.187\times10^{-3}\,mol\)

Let the number of moles of the solvent be x. 

According to Henry’s law,

p = KHx

Number of moles present in 5.00 × 10⁻² g of ethane

According to Henry’s law,

p = KHx

Hence, partial pressure of the gas shall be 0.764 bar.

14. What is meant by positive and negative deviations from Raoult's law and how is the sign of ∆_solH related to positive and negative deviations from Raoult's law?

Answer:

According to Raoult’s law, the partial vapour pressure of each volatile component in any solution is directly proportional to its mole fraction. The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The solutions that do not obey Raoult’s law (non-ideal solutions) have vapour pressures either higher or lower than that predicted by Raoult’s law. If the vapour pressure is higher, then the solution is said to exhibit positive deviation, and if it is lower, then the solution is said to exhibit negative deviation from Raoult’s law.

Vapour pressure of a two-component solution showing positive deviation from Raoult’s law

Vapour pressure of a two-component solution showing negative deviation from Raoult’s law

In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solution is zero. 

∆_solH = 0 

In the case of solutions showing positive deviations, absorption of heat takes place. 

\(\therefore\) ∆_solH = Positive 

In the case of solutions showing negative deviations, evolution of heat takes place. 

\(\therefore\) ∆_solH = Negative

15. 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 molar mass of the solute?

Answer:

Here, Vapour pressure of the solution at normal boiling point (p₁) = 1.004 bar 

Vapour pressure of pure water at normal boiling point (p₁ ⁰) = 1.013 bar 

Mass of solute, (w₂) = 2 g and Mass of solvent (water), (w₁) = 98 g 

Molar mass of solvent (water), (M₁) = 18 g mol⁻¹ 

According to Raoult’s law,

= 41.35 g mol⁻¹ 

Hence, the molar mass of the solute is 41.35 g mol⁻¹.

Answer 3

16. 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 g of octane?

Answer:

Vapour pressure of heptane (p₁ ⁰) = 105.2 kPa 

Vapour pressure of octane (p₂ ⁰) = 46.8 kPa

We know that, 

Molar mass of heptane (C₇H₁₆) = 7 × 12 + 16 × 1 = 100 g mol⁻¹

\(\therefore\) Number of moles of heptane = \(\frac{26}{100}\) mol = 0.26 mol 

Molar mass of octane (C₈H₁₈) = 8 × 12 + 18 × 1 = 114 g mol^−1​

​​​​​​\(\therefore\)  Number of moles of octane = \(\frac{35}{114}\) mol = 0.31 mol

Mole fraction of heptane, x₁ = \(\frac{0.26}{0.26+0.31}\) = 0.456

And, mole fraction of octane, x₂ = 1 − 0.456 = 0.544 

Now, partial pressure of heptane, p₁ = x₁p₁ ⁰ = 0.456 × 105.2 = 47.97 kPa 

Partial pressure of octane, p₂ = x₂p₂ ⁰ = 0.544 × 46.8 = 25.46 kPa 

Hence, vapour pressure of solution, p_total = p₁ + p₂ = 47.97 + 25.46 = 73.43 kPa

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

Answer:

1 molal solution means 1 mol of the solute is present in 100 g of the solvent (water). 

Molar mass of water = 18 g mol⁻¹ 

Therefore, number of moles present in 1000 g of water = \(\frac{1000}{18}\) = 55.56 mol 

Therefore, mole fraction of the solute in the solution is x₂ = \(\frac{1}{1+55.56}=0.0177\)

It is given that, Vapour pressure of water, p₁ ⁰ = 12.3 kPa 

Applying the relation,

\(\frac{p_1\,^0-p_1}{p_1 \,^0}= x_2 \) ⇒ \(\frac{12.3 - p_1}{12.3} = 0.0177\)

⇒ 12.3 − p₁ = 0.2177 ⇒ p₁ = 12.0823 = 12.08 kPa (approximately) 

Hence, the vapour pressure of the solution is 12.08 kPa.

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

Answer:

Let the vapour pressure of pure octane be p₁ ⁰ . 

