Ncert Solutions Chemistry Class 11th
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1.9. Calculate the atomic mass (average) of chlorine using the following data:
% Natural Abundance
Molar Mass
35Cl
75.77
34.9689
37Cl
24.23
36.9659
1.9. Calculate the atomic mass (average) of chlorine using the following data:
|
|
% Natural Abundance |
Molar Mass |
|
35Cl |
75.77 |
34.9689 |
|
37Cl |
24.23 |
36.9659 |
1.9. Average atomic mass = (Fractional abundance of 35Cl x molar mass of 35Cl) + (Fractional abundance of 37Cl x molar mass of 37Cl)
= (75.77/100 x 34.9689) + (24.23/100 x 36.9659)
= 26.4959 + 8.9568
= 35.4527
11.20. What happens when
(a) Borax is heated strongly,
(b) Boric acid is added to water,
(c) Aluminium is treated with dilute NaOH,
(d) BF3 is reacted with ammonia?
11.20. What happens when
(a) Borax is heated strongly,
(b) Boric acid is added to water,
(c) Aluminium is treated with dilute NaOH,
(d) BF3 is reacted with ammonia?
(a) When borax is heated strongly, it loses water and swells into the white mass, which on further heating melts to form a transparent glassy solid called borax glass and borax bead.
Na2B4O710H2O→ Na2B4O7+ 10H2O
Na2B4O7 → 2NaBO2+ B2O3
(b) When boric acid is added to water, it accepts electrons
1.8. Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively.
1.8. Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively.
1.8. Mass percentage of Fe = 69.9%
Mass percentage of O = 30.1%

11.19. Explain structures of diborane and boric acid.
11.19. Explain structures of diborane and boric acid.
Boric acid contains planar BO33- ions which are linked together through hydrogen bonding shown in the fig.

11.18. How is excessive content of CO2 responsible for global warming?
11.18. How is excessive content of CO2 responsible for global warming?
Excess of CO2 absorbs heat radiated by the earth. Some of it is dissipated into the atmosphere while the remaining part is radiated back to the earth. As a result, temperature of the earth increases. This is the cause of global warming.
1.7. How much copper can be obtained from 100 g of copper sulphate (CuSO4)? (Atomic mass of Cu= 63.5 amu).
1.7. How much copper can be obtained from 100 g of copper sulphate (CuSO4)? (Atomic mass of Cu= 63.5 amu).
1.7. 1 mole of CuSO4 contains 1 mole (1 g atom) of Cu
Molar mass of CuSO4= 63.5 + 32 + (4 x 16) = 159.5 g mol-1
Thus, Cu that can be obtained from 159.5 g of CuSO4 = 63.5 g
∴ Cu that can be obtained from 100 g of CuSO4 =63.5/159.5 * 100 = 39.81g
The highly poisonous nature of CO arises because of its ability to form a complex with haemoglobin, which is about 300 times more stable than the oxygen-haemoglobin complex. This prevents haemoglobin in the red blood corpuscles from carrying oxygen round the body and ultimately resulting in death.
11.16 Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3 is bubbled through. Give reasons.
11.16 Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3 is bubbled through. Give reasons.
Since, anhydrous HF is a covalent compound and weak acid due to high bond dissociation energy. AlF3does not dissolve in HF.
Whereas NaF is ionic compound.
3NaF + AlF3 → Na3 [AlF6]
Na3 [AlF6] + 3BF3 (g)→ AlF3 + 3Na+ [BF]–
11.15. If B-Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.
11.15. If B-Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.
B-Cl bond has dipole moment because of polarity. In BCl3, since the molecule is symmetrical (planar), the polarities cancel out and hence the dipole moment is zero
1.6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density 1.41 g mL-1 and the mass percent of nitric acid in it is being 69%.
1.6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density 1.41 g mL-1 and the mass percent of nitric acid in it is being 69%.
1.6. A mass percent of 69% means that 100 g of nitric acid solution contains 69 g of nitric acid by mass.
Molar mass of nitric acid HNO3= 1 + 14 + (3x16) = 63 gmol-1
