chemistry 12 solution chapter 2 exercise-2

CHEMISTRY (CHAPTAR 2)
NCERT EXERCISES WITH ANSWERS EX 2.2

Q – 2.1 Define the term solution. How many types of solution are formed? Write briefly about each type with an example.
Ans. A solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within creation limits.
Types of solution
A solution can be solid, liquid or a gas depending upon the physical state of the solvent.

 

S.No Solute Solvent Type of Sol. Examples
SOLID

1.

2.

3.

LIQUID

 

4.

5.

6.

GASEOUS

 

7.

8.

9.

SOLUTION

Solid

Liquid

Gas

SOLUTION

 

Solid

Liquid

Gas

SOLUTION

 

Solid

Liquid

Gas

(SOLID

Solid

Solid

Solid

(LIQUID

 

Liquid

Liquid

Liquid

(GASEOUS

 

Gas

Gas

Gas

SOLVENT)

Solid in Solid

Liquid in Solid

Gas in Solid

SOLVENT)

 

Solid in Liquid

Liquid in Solid

Gas in Liquid

SOLVENT)

 

Solid in Gas

Liquid in Gas

Gas in Gas

 

Alloys (brass, German silver, bronze, 22 carat gold etc.)

Hydrated salts, Amalgam of Hg with Na

Dissolved gases in minerals or H2 in Pd

 

 

Salt of glucose of sugar urea solution in water

Methanol of ethanol in water

Aerated drinks O2 in water

 

 

Iodine vapours in air, camphor in N2 gas

Humidity in air, chloroform mixed with N2 gas

Air (O2 + N2)

 

Q – 2.2. Suppose a solid solution is formed between tow substance, one whose particles are very large and the other whose particles are very small. What type of this solid solution is likely to be?
Ans. Interstitial solid solution.
Q – 2.3. Define the terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.
Ans. (i) Mole fraction – mole fraction of a constituent (solute as well as solvent) is the fraction obtained by dividing number of moles of that constituent by the total number of moles of all the constituents present in the solution.
(ii) Molality- Molality of a solution is defined as the number of moles of the solute dissolved in 1000 grams (1 kg) of the solvent.
(iii) Molarity – Molarity of a solution is defined as the number of moles of the solute dissolved per liter (or dm3) of solution.
(iv) Mass percentage – Mass percentage is one way of representing the concentration of an element in a compound or a component in a mixture. Mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100%.
Q – 2.4. Concentrated nitric acid used in the laboratory 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?
Ans. 68% nitric acid by mass means that
Mass of nitric acid = 68 g, Mass of solution = 100 g
Molar mass of HNO3 = 63 g mol-1
chemistry 2 chapter 2 ex 4 qes
Q – 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 the solution is 1.2 g ml-1, then what shall be the molarity of the solution?
Ans. 10 g glucose is present in 100 g solution, i.e., 90 g of water = 0.090 kg of water
chemistry 2 chapter 2 ex 5 qes
Q – 2.6. How many ml of a 0.1 M HCL are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of the two?

chemistry 2 chapter 2 ex 6.1 qeschemistry 2 chapter 2 ex 6.2 qes

Q – 2.7. A solution is obtained by mixing 300 g of 25% and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.?

