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found to have two mole of water of hydration per Ba2+ ion and the acid is mono-basic. What is molecular weight of anhydrous acid?
I g of a mixture containing equal no. of moles of carbonates of two alkali metals, required 44.4 mLof 0.5 N HCI for complete reaction. The atomic weight of one metal is 7, find the atomic weight of other metal.
Also calculate amount of sulphate formed on quantitative conversion of 1.0 g of the mixture in two sulphates.
What would be the molality of a solution obtained by mixing equal volumes of 30% by weight H2S04 (cl = 1.218 g mL- 1) and 70% by weight H2S04 (d= 1.610 g mL- 1)? If the resulting solution has density 1.425 g/mL, calculate its molarity.
A sample of fuming sulphuric acid containing H2S04, SO3 and S 02
weighing 1.0 g is found to require 23.47 mL of 1.0 N alkali for its neutralisation. A separate sample shows the presence of 1.5% S02. Find the percentage of free SO3, H2S04 and combined SO3 in the sample.
Calculate the ionic strength of a solution containing 0.2 M NaCI and 0.1 M Na2S04.
200 mL of a solution of mixture of NaOH and Na2C03 was first titrated with phenolphthalein and N/\0 HCI. 17.5 mL of HCI was required for the end point. After this methyl orange \Vas added and 2.5 mL of same HCI was again required for next end point. Find out amounts of NaOH and Na2C03 in mixture.
Two drops of phenolphthalein solution was added to 40.0 mL of an HCI solution. This solution was titrated with 0.10 M NaOH solution. When 30.0 mL of base had been added, part of the solution turned pink, but the colour disappeared upon mixing the solution. Addition of NaOH solution was continued dropwise until one drop addition produced a lasting pink colour. At this point, the volume of base added was 32.56 mL. Calculate:
(a) The concentration of HCI solution.
(b) Concentration of HCI in solution when 30.0 mL base had been added.
(c) pH of the solution when 30.0 mL base was added.
(d) pH of the solution when 32.56 mL base was added.
A sample supposed to be pure CaC03 is used to standardise a solution of HCI. The substance really was a mixture of MgC03 and BaC03, but the standardisation of HCI was accurate. Find the percentage of BaCO^ and MgC03 in mixture.
100 mL sample of hard water is passed through a column of the ion exchange resin RH2. The water coming off the column requires 15.17 mL of 0.0265 M NaOH for its titration. What is the hardness of water
r n 2+
as ppm ol C a
102
NUMIHK.AI PHYSICAL CHEMISTRY> Problem 56. A sea water sample has a density of l .03 g/cm1 and ,'..K% Nat '1 by mass.
A saturated solution of NaCl in water is 5.45 M Nat 'I. I low much water would have to be evaporated from 106 litre of sea water before NaCl would precipitate.
> Problem 57. One litre of a mixture containing BaF2 and H2SO4 was taken for analysis. 25 mL of this mixture was treated with 20.0 mL of 0.1 N KOH for complete neutralisation. Another 25 mL of the mixture was added to 100 mL of 0.05 N K2CO3 solution and precipitate was filtered off. The filtrate required 12 mL of 0.025 M oxalic acid solution using phenol-phthalein as indicator. Find the strength of BaF2 and H2SO4 in mixture.
> Problem 58. 5 g of CUSO45H2O is intended to be prepared by using CuO. and four times the stoichiometric amount of H2SO4. Assuming that 10% of the material is lost in crystallisation, what weight of oxide should be taken and how many litre or mL of a 5 M H2S04.
> Problem 59. A mixture contains 20 g of caustic soda, 20 g of sodium carbonate and 20 g of sodium bicarbonate in one litre. What will be the titre value if 55 mL of this mixture is used for titration against 1 N HCI if:
(a) First titrated with phenolphthalein.
(b) Methyl orange added after first end point.
(c) Methyl orange added from the very begining.
>• Problem 60. The reaction, Zn + CuS04 > Cu + ZnS04 goes to completion. In one experiment, 10 g of metallic zinc was added to 200 mL CuS04 solution.
After all the Cu was precipitated, it was found that not all the zinc had dissolved. After filtration, the total solid at the end of reaction was 9.81 g. Calculate the weight of Cu deposited and molarity of CUSO4 in original solution.
>• Problem 61. A sample of green crystals of nickel (II) sulphate heptahydrate was heated carefully to produce the bluish-green nickel (II) sulphate hexahydrate. What are the formulas of the hydrates? If 8.753 g of the heptahydrate produces 8.192 g of the hexahydrate, how many gram of anhydrous nickel (II) sulphate could be obtained?
>- Problem 62. A sample of metallic elements X, weighing 3.177 g. Combines with 0.6015 litre of 02 gas (at normal pressure and 20°C) to form the metal oxide with the formula XO. If the density of O2 gas under these conditions is 1.330 g/litre, what is the mass of this oxygen? The atomic weight of oxygen is 15.9994 amu. What is the atomic weight of X? What is the identity of X?
>- Problem 63. Copper sulphide reacts with nitric acid as 3CuS + 8HNO3 > 3CU(N03)2 + 3S(s) + 4H20( g ) + 2NO(G). In an ex-periment, the volume of moist NO gas at 27°C and one atm pressure was collected in a chamber 1642 mm x 760 mm x 30 111111 in dimensions.
Calculate the amount of copper sulphide taken and the volume of 8 M nitric acid required for the reaction. The vapour pressure ol pure water at 27°C is 27 111111 of Hg. (Cu = 63.5, S 32).