Then, the vapour pressure of the octane after dissolving the non-volatile solute is 80 100 p₁ ⁰ = 0.8p₁ ⁰ 

Molar mass of solute, M₂ = 40 g mol−1 and mass of octane, w₁ = 114 g 

Molar mass of octane, (C₈H₁₈), M₁ = 8 × 12 + 18 × 1 = 114 g mol⁻¹, 

Applying the relation,

Hence, the required mass of the solute is 8 g.

19. A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: 

(i) molar mass of the solute 

(ii) vapour pressure of water at 298 K.

Answer:

(i) Let, the molar mass of the solute be M g mol⁻¹

Now, the no. of moles of solvent (water), n₁ = \(\frac{90g}{18g\,mol^{-1}}= 5 \) mol and, the no. of moles of solute, n₂ = \(\frac{30\,g}{M\,g\,mol^{-1}}=\frac{30}{M} \) mol

p₁ = 2.8 kPa, 

Applying the relation:

After the addition of 18 g of water: n₁ = \(\frac{90g+18g}{18g\,mol^{-1}}=6\,mol\)

p₁ = 2.9 kPa, 

Again, applying the relation:

Dividing equation (i) by (ii), we have:

Therefore, the molar mass of the solute is 23 g mol⁻¹.

(ii) Putting the value of ‘M’ in equation (i), we have:

Hence, the vapour pressure of water at 298 K is 3.53 kPa.

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

Answer:

Here, ∆Tf = (273.15 − 271) K = 2.15 K 

Molar mass of sugar (C₁₂H₂₂O₁₁) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol⁻¹ 

5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 − 5)g = 95 g of water. 

Now, number of moles of cane sugar = \(\frac{5}{342}\) mol = 0.0146 mol

Therefore, molality of the solution, m = \(\frac{0.0146\,mol}{0.095\,kg}= 0.1537\, mol\,kg^{-1}\)

Applying the relation, ∆T_f = K_f × m

Molar of glucose (C₆H₁₂O₆) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol⁻¹ 

5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water. 

Therefore, number of moles of glucose = \(\frac5{180}mol=0.0278\,mol\)

Therefore, molality of the solution, m = \(\frac{0.0278\,mol}{0.095\,kg}=0.2926 \, mol \,kg^{-1}\)

Applying the relation, ∆T_f = K_f × m = 13.99 K kg mol⁻¹ × 0.2926 mol kg⁻¹ = 4.09 K (approximately)

Hence, the freezing point of 5% glucose solution is (273.15 − 4.09) K= 269.06 K.

21. Two elements A and B form compounds having formula 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 molar depression constant for benzene is 5.1 Kkg mol⁻¹. 

Calculate atomic masses of A and B.

Answer:

We know that, M₂ = \(\frac{1000\times w_2\times k_f}{\triangle T_f\times w_1}\)

Then,

Now, we have the molar masses of AB₂ and AB₄ as 110.87 g mol⁻¹ and 196.15 g mol⁻¹ respectively. 

Let the atomic masses of A and B be x and y respectively. 

Now, we can write:

x + 2y = 110.87     … (i)

x + 4y = 196.15     … (ii)

Subtracting equation (i) from (ii), we have

2y = 85.28 ⇒ y = 42.64 

Putting the value of ‘y’ in equation (1), we have x + 2 × 42.64 = 110.87 

⇒ x = 25.59 

Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.

Answer 4

22. At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

Answer:

Here, T = 300 K π = 1.52 bar 

R = 0.083 bar L K⁻¹ mol⁻¹ Applying the relation, π = CRT

⇒ \(C=\frac{\pi}{RT}=\frac{1.52\,bar}{0.083\,bar\,L\,K^{-1}\times 300K}= 0.061 \,mol\)

Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.

23. Suggest the most important type of intermolecular attractive interaction in the following pairs. 

(i) n-hexane and n-octane 

(ii) I₂ and CCl₄ 

(iii) NaClO₄ and water 

(iv) methanol and acetone 

(v) acetonitrile (CH₃CN) and acetone (C₃H₆O).

Answer:

(i) Van der Wall’s forces of attraction. 

(ii) Van der Wall’s forces of attraction. 

(iii) Ion-diople interaction. 

(iv) Dipole-dipole interaction. 

(v) Dipole-dipole interaction.

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

Answer:

n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in the noctane. 

