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The Solid State
1
Chapter1 The Solid State
VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)
1.
What type of substances exhibit antiferromagnetism? [Delhi 2008]
2.
What is the coordination number of each type of ions in a rock salt type crystal structure? [Delhi 2008, AI 2008]
3.
What is the total number of atoms per unit cell in a face centred cubic (fcc) structure? [Delhi 2008, AI 2008]
4.
Name an element with which silicon should be doped to give ntype of semiconductor. [Delhi 2008 C]
5.
What are F centres?
6.
Why is glass considered a super cooled liquid? [AI 2008 C]
[AI 2008 C]
19. “Crystalline solids are anisotropic in nature.” What does this statement mean? [Delhi SetI 2011, Foreign SetII, 2011] 20. Which stoichiometric defect in crystals increases the density of a solid? [Delhi SetII 2011] 21. What is meant by an ‘intrinsic semiconductor’? [ Foreign SetI 2011] 22. How many atoms are there in one unit cell of a body centred cubic crystal? [Foreign SetIII 2011] 23. What is meant by ‘doping’ in semiconductor? [Delhi SetI 2012] 24. Write a point of distinction between a metallic solid and an ionic solid other than metallic lustre. [Delhi SetII 2012]
Which point defect in crystals does not alter the density of relevant solid? [Delhi 2009 ] 25. How may the conductivity of an intrinsic semiconductor be increased? [AI SetI 2012] 8. How do metallic and ionic substances differ in conducting electricity? [AI 2009] 26. Which stoichiometric defect increases the density of a solid? [AI SetII 2012] 9. Which point defect of its crystals decreases the density of a [AI SetIII 2012] solid? [AI 2009, Delhi 2010] 27. What are ntype semiconductors? 7.
10. What is the number of atoms in a unit cell of facecentred cubic crystal? [AI 2009]
SHORT ANSWER TYPE QUESTIONS (2 OR 3 MARKS)
11. What is the percentage efficiency of packing in case of a simple cubic lattice? [AI 2009 C]
28. Explain the following properties giving suitable examples: (i) Ferromagnetism (ii) Paramagnetism [Delhi 2008]
12. Stability of a crystal is related to the magnitude of its melting point. How? [AI2009 C]
29. How would you account for the following? (i) Frenkel defects are not found in alkali metal halides. 13. Which point defect in crystals of a solid does not change the (ii) Schottky defects lower the density of related solids. density of the solid? [Delhi 2010] (iii) Impurity doped silicon is a semiconductor. 14. What type of interactions hold the molecules together in a [Delhi 2008] polar molecular solid? [AI 2010 ] 30. Ag crystallise in a fcc lattice. The edge length of its unit cell 15. What type of semiconductor is obtained when silicon is doped is 4.077 × 10 –8 cm and its density is 10.5 g cm –3 . Calculate on with arsenic? [AI 2010] this basis the atomic mass of Ag. [AI 2008] 16. Write a distinguishing feature of metallic solids. [AI 2010] 31. Explain the following terms with suitable examples : 17. Write a feature which will distinguish a metallic solid from (i) Schottky defect, (ii) Ferromagnetism. [Delhi 2008 C] an ionic solid. [Delhi 2010, Foreign 2010] 32. What is a semiconductor? Describe the two major types of 18. What is the number of atoms in a unit cell of a simple cubic semiconductors and contrast their conduction mechanism. crystal? [ Foreign 2010] [Delhi 2008 C]
2
Chapterwise XIIth CBSE BOARD 5 Yrs. Papers
10.5 g cm –3 , calculate the atomic mass of silver 33. Account for the following; (N A = 6.02 × 10 23 atoms mol –1 ). [AI 2010, Foreign 2010] (i) Fe 3 O 4 is ferrimagnetic at room temperature but becomes paramagnetic at 850 K. 44. Silver crystallizes in facecentered cubic unit cell. Each side of this unit cell has a length of 400 pm. Calculate the (ii) Zinc oxide on heating becomes yellow. radius of the silver atom. (Assume the atom just touch each (iii) Frenkel defect does not change the density of AgCl other on the diagonal across the face of the unit cell, that crystals. [AI 2008 C] is each face atom is touching the four corner atoms). 34. With the help of suitable diagrams, on the basis of band theory. [Delhi SetI 2011] Explain the difference between –3 45. The density of lead is 11.35 g cm and crystallize with fcc (i) a conductor and an insulator unit cell. Estimate the radius of lead atom. (ii) a conductor and a semiconductor. [AI 2008 C] (At. mass of lead = 207 g mol –1 and N A = 6.02 × 10 23 mol –1 ) 35. (a) In reference to crystal structure, explain the meaning of [Delhi SetIII 2011] coordination number. 46. Explain how we can determine the atomic mass of an (b) What is the number of atoms in a unit cell of (i) a face unknown metal if you know its density and the dimension centred cubic structure, (ii) a body centred cubic structure. of unit cell of its crystal? [AI SetI 2011] [Delhi 2009] 36. Iron has a body centred cubic unit cell with a cell edge of 47. Calculate the packing efficiency of a metal for a simple cubic lattice. [AI SetI 2011] 286.65 pm. The density of iron is 7.87 g cm –3 . Use this
37. 38.