MOLL AND EQUIVALENI CONCI IM
— — - — — — — — — Answers
103
I. n = 19 ; 2. 40% ;
3. 25% ; 4. 746.66 litre / kg ; 5. 2 : 1 ; 6. 166.66 mL, 59.03 litre ;
7. NH3 ; 8. 48.2 ;
9. 0.8 g, 2.24 litre 02 ; 10. C2H4 = 39.2%, CH4 = 60.8% ;
II. NH3 ; 12. 65.4%;
13. 59.37% ; 14. Al = 75.9%, Mg = 16.2%, Cu = 7.9 15. A12(S04)318H20 ; 16. 74.47 g ;
17. 1.1395 g ; 18. 53.50 ; 1 9 . 6 . 1 3 % ; 20. 0.5978 g ; 21. 40 mL ;
22. BaCI2-2H20 = 7.038 g , H20 = 42.962 g ;
2 3 . 0 . 1 8 3 ; 24. 40 g, 0.05 N ; 25. 18.33 m L ; 26. 86 ppm ; 27. x = 4.92, v = 3.94 ; 28. 62.58 ; 29. n — 1 ; 30. 157.8 mL ; 31. 29.78% ; 32. 20.78% ; 33. 122.16 g , Mg = 95.57% ; 34. 12.78 g ; 35. 82.08 g ; 36. 80% ; 37. (a) See solution, (b) 40.734% ; 38. 25 litre ;
39. 0.2 cm2 ;, 40. 2.386 x 10"5 litre ;
41. NaOH = 1.68 TV, Na2C03 = 0.82 N ; 42. 0.132 g , 0.0672 litre, 0.336% ; 43. 95.5%; 44. (i) 0.122, (ii) 3.362, 1.345;
45. C2H4
46. Na2C03 = 0.816 M , NaHC03 = 0.368 M ;
47. 122.31 : 48. 23, 1.3998 g ;
49. 7.61. 11.22; 50. 65.3%, 33.2%, 27.10% ; 51. 9.2 g/litre, 7.1 g/litre, 9.6 g/litre ;
52. NaOH = 0.06 g per 200 mL, Na2C03 = 0.0265 g per 200 mL ; 53. (a) 0.0814, (b) 3.66 x 10~3, (c) 2.4365, (d) 7 ;
54. BaC03 = 27.89%, MgC03 = 72.11% ;
55. 80.40 ; 56. 90.9 x 104 litre ;
57. BaF2 = 6.372 g/litre, H2S04 = 3.92 g/litre ; 58. w = 1.753 g, 4.41 mL ;
59. (a) 31.34 mL, (b) 16.93 mL, (c) 48.26 mL ; 60. 8.001 g , 0.63 M ; 61. 4.826 g ; 62. 0.80 g, 63.54 ;
63. 210.29 g. 0.734 litre.
74 NHMI HICAI PHYSICAL CHI MISTRY
Ratio of C : () 27.1
72,7 0.376
Thus, the ratio of amounts o i l ' reading with same mass of O 0.751:0.376 2 : 1
This is in agreement with law ol multiple proportions.
Solution 5. "2
Volume after reaction 2 0 12 litre
Volume after reaction = volume of H2 left 1 volume of HCI formed
= 2 + 12 = 14 litre
Solution 8. No. of g-atoms of Carbon = 1.008 g 180 g glucose has = N molecules
5.23 g glucose has = 5.23x6.023x10 23
MOLE AND EQUIVALENT CONCEPT
Solution 14. 22.4 litre water vapour at STP has = 6.023 x 1023 molecules
1 x 1 0 litre water vapours at STP has: 6.023 x 1023 x 10~3
22.4
2.69 x 1019 molecules Solution 15. 6.023 x 1023 molecules of NH3 has weight = 17 g
3.01 x io2 3 molecules of NH3 has weight =
Solution 16. 4 wheat grains are counted in 1 sec.
17x3.01x10 23
76 NUMERICAL PHYSICAL CHEMISTRY Also, v 111 g CaCI2 has = 2 TV ions of C P
2TV x 333
333 g CaCl2 has = — — — ions of CI
= 6 TV ions of CI"
Mol. wt. of BaCl2-2H20 = 244 g,
v 244 g BaCl2-2H20 = 36 g H20 = 2 mole of H20 2.x 366
366 g BaCl2-2H20 = ^ mole of H20 = 3 mole of HzO Solution 22. Mol. wt. of methane (CH4) = 16,
Solution 21.
16 g CH4 has molecules = TV
->A 13 U i . 6.023 xl02 3 x 24 24 g CH4 has molecules =
4 16
= 9.03 x l()2J molecules 16 g C H4 = TV atoms of carbon Also,
24 g CH4 has = 6.023 xl02 3 x 24 16
9.03 x 1023 atoms o f C and 16 g CH4 = 4 TV atoms o f H
24 g CH4 4 x 6.023 x 10 x 24 16 r»24
Solution 23. (a) (b)
= 3.612 x 10^ atoms of H Mass of iron = 20 g
Mass of 1.2 g-atom of N = 14 x 1.2= 16.8 g (c) Mass of 1 x 1023 atoms of C = 12x1 xlO23
6.023 x 10 ^ = 1 . 9 9 G (d) Mass of 1.12 I itre of 02 at STP =
Thus 20 g iron has maximum weight.
32 x 1.2
22.4 = 1.6g
3.168 Solution 24. 1 carat = 3.168 grains = " j ^ - g r a m
3.168x0.5
.-. 0.500 carat diamond = — — — g = 0.10 g
Thus weight of ring = 0.10 + 7.0 = 7.10 g = 7.10 x 10- 3 kg Solution 25. Density of Vanadium = 5.96 g/cm3 = 5.96 x ](T3 kg/cm'
5.96x10 . , j
~ 1(7^— 1 03 kg/m3