chemistry 2 chapter 2 ex 7 qes

Q – 2.8. An antifreeze solution is prepared from 222.6 g of ethylene glycol, C2H4(OH)2 and 200g 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?
Ans. Mass of the solute, C2H4(OH)2 =222.6 g,
Molar mass of C2H4(OH)2 = 62 g mol-1
chemistry 2 chapter 2 ex 8 qes
Q – 2.9. 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 the water sample.
chemistry 2 chapter 2 ex 9 qes
Q – 2.10. What role does the molecular interaction play in solution of alcohol and water?
Ans. There is strong hydrogen bonding in alcohol molecules as well as water molecules. On mixing, the molecular interaction are weakened. Hence, their solution will show positive deviations from ideal behavior. As a result, the solution will have higher vapour pressure and lower boiling point than that of water and alcohol.
Q – 2.11. Why do gases nearly always tend to be less soluble in liquid as the temperature is raised?
chemistry 2 chapter 2 ex 11 qes
Q – 2.12. State Henry’s law and mention some of its important applications.
Ans. Henry’s law – This is the most important factor influencing the solubility of a gas in a liquid at a particular temperature. A little thought clearly reveals that as we compress the gas over the liquid (i.e., we increase the pressure), the solubility will increase.
Quantitatively, the effect of pressure on the solubility of a gas in a liquid was studied by Henry (in 1803) and is called Henry’s law.
Applications of Henry’s law :
(i) in the production of carbonated beverages.
(ii) in the deep sea diving.
(iii) in the function of lungs.
(iv) For climbers or people living at high altitudes.
Q – 2.13. The partial pressure of ethane over a saturated solution containing 6.56 × 10-2 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?
chemistry 2 chapter 2 ex 13 qes
Q – 2.14. 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 ?
Ans. (a) Non- ideal solutions showing positive deviation- When a component B is added to another component A sometimes the partial pressure of a component A is found to be more than expected on the basis of Raoult’s law. A similar effect is observed for the other component B in the reversed mixing. The total vapour pressure of any solution is thus greater than that corresponding to an ideal solution of the same composition. Such behavior of solution is described as a positive deviations from Raoult’s law.
(b) Non- ideal solutions showing negative deviations – if for the two components A and B, the forces of interaction between the A and B molecules are more than the A – A and B – B forces of interaction, the escaping tendency of A and B types of molecules from the solution becomes less than from the pure liquids. In other words, for any composition of the solution, the partial vapour pressure of the solution will also be less than that expected from Raoult’s law.
chemistry 2 chapter 2 ex 14 qes
Q – 2.15. An aqueous solution of 2% non- volatile solute exerts a pressure of 1.0041 bar at the normal boiling point of the solvent. What id solvent. What is the molecular mass of the solute?
Ans. Vapor pressure of pure water at the boiling (p0) = 1 atm = 1.013 bar
Vapor pressure of solution ( ps) = 1.004 bar; Mass of solute (w2) = 2g
Mass of solution = 100 g ; Mass of solvent = 98 g
chemistry 2 chapter 2 ex 15 qes
Q – 2.16. Heptane and octane from ideal solution. At 373 K, the vapour pressure 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?chemistry 2 chapter 2 ex 16 qes
Q – 2.17. The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapor of 1 molal solution of a solute in it.
chemistry 2 chapter 2 ex 17 qes
Q – 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 to 80 %.
Ans. Reduction of vapour pressure to 80% means that is P0 = 100 mm,
than Ps = 80 mm.
Applying complete formula
chemistry 2 chapter 2 ex 18 qesNote that complete formula is required because concentration of solution is greater
than 5%. Complete formula can also be applied in the from.
chemistry 2 chapter 2 ex 18.1 qes
chemistry 2 chapter 2 ex 18.2 qes

Q – 2.19. A solution containing 30 g of a non – volatile solute exactly in 90 g water has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is then added to the solution, 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.

