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 –

Solution is the homogenous mixture of two or more substances which are chemically nonreactive. There are nine type of solutions given below –

Solid solutions –

(a) Gas in solid solution of hydrogen in palladium

(b) Liquid in solid amalgam e.g – Na – Hg

(c) Solid in solid gold ornaments (Cu /Ag with Au)

Liquid solution –

(a) Gas in liquid CO2 dissolved in water (aerated water).

(b) Liquid in liquid ethanol dissolved in water.

(c) Solid in liquid sugar dissolved in water, saline water etc.

Gasesous Solution –

(a) Gas in gas, air mixture of O2 and N2.

(b) Liquid in gas water vapor

(c) Solid in gas camphor in N2 gas, smoke etc.

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

Answer – The solution of hydrogen in palladium and dissolved gases in minerals.

3. Define the following terms –

(a) Mole fraction

(b) Molarity

(c) Molality

(d) Mass percentage Answer –

(a) Mole fraction – It is defined as the number of moles of solute to the total number of moles in solution. If A is the number of moles of solute dissolved in B moles of solvent. Mole fraction of solute, XA = nA / nA + nB

(b) Molarity – It is defined as the number of moles of solute present in one litre of the solution.

(c) Molality – It is defined as the number of moles of solute present in 1000 g of solvent.

Molality = Number of moles of the solute/weight of solvent in Kg

(d) Mass percentage –It is defined as the amount of solute in grams present in 100 grams of solution.

Mass percentage = (Mass of solute x 100) / Mass of solution

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


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


10% w/w solution of glucose in water means 10 g glucose and 90 g of water. Molar mass of glucose – 180 g/mol Molar mass of water = 18 g/mol Number of moles in 10g of glucose = 10/180 = 0.0555 moles Number of moles in 90 g of water = 90 /18 = 5 moles

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


The number of moles of components in the mixture =?

Consider – Amount of Na2CO3 in the mixture = xg

Amount of NaHCO3 in the mixture = (1 – x)g

Molar mass of Na2CO3 = 106 g/mol

Number of moles Na2CO3 = x/106 mol

Molar mass of NaHCO3 = 84 g/mol

Number of moles of NaHCO3 = 1 – x / 84 mol

According to the question, the mixture containing an equal quantity of Na2CO3 and NaHCO3 –

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.


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, then what shall be the molarity of the solution?


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

i. Express this in percent by mass.

ii. Determine the molality of chloroform in the water sample ?


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

Answer –

Alcohol and water have tendency to form the intermolecular hydrogen bonding. But when a solution of alcohol and water is formed, then interaction become weaker and less extensive than alcohol- alcohol and water – water interaction.Thus the molecules can easily escape and show positive deviation from ideal behavior. This increase the vapor pressure of the solution and lowers the boiling point of the resulting solution.

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

Answer –

The dissolution of gases is an exothermic reaction means energy is released in the form of heat. When the temperature is increases, heat is supplied and the equilibrium shifts in backward direction that decreases the solubility. This is according to the lechatlier’s principal. So, the gases become less soluble in liquids.

12. State Henry law and its important applications.


Henry law states that the partial pressure of a gas in the vapor phase is proportional to the mole fraction of the gas in the solution. According to Henry law – P = KH . x

P is the partial pressure of the gas in vapor phase.

x is the mole fraction of the gas.

KH is the Henry’s constant.

The important applications of Henry’s constant are given below –

( a) In deep sea diving.

Henry law states that the solubility of the gases increases, pressure increases. When a scuba diver dives deep into the sea, pressure increases that causes the nitrogen present in air to dissolve in blood in great amounts. As he comes back to the surface, the solubility of nitrogen decreases and dissolved gas is released that leads to the formation of nitrogen bubbles in blood. This results in the blockage of capillaries that leads to the formation of ‘bends’ or ‘decompression sickness’. So, oxygen tanks are used by scuba divers that filled with the air and diluted with the helium to avoid bends.

