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Modern Chemistry Chapter 12Solutions – Solute is the substance that is dissolved in a solution.Types of Mixtures • Solutions are homogeneous mixtures of two or more substances in a single phase.Section 1. Modern Chemistry 7 Solutions CHAPTER 12 REVIEW Solutions SECTION 3 SHORT ANSWER Answer the following questions in the space provided. Describe the errors made by the following students in making molar solutions. James needs a 0.600 M solution of KCl. He measures out 0.600 g of KCl and adds 1 L of water to the solid.
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions
Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions:
| Section Name | Topic Name |
| 2 | Solutions |
| 2.1 | Types of Solutions |
| 2.2 | Expressing Concentration of Solutions |
| 2.3 | Solubility |
| 2.4 | Vapour Pressure of Liquid Solutions |
| 2.5 | Ideal and Non-ideal Solutions |
| 2.6 | Colligative Properties and Determination of Molar Mass |
| 2.7 | Abnormal Molar Masses |
NCERT TEXTBOOK QUESTIONS SOLVED
2.1. Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Ans: Mass of solution = Mass of C6H6 + Mass of CCl4
= 22 g+122 g= 144 g
Mass % of benzene = 22/144 x 100 =15.28 %
Mass % of CCl4 = 122/144 x 100 = 84.72 %
2.2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Ans: 30% by mass of C6H6 in CCl4 => 30 g C6H6 in 100 g solution
.’. no. of moles of C6H6,(nC6h6) = 30/78 = 0.385
2.3.Calculate the molarity of each of the following solutions: (a) 30 g of CO(NO3)2.6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.
Ans: (a) Molar mass of CO(NO3)2.6H2O=310.7 g mol-1
no. of moles = 30/310.7 = 0.0966
Vol. of solution = 4.3 L
Molarity =0.0966/4.3 = 0.022M
(b) 1000 mL of 0.5M H2SO4 contain H2SO4 = 0.5 mole
30 mL of 0.5 M H2SO4 contain H2SO4
=0.5/1000 x 30 = 0.015 mole
Volume of solution = 500mL=0.5 L
Molarity = 0.015/0.5 = 0.03M
2.4.Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.
Ans: 0.25 Molal aqueous solution to urea means that
moles of urea = 0.25 mole
mass of solvent (NH2CONH2) = 60 g mol-1
.’. 0.25 mole of urea = 0.25 x 60=15g
Mass of solution = 1000+15 = 1015g = 1.015 kg
1.015 kg of urea solution contains 15g of urea
.’. 2.5 kg of solution contains urea =15/1.015 x 2.5 = 37 g
2.5. Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.
Ans:
2.6. H2 S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant.
Ans:Solubility of H2S gas = 0.195 m
= 0.195 mole in 1 kg of solvent
1 kg of solvent = 1000g
2.7. 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 CO2 pressure at 298 K.
Ans.:
2.8.The vapour pressure 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 total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Ans:
2.9. Vapour pressure of pure water at 298 K is 23.8 m m Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Ans:
2.10. Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.
Ans:
2.11. 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= 3.9 K kg mol-1.
Ans:
2.12. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.
Ans:
NCERT EXERCISES
2.1 Define the terra solution. How many types of solutions are formed? Write briefly about each type with an example.
Sol. A solution is a homogeneous mixture of two or more chemically non-reacting substances. Types of solutions: There are nine types of solutions.
Types of Solution Examples
Gaseous solutions
(a) Gas in gas Air, mixture of 02 and N2, etc.
(b) Liquid in gas Water vapour
(c) Solid in gas Camphor vapours in N2 gas, smoke etc.
Liquid solutions
(a) Gas in liquid C02 dissolved in water (aerated water), and 02 dissolved in water, etc.
(b) Liquid in liquid Ethanol dissolved in water, etc.
(c) Solid in liquid Sugar dissolved in water, saline water, etc.
Solid solutions
(a) Gas in solid Solution of hydrogen in palladium
(b) Liquid in solid Amalgams, e.g., Na-Hg
(c) Solid in solid Gold ornaments (Cu/Ag with Au)
2.2 Give an example of a solid solution in which the solute is a gas.
Sol. Solution of hydrogen in palladium and dissolved gases in minerals.