chemistry 2 chapter 2 ex 19 qes

chemistry 2 chapter 2 ex 19.1qes

Q – 2.20. 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 freezing point of pure water is 273.15 K.
Ans. 5% solution by mass means 5 g of solute is present in 100 g of solution
؞ Mass of solvent (water) = 95 g
chemistry 2 chapter 2 ex 20 qesQ – 2.21. Two elements A and B from compounds having molecular formula AB2 and BA4 . 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 atomic masses of A and B.
chemistry 2 chapter 2 ex 21 qesSuppose atomic masses of A and B are ‘a’ and ‘b’ respectively. Then
Molar mass of AB4 = a + 2b = 110.87 g mol-1 (i)
Molar mass of AB4 = a + 4b = 196.15 g mol-1 (ii)
Eqn. (ii) – Eqn. (i) gives 2 b = 85.28 or b = 42.64
Substituting in eqn. (i), we get a + 2 × 42.64 = 110.87 or a = 25.59
Thus, Atomic mass of A = 25.59 u,
Atomic mass of B = 42.64 u
Q – 2.22. At 300 K, 36 g glucose present per litre in its solution has an osmatic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bar at the same temperature, what should be its concentration?
chemistry 2 chapter 2 ex 22 qes
Q – 2.23. Suggest the most important type of intermolecular interaction in the following pairs:
(i) n-hexane and n-octane (ii) I2 and CCI4 (ii) NaClO4 and water (iv) methanol and
acetone (v) acetonitrile (CH3CN) and acetone (C3H6O).
Ans. (i) both are non-polar. Hence, intermolecular interaction in them will be London dispersion forces (Discussed in class XI).
(ii) Same as (i).
(iii) NaClO4 gives Na+ and CiO-4 ions in the solution while water is a polar molecule. Hence, intermolecular interaction in them will be ion-dipole interactions.
(iv) Both are polar molecules. Hence, interactions in them will be dipole-dipole interactions.
(v) Same as (iv).
Q – 2.24. Based on solute- solvent interaction, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN
Ans. (i) Cyclohexane and n-octane both are non- polar. Hence, they mix completely in all
proportions.
(ii) KCl is an ionic compound while n-octane is non-polar. Hence, KCl will be dissolve at
all in n-octane.
(iii) CH3OH and CH3CN both are polar but CH3CN is less polar than CH3OH. As the
solvent is non-polar, CH3CN will dissolve more than CH3OH in n-octane.
Thus, the order of solubility will be KCl < CH3OH< CH3CN<Cyclohexane.
Q – 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.
Ans. (i) Partially soluble because phenol has polar – OH group but aromatic phenyl, C6H5- group.
(ii) Insoluble because toluene is non-polar.
(iii) Highly soluble because formic acid can form hydrogen bonds with water.
(iv) Highly soluble because ethylene glycol can form hydrogen bonds with water.
(v) Insoluble because chloroform is an organic liquid.
(vi) Partially soluble because –OH group is polar but the large hydrogen part (C5H11) is
non-polar.
Q – 2.26. If the density of some lake water is 1.25 g ma- and contains 92 g of Na+ ions per kg of water, calculate the molality of ions in the lake.
chemistry 2 chapter 2 ex 26 qes
Q – 2.27. If the solubility product of CuS is 6 × 10-16, calculate the maximum molarity of CuS in aqueous solution.
Ans. Maximum molarity of CuS is aqueous solution = solubility of CuS in mol-1
If S is the solubility of CuS in mol-1, then
chemistry 2 chapter 2 ex 27 qesQ – 2.28. Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.
chemistry 2 chapter 2 ex 28 qes
Q – 2.29. Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic user. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10-3 m aqueous solution required for the above dose.
Ans. 1.5 × 10-3 m solution means that 1.5 × 10-3 mole of nalorphene is dissolved in 1 kg of water.
Molar mass of C19H21NO3 = 19 × 12 + 21 +14 + 48 =311 g mol-1
؞ 1.5 × 10-3 mole of C19H21NO3 = 1.5 × 10-3 × 311 g = 0.467 g = 467 mg
؞ Mass of solution = 1000 g + 0.467 g = 1000.467g
chemistry 2 chapter 2 ex 29 qesQ – 2.30. Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 ml of 0.15 M solution in methanol.
Ans. 0.15 M solution means that 0.15 mole of benzoic acid is present in 1 L,
i.e., 1000 ml of the solution.
Molar mass of benzoic acid (C6H5COOH) = 72 + 5 + 12 + 32 + 1 = 122 g mol-1