39.
40.
information to calculate Avogadro’s number (Atomic mass 48. Define the following terms in relation to crystalline of Fe = 56 g mol –1 ). [AI 2009 & Foreign, Delhi 2009] solids: (i) Unit cell What are Fcentres? Why are solids containing Fcentres (ii) Coordination number paramagnetic? [AI 2009 C] Give one example in each case. [AI SetII 2011] Silver crystallises with facecentred cubic unit cells. Each side of the unit cell has a length of 409 pm. What is the radius 49. Aluminium crystallises in a cubic closepacked structure. Radius of atom in the metal is 125 pm. of an atom of silver? (Assume that each face atom is touching (i) What is the length of the side of the unit cell? the four corner atoms). [AI 2009, AI 2010] (ii) How many unit cells are there in 1 cm 3 of aluminium? How are the following properties of crystals affected by [ Foreign SetI 2011] Schottky and Frenkel defects? (i) Density, (ii) Electrical conductivity. [Delhi 2009 C] 50. Silver crystallise with facecentred cubic unit cell. Each side of this unit cell has a length of 409 pm. What is the Copper crystallizes into an fcc lattice with edge length radius of silver atom? Assume the atoms just touch each 3.61 × 10 –8 cm. Calculate the density of copper. other on the diagonal across the face of the unit cell. [Given : Cu = 63.5 g mol –1 , N A = 6.022 × 10 23 mol –1 ] [ Foreign SetIII 2011] [Delhi 2009 C]
41. The well known mineral fluorite is chemically calcium 51. Tungsten crystallises in body centred cubic unit cell. If the edge of the unit cell is 316.5 pm, what is the radius of tungsten fluoride. It is known that in one unit cell of this mineral there atom? are 4 Ca 2+ ions and 8F – ions and that Ca 2+ ions are arranged in a fcc lattice. The F – ion fill all the tetrahedral holes in the OR face centred cubic lattice of Ca 2+ ions. The edge of the unit Iron has a body centred cubic unit cell with a cell dimension cell is 5.46 × 10 –8 cm in length. The density of the solid is of 286.65 pm. The density of iron is 7.874 g cm –3 . Use this –3 3.18 g cm . Use this information to calculate Avogadro’s information to calculate Avogadro’s number. (At. mass of Fe number. (Molar mass of CaF 2 = 78.08 g mol –1 ) = 55.845 u) [Delhi SetI 2012] [Delhi 2010, Foreign 2010] 52. Copper crystallises with face centred cubic unit cell. If the 42. The density of copper metal is 8.95 g cm –3 . If the radius of radius of copper atom is 127.8 pm, calculate the density of copper atom is 127.8 pm. What is copper unit cell a simple copper metal. (Atomic mass of Cu = 63.55 u and Avogadro’s cubic, a body centred cubic or a face centred cubic structure? number, N A = 6.02 × 10 23 mol –1 ) OR [Given : Atomic mass of Cu = 63.54 g mol –1 and 23 –1 Iron has a body centred cubic unit cell with the cell dimension N A = 6.02 × 10 mol ] [AI 2010, Delhi 2010] of 286.65 pm. Density of iron is 7.87 g cm –3 . Use this 43. Silver crystallises in fcc lattice. If the edge length of the unit information to calculate Avogadro’s number. (Atomic mass of cell is 4.07 ´ 10 –8 cm and density of the crystal is Fe = 56.0 u) [AI SetI 2012]
The Solid State
3
answers 1. Those substances show antiferromagnetism in which magnetic domains are aligned in opposite direction. 2. In rock salt type structure coordination number is +
6 : 6 Þ Coordination number of M = 6 Coordination number of A – = 6. 3. No. of atoms per unit cell in fcc 1 1 Z = ´ 8 (corners) + ´ 6 (face centres) 8 2 Z = 1 + 3 = 4.