2.3 Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage
Sol.(i) Mole fraction: It is defined as the ratio of the number of moles of the solute to the total number of moles in the solution. If A is the number of moles of solute dissolved in B moles of solvent, then Mole fraction of solute
(ii) Molality: It is defined as die number of moles of a solute present in 1000g (1kg) of a solvent.
NOTE: Molality is considered better way of expressing concentration of solutions, as compared to molarity because molality does not change with change in temperature since the mass of solvent does not vary with temperature,
(iii) Molarity: It is defined as the number of moles of solute present in one litre of solution.
NOTE: Molarity is the most common way of expressing concentration of a solution in laboratory. However, it has one disadvantage. It changes with temperature because volume of a solution alters due to expansion and contraction of the liquid with temperature.
(iv) Mass percentage: It is the amount of solute in grams present in 100g of solution.
2.4 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?
Sol. 68% nitric acid by mass means that 68g mass of nitric acid is dissolved in 100g mass of solution. Molar mass of HNO3= 63g mol-1
2.5 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 m L-1, then what shall be the molarity of the solution?
Sol. 10 percent w/w solution of glucose in water means 10g glucose and 90g of water.
Molar mass of glucose = 180g mol-1 and molar mass of water = 18g mol-1
2.6 How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2C03 and NaHCO3 containing equimolar amounts of both?
Sol. Calculation of no. of moles of components in the mixture.
2.7 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.
Sol.
2.8 An antifreeze solution is prepared from 222.6 g of ethylene glycol, (C2 H6O2 ) and200 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?
Sol.
2.9 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.
Sol. 15 ppm means 15 parts in million (106) by mass in the solution.
2.10 What role does the molecular interaction play in a solution of alcohol and water?
Sol. Alcohol and water both have strong tendency to form intermolecular hydrogen bonding. On mixing the two, a solution is formed as a result of formation of H-bonds between alcohol and H2 O molecules but these interactions are weaker and less extensive than those in pure H2O. Thus they show a positive deviation from ideal behaviour. As a result of this, the solution of alcohol and water will have higher vapour pressure and lower boiling point than that of water and alcohol.
2.11 Why do gases always tend to be less soluble in liquids as the temperature is raised?
Sol. When gases are dissolved in water, it is accompanied by a release of heat energy, i.e., process is exothermic. When the temperature is increased, according to Lechatlier’s Principle, the equilibrium shifts in backward direction, and thus gases becomes less soluble in liquids.
2.12 State Henry’s law and mention some important applications.
Sol. The effect of pressure on the solubility of a gas in a liquid is governed by Henry’s Law. It states that the solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas Mathematically, P = KHX where P is the partial pressure of the gas; and X is the mole fraction of the gas in the solution and KH is Henry’s Law constant.
Applications of Henry’s law
(i) In the production of carbonated beverages (as solubility of C02 increases at high pressure).
(ii) In the deep sea diving.
(iii) For climbers or people living at high altitudes, where low blood 02 causes climbers to become weak and make them unable to think clearly
2.13 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?
Sol.
2.14 According to Raoult’s law, what is meant by positive and negative deviaitions and how is the sign of ∆solH related to positive and negative deviations from Raoult’s law?
Sol. Solutions having vapour pressures more than that expected from Raoult’s law are said to exhibit positive deviation. In these solutions solvent – solute interactions are weaker and ∆solH is positive because stronger A – A or B – B interactions are replaced by weaker A – B interactions. Breaking of the stronger interactions requires more energy & less energy is released on formation of weaker interactions. So overall ∆sol H is positive. Similarly ∆solV is positive i.e. the volume of solution is some what more than sum of volumes of solvent and solute.
So there is expansion in volume on solution formation.
Similarly in case of solutions exhibiting negative deviations, A – B interactions are stronger than A-A&B-B. So weaker interactions are replaced by stronger interactions so , there is release of energy i.e. ∆sol H is negative.
2.15 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?
Sol.
2.16 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 35.0 g of octane?
Sol.
2.17 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
Sol. 1 molal solution of solute means 1 mole of solute in 1000g of the solvent.
2.18 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%.
Sol.
2.19 A solution containing 30g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18g of water is then added to the solution and the new of vapour pressure becomes 2.9 kPa at 298 K. Calculate
(i) molar mass of the solute.
(ii) vapour pressure of water at 298 K.
Sol. Let the molar mass of solute = Mg mol-1
2.20 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.