؞ 0.15 mole of benzoic acid = 0.15 × 122 g = 18.3 g
؞ Thus, 1000 ml of the solution contain benzoic acid = 18.3 g
chemistry 2 chapter 2 ex 30 qesQ – 2.31. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid trifluoroacetic acid increase in the order given above. Explain briefly.
Ans. The depression in freezing points are in the order :
acetic acid < trichloroacetic acid < trifloroacetic acid
chemistry 2 chapter 2 ex 31 qesFluorine, being most electronegative, has the highest electron withdrawing inductive effect. Consequently, trifluoroacetic acid is the strongest aid while acetic acid is the weakest acid. Hence, trifluoroacetic acid ionizes to the largest extent while acetic acid ionizes to the minimum extent to give ions in their solutions in water. Greater the ions produced, grater is the depression in freezing point. Hence, the depression in freezing point is maximum for the fluoroacetic acid and minimum for acetic acid.
Q – 2.32. Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10-3, Kƒ = 1.86 K kg mol-1.
Ans. Molar mass of CH3CH2CHClCOOH = 15 + 14 + 13 + 35.5 + 45 = 122.5 g mol-1
chemistry 2 chapter 2 ex 32.1 qeschemistry 2 chapter 2 ex 32.2 qesQ – 2.33. 19.5 g of CH2FCOOH is dissolved in 500 g of water. the depression in the freezing point observed is 1.00C. Calculate the van’t Hoff factor and dissociation constant of floroacteic acid. Kƒ for water is 1.86 K kg mol-1
chemistry 2 chapter 2 ex 33 qesCalculation of dissociation constant. Suppose degree of dissociation at the given
concentration is α.
chemistry 2 chapter 2 ex 33.1 qesQ – 2.34. Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour
pressure of water at 293 K when 25 g glucose is dissolved in 450 g of water.
Ans. Here, P0 = 17.535 mm, w2 = 25 g, w1 = 450 g
For solute (glucose, C6H12O6), M2 = 180 g mol-1,
chemistry 2 chapter 2 ex 34 qesQ – 2.35. Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
chemistry 2 chapter 2 ex 35 qes
Q – 2.36. 100 g 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.
chemistry 2 chapter 2 ex 36 qesSubstituting this value in eqn. (i), we get PA = 0.114 × 280.7 torr = 32 torr.
Q – 2.37. 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 mixtures is:
100 × 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
Polt this data also on the same graph paper,Indicate whether it has positive deviation or negative deviation from the ideal solution.
Ans. 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                         632.8     603.0       579.5       562.1      580.4       599.5      615.3    641.8
Q – 2.38. Benzene and toluene form ideal solution over the entire range of compassion. 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 the vapour phase if 80 g of benzene is mixed with 100 g of toluene.
Ans. Molar mass of benzene (C6H6) = 78 g mol-1,
Molar mass of toluene C6H5CH3) = 92 g mol-1
chemistry 2 chapter 2 ex 38 qesMole fraction of toluene = 1 – 0.486 = 0.514
P0Benzene = 50.71 mm, P0Toluene = 32.06 mm
Applying Raoult’s law,
PBenzene = xBenzene × P0Benzene = 0.486×50.71 mm = 24.65 mm
PTotal = xToluene × P0Toluene = 0.514 × 32.06 mm = 16.48 mm
؞ Mole fraction of benzene in law vapoure phase
chemistry 2 chapter 2 ex 38.1 qesQ – 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. 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.
Ans. Total pressure of air in equilibrium with water = 10 atm
As air contains 20% oxygen and 79% nitrogen by volume,
chemistry 2 chapter 2 ex 39 qesQ – 2.40. Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmatic pressure is 0.75 atm at 270C.
chemistry 2 chapter 2 ex 40 qesMolar mass of CaCl2 =40 + 2 × 35.5 = 111 g mol-1
؞ Amount dissolved = 0.0308 × 111 g = 3.42 g.
Q – 2.41. Determine the osmatic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 250C, assuming that it is completely dissociated.
Ans. K2SO4 dissolved = 25 mg = 0.025 g
Volume of solution = 2 L, T = 250C =298 K
As K2SO4 dissociates completely as K2SO4 → 2K+ + SO42- ,
i.e., ions produced = 3 , ؞ I = 3
chemistry 2 chapter 2 ex 41 qes
Updated: September 19, 2020 — 6:18 pm

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