23. Addition of an appropriate amount of suitable impurity in a crystalline solid is called doping. Doping is done to increasing conductivity of intrinsic semiconductors. 24. Ionic crystals are insulators because their ions are not free to move. Whereas metallic crystals conduct electricity due to the presence of sea of mobile electrons. 25. Conductance of a semiconductor increases with rise in temperature.
4. Phosphorus.
26. Interstitial defect.
5. Voids created by anion vacancy and occupied by free electrons are called F centres. F centres cause colour to the crystal
27. A semiconductor in which doped impurity has more valence electrons than the pure element is called ntype semiconductor. e.g. Ge or Si (Group14) doped with P or As(group15).
6. Glass is an amorphous solid. It has tendency to flow very 28. (i) Ferromagnetism : Materials which are strongly attracted slowly like liquids. Hence, glass is considered as super by magnetic field are called ferromagnetic materials cooled liquid. and the property thus exhibited is caused 7. Frenkel defect. ferromagnetism. 8. Conductivity of metals is due to the motion of free electrons e.g., Fe, Co, Ni show ferromagnetism at room temperature. whereas conductivity of ionic substance is due to movement (ii) Paramagnetism : Substances which are weakly attracted of ions in molten or solution state. by a magnetic field are called paramagnetic substances 9. Schottky defect. and the property is called paramagnetism. e.g. O 2 , S 2 , Cu 2+ , Mn 2+ , etc. 1 1 10. Z = 8 ´ + 6 ´ = 4. 29. (i) This is due to the fact that alkali metal ions have large 8 2 11. Packing efficiency of simple cubic lattice = 52.4%. size which can’t fit into the interstitials sites. (ii) Schottky defects occur when cations and anions are 12. Stability of a crystal increases with increasing magnitude missing from their lattice site. Mass of unit cell decreases of melting point. which decreases the density of the solid. 13. Frenkel defect. (iii) The conductivity of intrinsic semiconductor like silicon 14. Dipoledipole interactions. is too low to be of practical use. This conductivity is 15. ntype semiconductor. increased by adding an appropriate amount of suitable impurity like Al or As which is electron deficient or 16. Malleable, ductile, lustrous and conductor of heat and electron rich. So the electrical conductivity of silicon electricity. is increased. 17. (i) Metallic solids are malleable and ductile whereas ionic 30. Given a = 4.077 × 10 –8 cm, d = 10.5 g cm –3 solids are hard and brittle. For fcc lattice, Z = 4 (ii) Metallic solids are conductors but ionic solids are insulators. Z ´ M Using d = 3 1 a ´ N A 18. For a simple cubic crystal Z = 8 ´ = 1. 8 10.5 ´ (4.077 ´ 10 -8 ) 3 ´ 6.022 ´ 10 23 or M = 19. Crystalline solids are anisotropic in nature means some of 4 their physical properties like electrical conductivity, = 107.09 g mol –1 . refractive index, etc., are different in different directions. 31. (i) Schottky defect : The defect in which cations and anions are missing in the stoichiometric ratio of 20. Interstitial defect increases density of solid. compound is called Schottky defect. Schottky defect is 21. Pure substance which acts as semiconductor is called actually vacancy defect in ionic solids. In this defect intrinsic semiconductor. e.g. Si and Ge at high temperature. electrical neutrality is maintained. In this defect density 22. 2 of solid decreases.
Chapterwise XIIth CBSE BOARD 5 Yrs. Papers
4
Schottky defect is shown by ionic solids in which the cation and anion are of almost similar sizes. Examples : NaCl, KCl, CsCl, AgBr, etc.
Conduction band
NaCl has one Schottky defect per 10 16 lattice points or 10 6 Schottky pairs per cm 3 of solid at room temperature.