The order of increasing polarity is: Cyclohexane < CH₃CN < CH₃OH < KCl 

Therefore, the order of increasing solubility is: KCl < CH₃OH < CH₃CN < Cyclohexane

25. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? 

(i) phenol 

(ii) toluene 

(iii) formic acid 

(iv) ethylene glycol 

(v) chloroform 

(vi) pentanol.

Answer:

(i) Phenol (C₆H₅OH) has the polar group −OH and non-polar group −C₆H₅. Thus, phenol is partially soluble in water. 

(ii) Toluene (C₆H₅−CH₃) has no polar groups. Thus, toluene is insoluble in water. 

(iii) Formic acid (HCOOH) has the polar group −OH and can form H-bond with water. Thus, formic acid is highly soluble in water

(iv) Ethylene glycol  has polar −OH group and can form H−bond. Thus, it is highly soluble in water. 

(v) Chloroform is insoluble in water. 

(vi) Pentanol (C₅H₁₁OH) has polar −OH group, but it also contains a very bulky nonpolar −C₅H₁₁ group. Thus, pentanol is partially soluble in water.

26. If the density of some lake water is 1.25 g mL⁻¹ and contains 92 g of Na⁺ ions per kg of water, calculate the molality of Na⁺ ions in the lake.

Answer:

Number of moles present in 92 g of Na⁺ ions = \(\frac{92g}{23\,g\,mol^{-1}}=4\,mol\)

Therefore, molality of Na⁺ ions in the lake = \(\frac{4\,mol}{1\,kg}= 4m\) 

27. If the solubility product of CuS is 6 × 10⁻¹⁶, calculate the maximum molarity of CuS in aqueous solution.

Answer:

Solubility product of CuS, Ksp = 6 × 10⁻¹⁶ 

Let s be the solubility of CuS in mol L⁻¹.

\(CuS\leftrightarrow Cu^{2+}+S^{2-}\)

\(K_{sp}=\overset s{[Cu^{2+}]}\overset s{[S^{2-}]}\)

Now,

= s × s = s² 

Then, we have, K_sp = s² = 6 × 10⁻¹⁶

⇒ s = \(\sqrt{6\times10^{-16}}\) = 2.45 × 10⁻⁸ mol L⁻¹ 

Hence, the maximum molarity of CuS in an aqueous solution is 2.45 × 10⁻⁸ mol L⁻¹.

28. Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN.

Answer:

6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN. 

Then, total mass of the solution = (6.5 + 450) g = 456.5 g

Therefore, mass percentage ofC₉H₈O₄ = \(\frac{6.5}{456.5}\times100\%=1.424 \%\)

29. Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10⁻³m aqueous solution required for the above dose.

Answer:

The molar mass of Nalorphene (C₁₉H₂₁NO₃) is given as: 1

9 × 12 +12 × 1 + 1 × 14 + 3 × 16 = 311 g mol⁻¹

In 1.5 × 10⁻³ m aqueous solution of nalorphene, 1 kg (1000 g) of water contains 1.5 × 10⁻³ mol 

= 1.5 × 10⁻³ × 311 g = 0.4665 g 

Therefore, total mass of the solution = (1000 + 0.4665) g = 1000.4665 g

This implies that the mass of the solution containing 0.4665 g of nalorphene is 1000.4665 g.

Therefore, mass of the solution containing 1.5 mg of nalorphene is: \(\frac{1000.4665\times1.5\times10^{-3}}{0.4665}g = 3.22g\)

Hence, the mass of aqueous solution required is 3.22 g.

30. Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol.

Answer:

0.15 M solution of benzoic acid in methanol means, 

1000 mL of solution contains 0.15 mol of benzoic acid

Therefore, 250 mL of solution contains = \(\frac{0.15\times250}{1000}mol\,of\,benzoic\,acid=0.0375\,mol\,of\,benzoic\,acid\)

Molar mass of benzoic acid (C₆H₅COOH) = 7 × 12 + 6 × 1 + 2 × 16 = 122 g mol⁻¹ 

Hence, required benzoic acid = 0.0375 mol × 122 g mol⁻¹ = 4.575 g

31. 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:

Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H+ ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point. Hence, the depression in the freezing point increases in the order: 

Acetic acid < trichloroacetic acid < trifluoroacetic acid

Answer 5

32. 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 × 10⁻³, K_f = 1.86 K kg mol⁻¹.

