NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions are part of NCERT Solutions for Class 12 Chemistry. Here we have given NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions.
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NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions
NCERT IN-TEXT QUESTIONS
Calculate the mass percent of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Calculate the mole fraction of benzene in a solution containing 30% by mass of it in carbon tetrachloride.
Let us start with 100 g of the solution in which
Mass of benzene = 30 g
Mass of carbon tetrachloride = 70 g
Molar mass cf benzene (C6H6) = 6 x 12 + 6 x 1 = 78g mol-1.
Calculate the molarity of each of the following solutions
(a) 30 g of Co(NO3)26H2O in 4·3 L of solution
(b) 30 mL of 0-5 M H2SO4 diluted to 500 mL.
Calculate the mass of urea (NH2CONH2) required to prepare 2-5 kg of 0-25 molal aqueous solution.
(b) molarity and
(c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI solution is 1·202 g mL-1.
Step I. Calculation of molality of solution
Weight of KI in 100 g of the solution = 20 g
Weight of water in the solution = 100 – 20 = 80 g = 0-08 kg
Molar mass of KI = 39 + 127 = 166 g mol-1.
Step II. Calculation of molarity of solution
Step III. Calculation of mole fraction of Kl
H2S a toxic gas with rotten egg smell, is used for the qualitative analysis. If the solubility of H2S in water at S.T.P is 0·195 m ; calculate Henry’s law constant.
Step I. Calculation of mole fraction of H2S
Step II. Calculation of Henry’s Law constant
According to Henry’s Law,
Henry’s Law constant for CO2 in water is 1·67 x 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2·5 atm pressure of CO2 at 298 K. (D.S.B. 2008 Supp.)
Step I. Calculation of number of moles of CO2.
According to Henry’s Law,
( has been neglected as the gas is very little soluble in water)
∴ = x (27·78 mol) = (1·52 x 10-3) x (27·78 mol) = 0·0422 mol
Step II. Mass of CO2 dissolved in water = (0·0422 mol) x (44 g mol-1) = 1·857 g.
The vapour pressures of pure liquids A and B are 450 mm and 700 mm of Hg respectively at 350 K. Calculate the composition of the liquid mixture if total vapour pressure is 600 mm of Hg. Also find the composition in the vapour phase.
Vapour pressure of pure liquid A () = 450 mm
Vapour pressure of pure liquid B () = 700 mm
Total vapour pressure of the solution (P) = 600 mm
The vapour pressure of pure water at 298 K is 23·8 mm Hg. 50 g of Urea (NH2CONH2) is added to 850 g of water. Calculate the vapour pressure of water for this solution and also its relative lowering in vapour pressure.
Step I. Calculation of vapour pressure of water for the solution
According to Raoult’s Law,
Step II. Calculation of relative lowering in vapour pressure
Boiling point of water at 750 mm Hg is 99·63°C. How much sucrose is to be added to 500 g of water so that it may boil at 100°C ? (K6 for water = 0·52 K kg mol-1).
Calculate the mass of ascorbic acid (vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1·5°C. (Kf for CH3COOH) = 3·9 K kg mol-1)
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 solution at 37°C.
Define the term solution. What kinds of solutions are possible ? Write briefly about each type of solution with an example.
A solution may contain two or more substances also called components or constituents. A solution which has two components is known as binary solution (e.g. a solution of NaCl in water) while a solution with three components is called ternary solution (e.g. a solution of NaCl and KCl in water). Similarly even more than three components may also be present.
TYPES OF SOLUTION
Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small. What type of solid solution is this likely to be ?
The solution likely to be formed is interstitial solid solution.
Define the following terms :
(i) Mole fraction
(iv) Mass percentage.
(i) Mole fraction :
The ratio of the number of moles of one component to the total number of moles of all the components present in the solution.
For a binary solution made up of components A and B,
The number of gram moles of the solute dissolved in 1000 g (or kg) of the solvent.
The number of gram formula mass of the solute dissolved per litre of the solution.
(iv) Mass percentage:
The number of parts by mass of one component (solute or solvent) per 100 parts by mass of the solution. If A and B are the two components of a binary solution,
Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of acid if the density of the solution is 1·504 g mL-1 ?
Mass of HNO3 in solution = 68 g
Molar mass of HNO3 = 63 g mol-1
Mass of solution = 100 g
Density of solution = 1·504 g mL-1
A solution of glucose in water is labelled as 10 percent 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-1, then what should be the molarity of the solution ? (C.B.S.E. 2013, Manipur Board 2015)
How many mL of a 0·1 M HCl are required to react completely with lg mixture of Na2CO3 and NaHCO3 containing equimolar amounts of two ?
Calculate the percentage composition in terms of mass of a solution obtained by mixing 300 g of a 25% and 400 g of a 40% solution by mass.
An antifreeze solution is prepared from 222·6 g of ethylene glycol C2H4(OH)2 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? (C.B.S.E. Delhi 2007)
A sample of drinking water was found to be severely contaminated with chloroform(CHCl3) supposed to be carcinogen. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in water sample.