Valence band
– + – + – + – +
+ – + – + – + –
– + – + – +
– + – + – + + –
– + – + – + + – + – + – + – + – + – + – + –
Conductor
+ – + – + – + –
Conduction band Small forbidden zone Valence band Semiconductor
35. (a) The number of lattice points touching one lattice point in a crystal lattice is called coordination number. (b) Number of atoms in a unit cell of
1 1 + 6 ´ = 4. 8 2 (ii) Refer Ans. 28 (i). 1 (ii) bcc structure, Z = 8 ´ + 1 ´ 1 = 2. 32. Semiconductors : In semiconductors the energy gap 8 between valence band and conduction band is small. 36. Given : For bcc, Z = 2 Therefore, some electrons may jump from valence band to a = 286.65 pm = 2.87 × 10 –8 cm conduction band and show some conductivity. d = 7.87 g cm –3 , N A = ? Semiconductors are of two types : Z ´ M Using formula d = (i) Intrinsic semiconductors N A ´ a 3 (ii) Extrinsic semiconductors Z ´ M 2 ´ 56 g mol -1 (i) Intrinsic semiconductors : A pure element which = or, N A = d ´ a 3 7.87 ´ (2.87 ´ 10 -8 cm) 3 behaves as semiconductor is called intrinsic 23 semiconductor. or,N A = 6.022 × 10 . (ii) Extrinsic semiconductor : A substance which behaves 37. Electrons trapped in anion vacancy of a crystal are called as semiconductor by adding some impurities is called Fcentres. Solids containing Fcentres are paramagnetic extrinsic semiconductor. Addition of appropriate amount due to the presence of unpaired electron. of suitable impurities increases conductivity of 38. Given : Structure fcc, a = 409 pm, r = ? semiconductors. This process of addition of impurities a 409 pm is called doping. Extrinsic semiconductors are formed Using formula r = = = 144.6 pm. by doping impurity of lower or higher groups. 2 2 2 ´ 1.414 (a) ntype extrinsic semiconductor 39. (i) Density : It decreases due to Schottky defect but remains (b) ptype extrinsic semiconductor. same in Frenkel defect. (ii) Electrical conductivity : It increases in both Schottky 33. (i) On heating at 850 K the magnetic domains undergo and Frenkel defects. realignment and become paramagnetic. (ii) Zinc oxide when heated loses some oxide ion creating 40. Given : for fcc Z = 4, a = 3.61 × 10 –8 cm, d = ? anion vacancy. This defect arises yellow colour. Z ´ M Using formula d = (iii) In Frenkel defect, no ions are missing from the crystal N A ´ a 3 as a whole. Hence, density does not change. 4 ´ 63.5 g mol -1 34. (i) In conductor valence band and conduction band overlap or, d = 6.022 ´ 1023 mol- 1 ´ (3.61 ´ 10 -8 cm)3 with each other whereas in insulator there is a large or,d = 8.965 g cm –3 . forbidden zone between the two. Conduction band Valence band Conductor
(i) fcc structure, Z = 8 ´
Conduction band Large forbidden zone Valence band Insulator
(ii) In conductor, valence band and conduction band overlap together whereas in semiconductor there is small forbidden band between the two.
41. Given for fcc lattice Z = 4, d = 3.18 g cm –3 a = 5.46 × 10 –8 cm, N A = ? Using formula d =
Z ´ M N A ´ a 3
4 ´ 78.08 g mol -1 3.18 g cm -3 ´ (5.46 ´ 10 -8 cm) 3 or,N A = 6.033 × 10 23 mol –1 .
or, N A =
The Solid State
5
42. Let us assume Z = 4 (copper has fcc) Z ´ M d = 3 a ´ N A 4 ´ 63.54 8.95 = 3 a ´ 6.02 ´ 10 23 a 3 = 47.145 × 10 –24 cm 3
494.9 494.9 2 494.9 ´ 1.414 pm = pm = pm 4 4 2 2 \ r = 174.95 pm
\ r =
46. We can determine the atomic mass of an unknown metal by using the formula of density of its unit cell. d (density) =
a = 3 47.145 ´ 10 -24 a = 3.612 × 10 –8 cm = 361.2 pm
By knowing density d, dimension of unit cell we can calculate M, the atomic mass of metal as for a given unit cell, Z is fixed and N A is universal constant.
for fcc, 4r = 2 a It shows Cu has fcc structure, a 361.2 r= = =127.7 pm. 2 2 2 ´ 1.141 43. Given : a = 4.07 × 10 –8 cm, d = 10.5 g cm –3 M = ?, N A = 6.02 × 10 23 mol –1 , Z = 4 (for fcc) Z ´ M N A ´ a 3 d ´ N A ´ a 3 or, M = Z 10.5 g cm - 3 ´ 6.022 ´ 10 23 ´ (4.07 ´ 10 -8 cm) 3 or, M = 4 or,M = 106.57 g mol –1 .