Answer:

Molar mass of CH₃CH₂CHClCOOH = 15 + 14 + 13 + 35.5 +12 + 16 + 16 + 1 = 122.5 g mol^–1 

No. of moles present in 10 g of CH₃CH₂CHClCOOH = \(\frac{10g}{122.5\,g\,mol^{-1}}= 0.0816\, mol\)

It is given that 10 g of CH₃CH₂CHClCOOH is added to 250 g of water.

\(\therefore\) Molality of the solution, \(\frac{0.0186}{250}\times1000=0.3264\,mol\,kg^{-1}\)

Let α be the degree of dissociation of CH₃CH₂CHClCOOH. 

CH₃CH₂CHClCOOH undergoes dissociation according to the following equation:

Since α is very small with respect to 1, 1 − α ≈ 1

Again,

Total moles of equilibrium = 1 − α + α + α 

= 1 + α

Hence, the depression in the freezing point of water is given as:

 33. 19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

Answer:

It is given that:

We know that:

Therefore, observed mass of CH₂FCOOH, (M₂)_obs = 72.54 g mol^–1 

The calculated mass of CH₂FCOOH, (M₂)cal = 14 + 19 + 12 + 16 + 16 + 1 = 78 g mol^–1

Therefore, the van’t Hoff factor, \(i=\frac{(M_2)cal}{(M_2)obs}\)

\(=\frac{78\,g\,mol^{-1}}{72.54\,g\,mol^{-1}}\)

= 1.0753

Let a be the degree of dissociation of CH₂FCOOH.

Now, the value of K_a is given as:

Taking the volume of the solution as 500 mL, we have the concentration:

34. Vapour pressure of water at 293 Kis 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:

Vapour pressure of water, \(p^0_1\) = 17.535 mm of Hg 

Mass of glucose, w₂ = 25 g 

Mass of water, w₁ = 450 g

We know that, 

Molar mass of glucose (C₆H₁₂O₆), M₂ = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol⁻¹

Molar mass of water, M₁ = 18 g mol⁻¹ 

Then, number of moles of glucose, n₂ = \(\frac{25g}{180\,g\,mol^{-1}}=0.139\,mol\)

And, number of moles of water, n₁ = \(\frac{450g}{18g\,mol^{-1}}= 25 \,mol\)

We know that,

⇒ 17.535 − p₁ = 0.097 

⇒ p₁ = 17.44 mm of Hg 

Hence, the vapour pressure of water is 17.44 mm of Hg.

35. Henry’s law constant for the molality of methane in benzene at 298 Kis 4.27 × 10⁵ mm Hg. Calculate the solubility of methane in benzene at 298 Kunder 760 mm Hg.

Answer:

Here, p = 760 mm Hg k_H = 4.27 × 105 mm Hg 

According to Henry’s law, p = k_Hx

Hence, the mole fraction of methane in benzene is 178 × 10⁻⁵.

36. 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:

Number of moles of liquid A, n_A = \(\frac{100g}{140\,g\,mol^{-1}} = 0.714 \,mol\)

Number of moles of liquid B, n_B = \(\frac{1000g}{180\,g\,mol^{-1}} = 5.556\,mol\)

Then, mole fraction of A, x_A = \(\frac{n_A}{n_A+n_B}=\frac{0.714}{0.714+5.556}= 0.114\)

And, mole fraction of B, x_B = 1 − 0.114 = 0.886 

Vapour pressure of pure liquid B, p_B 0= 500 torr 

Therefore, vapour pressure of liquid B in the solution, 

p_B = \(p^0_B\) x_B = 500 × 0.886 = 443 torr 

Total vapour pressure of the solution, p_total = 475 torr 

So, vapour pressure of liquid A in the solution,

p_A = p_total − p_B = 475 − 443 = 32 torr

Now, p_A = \(p^0_A\) x_A 

_⇒ \(p^0_A\) = \(\frac{p_A}{x_A}=\frac{32}{0114}\) = 280.7 torr

Hence, the vapour pressure of pure liquid A is 280.7 torr

Answer 6

37. Vapour pressure of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot p_total’ p_chloroform’ and p_acetone as a function of x_acetone. The experimental data observed for different compositions of mixture is.