What role does the molecular interaction play in solution of alcohol in water ?
In case of alcohol as well as water, the molecules are interlinked by intermolecular hydrogen bonding. However, the hydrogen bonding is also present in the molecules of alcohol and water in the solution but it is comparatively less than both alcohol and water. As a result, the magnitude of attractive forces tends to decrease and the solution shows positive deviation from Raoult’s Law. This will lead to increase in vapour pressure of the solution and also decrease in its boiling point.
Why do gases nearly always tend to be less soluble in liquids as the temperature is raised ?
The dissolution of a gas in a liquid is exothermic in nature because the gas contracts in volume.
Gas + Liquid ⇌ Dissolved gas ; ∆H = – ve
An increase in temperature will favour the reverse process since it is of endothermic nature. Therefore, the solubility of the gas in solution decreases with the rise in temperature.
State Henry’s law and mention some of its important applications.
Henry’s law: The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas in equilibrium with the liquid at that temperature.
The partial pressure of a gas in vapour phase is proportional to the mole fraction of the gas (x) in the solution. p = KHX
where KH is Henry’s law constant.
Applications of Henry’s law :
(i) In order to increase the solubility of CO2 gas in soft drinks and soda water, the bottles are normally sealed under high pressure. Increase in pressure increases the solubility of a gas in a solvent according to Henry’s Law. If the bottle is opened by removing the stopper or seal, the pressure on the surface of the gas will suddenly decrease. This will cause a decrease in the solubility of the gas in the liquid i.e. water. As a result, it will rush out of the bottle producing a hissing noise or with a fiz.
(ii) As pointed above, oxygen to be used by deep sea divers is generally diluted with helium inorder to reduce or minimise the painfril effects during decompression.
(iii) As the partial pressure of oxygen in air is high, in lungs it combines with haemoglobin to form oxyhaemoglobin. In tissues, the partial pressure of oxygen is comparatively low. Therefore, oxyhaemoglobin releases oxygen in order to carry out cellular activities.
The partial pressure over a saturated solution containing 6·56 x 10-2 g of ethane is 1 bar. If the solution contains 5·0 x 10-2 g of ethane, what shall be the partial pressure of the gas ?
According to Henry’s law,
The mass of the gas (m) dissolved in solution ∝ Partial pressure (p) (At constant temperature)
What is meant by positive and negative deviations from Raoult’s law and how is the sign of ∆Hsol related to positive and negative deviations from Raoult’s law ?
(a) Positive Deviation:
(i) ∆Vmixing is positive: This is quite likely also because in ge presence of weak forces of interaction, the interaction, the volume of the solution is bound to increase.
(ii) ∆Hmixing is positive: Energy is needed to form the solution because the components of the solution have to be brought closer to form the solution. Thus, the process of mixing is of endothermic nature.
(b) Negative Deviation:
(i) ∆Vmixing is negative: Because of the increased forces of interaction, the molecules of the two components will come closer and as a result, there is a decrease in the volume of the solution.
(ii) ∆Hmixing is negative: Energy is expected to be released because of the increase in the forces of interaction. Therefore, the process of mixing is exothermic in nature or ∆Hmixing is negative.
An aqueous solution of 2 percent non-volatile solute exerts a pressure of 1·004 bar at the boiling point of the solvent. What is the molecular mass of the solute ?
According to Raoult’s Law,
Heptane and octane form ideal solution. At 373 K, the vapour pressure of the two liquid components are 105·2 k Pa and 46·8 k Pa respectively. If the solution contains 25 g of heptane and 35 g of octane, calculate :
(i) Vapour pressure exerted by heptane
(ii) Vapour pressure exerted by octane
(iii) Vapour pressure exerted by the solution
(iv) Mole fraction of octane in the vapour phase. (C.B.S.E. Sample Paper, 2010)
The vapour pressure of water is 12·3 kPa at 300 K. Calculate the vapour pressure of 1 molal solution in it.
1 molal solution implies one mole of the solute dissolved in 1000 g (1 kg) of solvent i.e. water.
Calculate the mass of a non-volatile solute (molecular mass 40 g mol-1) that should be dissolved in 114 g of octane to reduce its pressure to 80%. (C.B.S.E. Outside Delhi 2008)
According to Raoult’s Law,
A solution containing 30 g of a non-volatile solute exactly in 90 g of water has a vapour pressure of 2·8 k Pa at 298 K. Further 18 g of water is then added to the solution and the new vapour pressure becomes 2·9 k Pa at 298 K. Calculate
(i) Molecular mass of the solute.
(ii) Vapour pressure of water at 298 K. (C.B.S.E. Outside Delhi 2005)
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose in water if the freezing point of pure water is 273·15 K. (C.B.S.E. Delhi 2008)
Freezing point temperature of glucose solution = (273·15 – 4·085) K = 269·07 K.