44. a = 400 pm
Z ´M Z ´ M Þ a 3 = a3 ´ N A d ´ N A
For fcc unit cell, Z = 4 Given, M = 207 g mol –1 , N A = 6.02 × 10 23 mol –1 d = 11.35 g cm –3 Substituting these value in equation (i), we get a 3 =
4 ´ 207 g mol -1 11.35 g cm -3 ´ 6.0 ´ 1023 mol -1
a3 =
4 ´ 207 ´ 10 cm 3 11.35 ´ 6.02 ´ 10 24
æ 8280 ö Þ a = ç è 11.35 ´ 6.02 ÷ø
1/3
´ 10 -8 cm
\ a = 4.949 × 10 –8 cm Þ a = 494.9 pm For fcc, r =
For example, unit cell of a cubic crystal is a = b = c, a = b = g = 90°.
a
2 2 400 400 2 400 2 = ´ = = 100 2 \ r = 4 2 2 2 2 2 Þ r = 100 × 1.414 = 141.4 pm
45. d =
Z ´ volume of one atom ´ 100 Volume of cubic unit cell 4 Z ´ pr 3 3 = ´ 100 a 3 For a simple cubic lattice, a = 2r and Z = 1 4 1 ´ pr 3 p \ Packing efficiency = 3 3 ´ 100 = ´ 100 = 52.4% (2r ) 6 48. (i) Unit cell : A unit cell is smallest portion of a crystal lattice which, when repeated in different direction, generates the entire lattice.
47. Packing efficiency =
Using formula d =
For fcc, r =
Z (No. of atoms per unit cell) ´ M (atomic mass) a 3 (cell edge) ´ N A (Avogadro number)
a 2 2
c
a b
...(i)
(ii) Coordination number : The number of nearest neighbours of any constituent particle in a packing is called its coordination number. The coordination number of an atom in the bcc structure is 8. 49. (i) For fcc (or ccp), a = 2 2r = 2 ´ 1.414 ´ 125 pm = 354 pm (ii) a = 354 pm = 3.54 × 10 –8 cm Volume of one unit cell = a 3 = (3.54 × 10 –8 cm) 3 = 4.436 × 10 –23 cm 3 Total volume Number of unit cells = Volume of one unit cell 1 cm 3 = = 2.25 ´ 10 22 4.44 ´ 10-23 cm 3 50. As the atoms just touch each other on the diagonal across the face of unit cell, therefore b 2 = a 2 + a 2 = 2a 2 , b = 2 a ...(i) Also, b = r + 2r + r = 4r ...(iii)
Chapterwise XIIth CBSE BOARD 5 Yrs. Papers
6
From (i) and (ii), we get 4r = 2 a Þ r =
2 1.414 ´ 409 pm a ; r = = 144.58 pm 4 4
52. Given : Structure fcc, hence Z = 4, r = 127.8 pm, M = 63.55 u, N A = 6.02 × 10 23 mol –1 , d = ?
r a
Z ´ M 2 ´ 56 g mol -1 = d ´ a 3 7.87 ´ (2.87 ´ 10 -8 cm) 3 or, N A = 6.022 × 10 23 .
or, N A =
b
Using formula, d =
4 ´ 63.55 g mol -1 6.022 ´ 10 mol -1 ´ (2 2 ´ 127.8 ´ 10 -10 ) 3 = 8.965 g/cm 3 For fcc structure, a = 2 2 r . OR
we get d =
a
51. Given : Structure = bcc, a = 316.5 pm, r = ? Using formula, r =
3a 1.732 ´ 316.5 = = 137.04 pm 4 4 OR
Given : For bcc, Z = 2 a = 286.65 pm = 2.87 × 10 –8 cm d = 7.87 g cm –3 , N A = ? Using formula d =
Z ´ M N A ´ a 3
Z ´ M N A ´ a 3 23
Given : For bcc, Z = 2 a = 286.65 pm = 2.87 × 10 –8 cm, d = 7.87 g cm –3 , N A = ? Using formula d =
Z ´ M N A ´ a 3
Z ´ M 2 ´ 56 g mol -1 = d ´ a 3 7.87 ´ (2.87 ´ 10 -8 cm) 3 23 or, N A = 6.022 × 10 .