Sol. Mass of sugar in 5% (by mass) solution means 5gin 100g of solvent (water)
2.21 Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20g 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 K kg mol-1. Calculate atomic masses of A and B.
Sol.
2.22 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?
Sol.
2.23 Suggest the most important type of intermolecular attractive interaction in the following pairs:
(i) n-hexane and n-octane
(ii) I2 and CCl4.
(iii) NaCl04 and water
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H60)
Sol. (i) Both w-hexane and n-octane are non-polar. Thus, the intermolecular interactions will be London dispersion forces.
(ii) Both I2 and CCl4 are non-polar. Thus, the intermolecular interactions will be London dispersion forces.
(iii) NaCl04 is an ionic compound and gives Na+ and Cl04– ions in the Solution. Water is a polar molecule. Thus, the intermolecular interactions will be ion-dipole interactions.
(iv) Both methanol and acetone are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.
(v) Both CH3CN and C3H6O are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.
2.24 Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octanc and explain.
Cyclohexane, KCl, CH3OH, CH3CN.
Sol. (a) Cyclohexane and n-octane both are non-polar. They mix completely in all proportions.
(b) KCl is an ionic compound, KCl will not dissolve in n-octane.
(c) CH3OH is polar. CH3OH will dissolve in n-octane.
(d) CH3CN is polar but lesser than CH3OH. Therefore, it will dissolve in n-octane but to a greater extent as compared to CH3OH. Hence, the order is KCl < CH3OH < CH3CN < Cyclohexane.
2.25 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
Sol. (i) Phenol (having polar – OH group) – Partially soluble.
(ii) Toluene (non-polar) – Insoluble.
(iii) Formic acid (form hydrogen bonds with water molecules) – Highly soluble.
(iv) Ethylene glycol (form hydrogen bonds with water molecules) Highly soluble.
(v) Chloroform (non-polar)- Insoluble.
(vi) Pentanol (having polar -OH) – Partially soluble.
2.26 If the density of some lake water is 1.25 g mL-1 and contains 92g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake.
Sol.
2.27 If the solubility product of CuS is 6 x 10-16, calculate the maximum molarity of CuS in aqueous solution.
Sol.
2.28 Calculate the mass percentage of aspirin (C9H8O4 in acetonitrile (CH3CN) when 6.5g of CHO is dissolved in 450 g of CH3CN.
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Solution:
2.29 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 x 10-3 m aqueous solution required for the above dose.
Solution:
2.30 Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol 0.15 M solution means than 0.15 mole of benzoic acid is dissolved in 1L of solution.
Solution:
2.31 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.
Solution:
Fluorine being more electronegative than chlorine has the highest electron withdrawing inductive effect. Thus, triflouroacetic acid is the strongest trichloroacetic acid is second most and acetic acid is the weakest acid due to absence of any electron withdrawing group. Thus, F3CCOOH ionizes to the largest extent while CH3COOH ionizes to minimum extent in water. Greater the extent of ionization greater is the depression in freezing point. Hence, the order of depression in freezing point will be CH3COOH < Cl3CCOOH < F3CCOOH.
2.32 Calculate the depression in the freezing point of water when 10g of CH3CH2CHClCOOH is added to 250g of water. Ka = 1.4 x 1o-3 Kg = 1.86 K kg mol-1.
Solution:
2.33 19.5g of CH2FCOOH is dissolved in 500g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’s Hoff factor and dissociation constant of fluoroacetic acid.
Solution:
2.34 Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25g of glucose is dissolved in 450g of water.
Solution:
2.35 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 Hg.
Solution:
2.36 100g of liquid A (molar mass 140 g mol-1) was dissolved in 1000g of liquid B (molar mass 180g 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.
Solution:
2.37 Vapour pressures 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, Pchlroform and Pacetone as a function of χacetone. The experimental data observed for different compositions of mixtures is:
Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.
Solution:
2.38 Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure 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 vapour phase if 80g of benzene is mixed with 100g of toluene.
Solution:
2.39 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 x 107 mm and 6.51 x 107 mm respectively, calculate the composition of these gases in water.
Solution:
2.40 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.
Solution:
2.41 Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25°C, asstiming that it is completely dissociated.
Chemistry Chapter 12 Solutions Answers
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Modern Chemistry Chapter 12 Answers
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