100 \(\times\) xacetone	0	11.8	23.4	36.0	50.8	58.2	64.5	72.1
pacetone / mm Hg	0.	54.9	110.1	202.4	322.7	405.9	454.1	521.1
pchloroform / mm Hg	632.8	548.1	469.4	359.7	257.7	193.6	161.2	120.7

Answer:

From the question, we have the following data

100 \(\times\) xacetone	0	11.8	23.4	36.0	50.8	58.2	64.5	72.1
pacetone / mm Hg	0.	54.9	110.1	202.4	322.7	405.9	454.1	521.1
pchloroform / mm Hg	632.8	548.1	469.4	359.7	257.7	193.6	161.2	120.7
ptotal(mm Hg)	632.8	603.0	579.5	562.1	580.4	599.5	615.3	641.8

 

It can be observed from the graph that the plot for the p_total of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.

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

Answer:

Molar mass of benzene (C₆H₆) = 6 × 12 + 6 × 1 = 78 g mol⁻¹

Molar mass of toluene (C₆H₅CH₃) = 7 × 12 + 8 × 1 = 92 g mol⁻¹

Now, no. of moles present in 80 g of benzene = \(\frac{80}{78}\) = 1.026 mol 

And, no. of moles present in 100 g of toluene = \(\frac{100}{92}\) =  1.087 mol

 Mole fraction of benzene, x_b = \(\frac{1.026}{1.026 + 1.087}\) = 0.486

And, mole fraction of toluene, x_t = 1− 0.486 = 0.514  

It is given that vapour pressure of pure benzene \(p^0_b\) = 50.71 mm Hg

And, vapour pressure of pure toluene, \(p^0_t\) = 32.06 mm Hg 

Therefore, partial vapour pressure of benzene, p_b = x_b. \(p^0_b\) = 0.486 × 50.71 = 24.645 mm Hg

And, partial vapour pressure of toluene, p_t = x_t . \(p^0_t\) = 0.514 × 32.06 = 16.479 mm Hg 

Hence, mole fraction of benzene in vapour phase is given by:

39. The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 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 × 10⁷ mm and 6.51 × 10⁷ mm respectively, calculate the composition of these gases in water.

Answer:

Percentage of oxygen (O₂) in air = 20 %

Percentage of nitrogen (N₂) in air = 79% A

lso, it is given that water is in equilibrium with air at a total pressure of 10 atm, i.e., (10 × 760) mm Hg = 7600 mm Hg

Therefore, 

Partial pressure of oxygen, \(p_{O_2}\) = \(\frac{20}{100}\) x 7600 mm Hg = 1520 mm Hg 

Partial pressure of nitrogen, \(p_{N_2} \) = \(\frac{79}{100}\) × 7600 mm Hg = 6004 mm Hg

Now, according to Henry’s law: p = K_H.x 

For oxygen: \(p_{O_2}=K_H.x_{O_2}\)

For nitrogen: \(p_{N_2}=K_H.x_{N_2}\)

Hence, the mole fractions of oxygen and nitrogen in water are 4.61 ×10⁻⁵ and 9.22 × 10⁻⁵ respectively.

40. 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°C.

Answer:

We know that,

Here, R = 0.0821 L atm K^-1 mol^-1 and M = 1 × 40 + 2 × 35.5 = 111g mol^-1

Therefore, w = \(\frac{0.75\times111\times2.5}{2.47\times0.0821\times300}\) = 3.42 g

Hence, the required amount of CaCl₂ is 3.42 g.

Answer 7

 41. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 liter of water at 25° C, assuming that it is completely dissociated.

Answer:

When K₂SO₄ is dissolved in water, K⁺ and \(SO^{2-}_4\) ions are produced. 

K₂SO₄ → 2K⁺  + \(SO^{2-}_4\)

Total number of ions produced = 3, therefore i =3 

Given, w = 25 mg = 0.025 g, V = 2 L, T = 25°C = (25 + 273) K = 298 K 

Also, we know that: 

R = 0.0821 L atm K^-1 mol^-1 and M = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol^-1 

Appling the following relation,