or, N A =
JJJ
Solutions
7
Chapter2 Solutions
at 0.320°C, what would be the value of van’t Hoff factor? [K f for water is 1.86°C mol –1 ] [Delhi 2009]
VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)
State the main advantage of molality over molarity as the 13. 100 mg of a protein is dissolved in just enough water to make unit of concentration. [Delhi 2009 C] 10.0 mL of solution. If this solution has an osmotic pressure 2. What is meant by ‘reverse osmosis’? [AI SetIII 2011] of 13.3 mm Hg at 25°C, what is the molar mass of the protein? [R = 0.0821 L atm mol –1 K –1 and 760 mm Hg = 1 atm] 3. State Henry’s law about partial pressure of gas in a mixture. [AI 2009, Delhi 2009] [Foreign SetI 2011] 1.
4.
State Raoult’s law in its general form with respect to solution [Foreign SetIII 2011]
SHORT ANSWER TYPE QUESTIONS (2 OR 3 MARKS)
14. Calculate the amount of sodium chloride which must be added to one kilogram of water so that the freezing point of water is depressed by 3 K. (Given : K f = 1.86 K kg mol –1 , Atomic mass : Na = 23.0, Cl = 35.5) [Delhi 2009 C]
State Henry’s law correlating the pressure of a gas and its solubility in a solvent and mention two applications for the 15. A solution of urea in water has a boiling point of law. [AI 2008, Delhi 2008] 373.128 K. Calculate the freezing point of the same solution. [Given : For water K f = 1.86 K m –1 , K b = 0.52 K m –1 ] 6. Calculate the temperature at which a solutions containing [Delhi 2009 C] 54 g of glucose. (C H O ) in 250 g of water will freeze. (K 5.
6 12
6
of water = 1.86 K mol –1 kg). 7.
8.
f
[AI 2008]
State Raoult’s law for solutions of volatile liquids. Taking suitable examples explain the meaning of positive and negative deviations from Raoult’s law. [AI 2008, Delhi 2008] Define the term osmotic pressure. Describe how the molecular mass of a substance can be determined by a method based on measurement of osmotic pressure. [AI 2008, Delhi 2008]
16. What is meant by positive and negative deviations from Raoult’s law and how is the sign of D mix H related to positive and negative deviation from Raoult’s case. [AI 2009 C] 17. Calculate the freezing point of a solution containing 18 g glucose, C 6 H 12 O 6 and 69.6 g sucrose, C 12 H 22 O 11 in 200 g of water. The freezing point of pure water is 273 K and K f for water is 1.86 K m –1 . [AI 2009 C]
18. A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 of benzene has a boiling 9. The boiling point elevation of 0.30 g acetic acid in 100 g point of 80.31°C. Determine the molar mass of this benzene is 0.0633 K. Calculate the molar mass of acetic acid compound. (Boiling point of pure benzene = 80.10°C and K b from this data. What conclusion can you draw about the for benzene = 2.53°C Kg mol –1 ) [Delhi 2010] molecular state of the solute in the solution? [Given : K b for benzene = 2.53 K kg mol –1 ]
[AI 2008 C]
10. Distinguish between the terms molality and molarity. Under what conditions are the molarity and molality of a solution nearly the same? [AI 2008 C, Delhi 2009, AI 2010]
19. A solution of glycerol (C 3 H 8 O 3 : molar mass = 92 g mol –1 ) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C. What mass of glycerol was dissolved to make this solution? K b for water = 0.512 K kg mol –1 . [Delhi 2010, SetII 2012]
11. The freezing point of a solution containing 0.2 g of acetic 20. Nonideal solution exhibit either positive or negative from acid in 20.0 g of benzene is lowered by 0.45°C. Calculate Raoult’s law. What are these deviation and why are they (i) the molar mass of acetic acid from this data. caused? Explain with one example for each type. (ii) van’t Hoff factor. –1 [Delhi 2010 C] [For benzene, K f = 5.12 K kg mol ] What conclusion can you draw from the value of van’t Hoff 21. Define the terms, ‘osmosis’ and ‘osmotic pressure’. What is factor obtained ? [AI 2008 C] the advantage of using osmotic pressure as compared to other colligative properties for the determination of molar masses 12. Calculate the freezing point depression expected for 0.0711 of solutes in solutions? [AI 2010] m aqueous solution of Na 2 SO 4 . If this solution actually freezes