Two elements A and B form compounds having molecular formulae AB2 and AB4. When dissolved in 20 g of benzene, 1 g of AB2 lowers the freezing point by 2·3 K whereas 1 g of AB4 lowers it by 1·3 K. Molal depression constant for benzene is 5·1 K kg mol-1. Calculate atomic masses of A and B. (C.B.S.E. Delhi 2004)
At 300 K, 36 g glucose present per litre in its solution has osmotic pressure of 4·98 bar. If the osmotic pressure of the solution is 1·52 bar at the same temperature, what would be its concentration ?
Suggest the most important type of intermolecular attractive interactions in the following pairs :
(i) n-hexane and n-octane
(ii) I2 and CCl4
(iii) NaClO4 and water (H2O)
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H6O).
(i) London’s forces
(ii) London’s forces
(iii) Ion-dipole interactions
(iv) Intermolecular hydrogen bonding
(v) Dipole- dipole interactions.
Based on solute solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.
n-octane (C8H18) is a non-polar liquid and solubility is governed by the principle that like dissolve like. Keeping this in view, the increasing order of solubility of different solutes is:
KCl < CH3OH < CH3C=N < C6H12 (cyclohexane).
Among the following compounds, identify which are insoluble, partially soluble and highly soluble in water ?
(iii) formic acid
(iv) ethylene glycol
(i) phenol (C6H5OH): Is partially soluble in water due to weak dipole-dipole interactions in the molecules of phenol and water.
(ii) toluene (C7H8) : Is insoluble in water because it is an aromatic hydrocarbon (non-polar) while water is polar in nature.
(iii) formic acid (HCOOH) : Is highly soluble in water since it can form hydrogen bending with water.
(iv) ethylene glycol (HOCH2CH2OH) : Is highly soluble in water since it can form hydrogen bonding with water.
(v) chloroform (CHCl3): Is insoluble in water because it is an organic heavy liquid and forms a separate layer.
(vi) pentanol (C5H11OH) : In partially soluble in water because the bulky C5H11 group decreases its extent of hydrogen bonding with water.
If the density of lake water is 1·25 g mL-1, and it contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake. (C.B.S.E. Outside Delhi 2008)
If the solubility product of CuS is 6 x 10-16, calculate the maximum molarity of CuS in aqueous solution.
Dissociation of CuS in aqueous solution is :
By definition Ksp corresponds to the product of ionic concentration of the salt in saturated solution and it represents maximum molarity of the salt. Therefore, maximum molarity of the salt = 2· 45 x 10-8 M.
Calculate the mass percent of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6·5 g of aspirin is dissolved in 450 g of CH3CN.
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 x 10-3 m aqueous solution required for the above doze ?
Calculate the amount of benzoic acid (C5H5COOH) required for preparing 250 mL of 0· 15 M solution in methanol.
The depression in freezing point of water observed for the same amount of acetic acid, trichloroactetic acid and trifluoroacetic acid increases in the order given above. Explain. (C.B.S.E. 2008 Supp.)
The depression in freezing point of a solute in water depends upon the number of particles or ions furnished by it in solution or upon its degree of dissociation (α). All the three organic acids ionise in aqueous solution. However, the relative order of acidic strengths is as given below.
This is linked with the electronegativity of the halogen atoms present. Fluorine (F) is more electronegative than (Cl). Under the circumstances, trifluoroacetic acid gives maximum ions in solution since it is the strongest acid. Consequently, the depression in freezing point (∆Tf) is the maximum in this case and is the least for acetic acid which is the weakest acid.
Calculate the depression in freezing point of water when 10 g of CH3CH2CH(Cl)COOH is added to 250 g of water. Ka = 1·4 x 10-3; Kf = 1·86 K kg mol-1. (C.B.S.E. 2008 Supp.)
Step I. Calculation of degree of dissociation of acid Mass of acid = 10 g
19·5 g of CH3FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1·0°C. Calculate Van’t Hoff factor and dissociation constant of the acid : Kf = 1·86 K kg mol-1.
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.
According to Raoult’s Law,
Henry’s Law constant for the molality of methane in benzene at 298 K is 4·27 x 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm of Hg. (C.B.S.E. 2013)
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.
Vapour pressures of pure acetone and chloroform at 328 K are 632·8 mm Hg and 741·8 mm Hg respectively. Assuming that they form ideal solution over die entire range of composition, plot ptotal. pchloroform, and pacetone as a function of xacetone. The experimental data observed for different compositions of mixture is :
Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative the ideal solution.
From the available information :
Since the plot or graph dips downwards, the solution shows negative deviation from Raoult’s Law.
Benzene and naphthalene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50·71 mm Hg and 32·06 mm Hg respectively. Calculate the mole pure fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of naphthalene.
Air is a mixture of 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 at 298 K are 3·30 x 107 mm and 6·51 x 107 mm respectively, calculate the composition of these gases in water.
Determine the amount of CaCl2 (i = 2·47) dissolved in 2·5 litre of water so that its osmotic pressure is 0·75 atm at 27°C.
According to Van’t Hoff equation :
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25°C, assuming that it is completely dissociated. (C.B.S.E. 2013)
Step I. Calculation of Van’t Hoff factor (i)
K2SO4 dissociates in water as :
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