( b) The people living at high altitude, concentration of oxygen in their blood and tissue is low. low blood O2 causes causes climbers to become weak and make them unable to think clearly. This is because at high altitude, the partial pressure of the oxygen is less than at ground level. These are the symptoms of anoxia.

(c) Bottles are sealed under high pressure to increase the solubility of CO2 in soft drinks and soda water.

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


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 pressure of each volatile component in any solution in vapor phase is directly proportional to its mole fraction. The solutions which obey Raoult’s law are known as ideal solutions. The solutions that do not obey Raoult’s law are known as non-ideal solution. If the solution having vapor pressure is higher, then it is said to exhibit positive deviation, and if it is lower vapor pressure than expected, then the solution is said to exhibit negative deviation from Raoult’s law

Vapor pressure of two-component solution showing positive deviation In case of positive deviation, absorption of heat takes place. Hsol = positive

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?


16. Heptane and octane form an ideal solution. At 373k, the vapor pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapor pressure of a mixture of 26.0 g of heptanes and 35 g of octane?


17. The vapor pressure of water is 12.3 kPa at 300k. Calculate vapor pressure of 1 molal solution of non – volatile solute in it.


18. Calcualte the mass of non – volatile solute (molar mass 40 g/mol) which should be dissolved in 114 g octane to reduce its vapor pressure to 80%.


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

(i) Molar mass of solute
(ii) Vapor pressure of water at 298K


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


21. Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene, 1g of AB2 lowers the freezing point by 2.3K whereas 1g of AB4 lowers it by 1.3K. The molar depression constant for benzene is 5.1 K kg/mol. Calculate atomic mass of A and B.


Then, x will be = 25.59

22. At 300 K, 36g of glucose present in a litre of its solution has an osmotic pressure of 4.08 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?


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).


(i) Both are non – polar. So, the intermolecular forces present between them is vanderwaal forces of attraction.

(ii) Both are non – polar. So, the intermolecular forces present between them is vanderwaal forces of attraction.

(iii) NaCl04 is an ionic compound and water is a polar molecule. Thus, the intermolecular forces between them will be ion-dipole interactions.

(iv) Both are polar molecules.So, intermolecular forces of attraction between them will be dipole-dipole interactions.

(v) Both are polar molecules. Thus, intermolecular forces of attraction between them will be dipole-dipole interactions.

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


n-octane is a non-polar solvent. So. the solubility of non-polar solute is more than polar solute in the n-octane. The order of increasing polarity is: \

Cyclohexane < CH3CN < CH3OH < KCl

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.


(i) Phenol (C6H5OH) has the polar group – OH => partially soluble in water.

(ii) Toluene (non polar) => insoluble in water.

(iii) Formic acid (HCOOH) can form H-bond with water => highly soluble in water.

(iv) Ethylene glycol can form H−bond => highly soluble in water.

(v) Chloroform => insoluble in water. (vi) Pentanol (having polar -OH) => Partially soluble.

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.


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


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


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.


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


Required Benzoic acid = 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.


Being most electronegative element , Flourine withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid (F3CCOOH) lose H+ ions easily to a larger extent whereas acetic acid ionizes in water to minimum extent. So, the more ions produced, the greater is the depression of the freezing point. Hence, the order increases – Acetic acid < trichloroacetic acid < trifluoroacetic acid

32. Calculate the depression in the freezing point of water when 10g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 x 1o-3 Kg = 1.86 K kg mol-1


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.


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


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.


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.


= 280.7 torr

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 ptotal’ pchloroform’ and pacetone as a function of xacetone. The experimental data observed for different compositions of mixture is.


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.


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 298K. The water is in equilibrium with air at a pressure of 10 atm. At 298 Kif the Henry’s law constants for oxygen and nitrogen are 3.30 × 107 mm and 6.51 × 107 m respectively, calculate the composition of these gases in water.


40. Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litres of water such that its osmotic pressure is 0.75 atm at 27°C.


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.


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