8
Chapterwise XIIth CBSE BOARD 5 Yrs. Papers
LONG ANSWER TYPE QUESTIONS (5 MARKS) 22. What mass of NaCl (molar mass = 58.5 g mol –1 ) be dissolved in 65 g water of lower the freezing point by 7.5°C ? The freezing 33. (a) The depression in freezing point of water observed for point depression constant K f for water is 1.86 K kg mol –1 . the same molar concentrations of acetic acid, Assume van’t Hoff factor for NaCl is 1.87. [AI 2010] trichloroacetic acid and trifluoroacetic acid increases
23. What mass of ethylene glycol (molar mass = 62.0 g mol –1 ) must be added to 5.50 kg of water to lower the freezing point of water from 0°C to – 10°C? [K f for water = 1.86 K kg mol –1 ] [AI 2010]
in the order as stated above. Explain. (b) Calculate the depression in freezing point of water when 20.0 g of CH 3 CH 2 CHClCOOH is added to 500 g of water. [Given K a = 1.4 × 10 –3 , K f = 1.86 K kg mol –1 ] [Delhi 2008 C]
24. 15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at – 0.34°C. 34. (a) What is meant by : What is the molar mass of the substance. –1 (i) Colligative properties (K f for water = 1.86 K kg mol ) (ii) Molality of a solution. [AI 2010, Delhi SetIII 2012] (b) What concentration of nitrogen should be present in a 25. State the following : glass of water at room temperature. Assume a (i) Raoult’s law in its general form in reference to solutions. temperature of 25°C, a total pressure of 1 atmosphere (ii) Henry’s law about partial pressure of a gas in a mixture. and mole fraction of nitrogen in air of 0.78. [AI SetI 2011] [K H for N 2 = 8.42 × 10 –7 M/mm Hg] [AI 2009] 26. A solution prepared by dissolving 8.95 mg of a gene fragment 35. (a) Differentiate between molarity and molality for a in 35.0 mL of water has an osmotic pressure of 0.335 torr at solution. How does a change in temperature influence 25°C. Assuming that the gene fragment is a nonelectrolyte, their values? calculate its molar mass. [AI SetI 2011] (b) Calculate the freezing point of an aqueous solution 27. What mass of NaCl must be dissolved in 65.0 g of water to lower the freezing point of water by 7.50°C? The freezing point depression constant (K f ) for water is 1.86°C/m. Assume van’t Hoff factor for NaCl is 1.87. (Molar mass of NaCl = 58.5 g). [AI SetII 2011] 28. Differentiate between molality and molarity values for a solution. What is the effect of change in temperature on molarity and molality values? [AI SetIII 2011] 29. Differentiate between molarity and molality of a solution. Explain how molarity value of a solution can be converted into its molality? [Foreign SetI 2011]
containing 10.50 g of MgBr 2 in 200 g of water. (Molar mass of MgBr 2 = 184 g) (K f for water = 1.86 K kg mol –1 ) OR (a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain. (b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.00 g of water. (K b for water = 0.512 K kg mol –1 ), (Molar mass of NaCl = 58.44 g) [Delhi SetI 2011]
36. (a) State the following : (i) Henry’s law about partial pressure of a gas in a mixture (ii) Raoult’s law in its general form in reference to solutions. (b) A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25°C. Assuming the gene fragment is 31. A 1.00 molal aqueous solution of trichloroacetic acid a nonelectrolyte, determine its molar mass. (CCl 3 COOH) is heated to its boiling point. The solution has the boiling point of 100.18°C. Determine the van’t Hoff factor OR –1 for trichloroacetic acid. (K b for water = 0.512 K kg mol ). (a) Differentiate between molarity and molality in a solution. What is the effect of temperature change on OR molarity and molality in a solution? Define the following terms: (b) What would be the molar mass of a compound if (i) Mole fraction (ii) Isotonic solutions 6.21 g of it dissolved in 24.0 g of chloroform to form (iii) van’t Hoff factor (iv) Ideal solution a solution that has a boiling point of 68.04°C. The [Delhi SetI 2012] boiling point of pure chloroform is 61.7°C and the 32. Calculate the amount of KCl which must be added to 1 kg of boiling point elevation constant, K b for chloroform is water so that the freezing point is depressed by 2 K (the K f 3.63°C/m. [Delhi SetII 2011] for water = 1.86 K kg mol –1 ). [Delhi SetI 2012] 30. A 0.561 m solution of an unknown electrolyte depresses the freezing point of water by 2.93°C. What is van’t Hoff factor for this electrolyte? The freezing point depression constant (K f ) for water is 1.86°C kg mol –1 . [Foreign SetI 2011]
Solutions
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37. (a) Define the following terms: (i) Mole fraction (ii) Ideal solution (b) 15.0 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution freezes at –0.34°C. What is the molar mass of the material? (K f for water = 1.86 K kg mol –1 ) OR
(a) Explain the following: (i) Henry’s law about dissolution of a gas in a liquid. (ii) Boiling point elevation constant for a solvent. (b) A solution of glycerol (C 3 H 8 O 3 ) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C. What mass of glycerol was dissolved to make this solution? (K b for water = 0.512 K kg mol –1 ). [AI SetI 2012]
a n s w e r s type of behaviour of solution is called positive deviation from Raoult’s law. e.g. Ethanol water. (ii) Negative deviation from Raoult’s law : When the If a pressure larger than the osmotic pressure is applied total vapour pressure is less than corresponding vapour on the solution side, the solvent starts to flow from the pressure, expected on the basis of Raoult’s law. This solution into the pure solvent through the semipermeable type of behaviour of solution is called negative deviation membrane. This phenomenon is called reverse osmosis. from Raoult’s law. e.g. Acetone chloroform. Henry’s law : It states that the partial pressure of the gas 8. p = CRT in vapour phase (p) is proportional to the mole fraction of æ W ö n W ´ R ´ T the gas (x) in the solution. n = B ÷ or, p = RT Þ p = B ç M B ø V M B ´ V è p = K H x W B = mass of the solute. Where K H is Henry’s law constant, p = Partial pressure of M B = molecular mass of the solute. gas and x = mole fraction. A known mass of a substance is added to known volume Raoult’s law : It states that for any solution the pressure of a solution and osmotic pressure of solution is determined. of each volatile component of the solution is directly Using formula, molecular mass can be calculated. proportional to its mole fraction. P A = X A .P° A and P B = X B P° B and P T = P A + P B 9. Given, w 1 = 100 g, w 2 = 0.30 g It states that at constant temperature the solubility of a gas DT b = 0.0633 K and K b = 2.53 K kg mol –1 in a liquid is directly proportional to the pressure of the gas. K ´ w 2 ´ 1000 Molar mass of acetic acid = ( M 2 ) = b P = K H .c DTb ´ w1 Applications : 2.53 ´ 0.30 ´ 1000 7.59 = = = 119.90 (i) In the production of carbonated beverages. 0.0633 ´ 100 0.0633 (ii) In the deep sea diving. Actual molar mass of acetic acid = CH 3 COOH (iii) In the functioning of lungs. = 12 + 3 + 12 + 16 + 16 + 1= 60 g mol –1 Molecular mass of glucose, M 2 = 72 + 12 + 96 = 180 Here, there is difference between calculated molar mass
1. Molality is independent of temperature, whereas molarity is a function of temperature. 2.
3.
4.
5.
6.
K f ´ W 2 ´ 1000 1.86 ´ 54 ´ 1000 = M 2 ´ W1 180 ´ 250 100440 = = 2.23 45000 Freezing point of solution = (0 – 2.23)°C = – 2.23°C. DT f =
7. It states that for a solution of volatile liquids the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. Nonideal solutions show positive and negative deviation from Raoult’s law. (i) Positive deviation from Raoult’s law : When the total vapour pressure is greater than corresponding vapour pressure expected on the basic of Raoult’s law. This
and actual molar mass 60 Normal molar mass = = 0.5 Abnormal molar mass 119.90 Here, i c A.P ° + c B.P ° A A 60 For positive deviations from Raoult’s law = = 0.52 113.77 D mix H = +ve, DV mix = +ve, DP mix = +ve Here, i
