Class 11th

1.7. How much copper can be obtained from 100 g of copper sulphate (CuSO₄)? (Atomic mass of Cu= 63.5 amu).

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1.7. 1 mole of CuSO₄contains 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

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11.17 Suggest a reason as to why CO is poisonous.

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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.

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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.

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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]^–

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11.15. If B-Cl bond has a dipole moment, explain why BCl₃ molecule has zero dipole moment.

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B-Cl bond has dipole moment because of polarity. In BCl_3, since the molecule is symmetrical (planar), the polarities cancel out and hence the dipole moment is zero

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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%.

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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

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11.14. Suggest reasons why the B-F bond lengths in BF₃ (130 pm) and BF₄^– (143 pm) differ.

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The difference in bond length is due to the difference in the state of hybridisation. In BF₃ 'B' is sp² hybridised and in BF₄^– 'B' is sp³ hybridised.

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1.5. Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^-1.

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1.5

Since the molar mass of sodium acetate is 82.0245 g mol^-1.

⇒ 1 mol of CH₃COONa has a mass = 82.0245 g

∴Mass of sodium acetate (CH₃COONa) required to make 500 ml of 0.375 molar aqueous solution

= 0.375 x 0.5 mol x 82.0245 g mol^-1

= 15.3795 g = 15.380 g

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11.13.

Rationalise the given statements and give chemical reactions:
Lead (II) chloride reacts with Cl₂to give PbCl₄.
Lead (IV) chloride is highly unstable towards heat.
Lead is known not to form an iodide PbI₄.

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PbCl₂+ Cl₂→ PbCl₄.
This is because Pb can show +2 oxidation state more easily than +4 due to inert pair effect.
PbCl₄→ PbCl₂ + Cl₂
Because Pb²⁺ is more stable than Pb⁴⁺ due to inert pair effect.
PbI₄does not exist because I^- ion being a powerful reducing agent reduces Pb⁴⁺ ion to Pb²⁺ ion in solution.

Pb⁴⁺ + 2I^– → Pb²⁺ + I₂

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11.12. Explain the difference in properties of diamond and graphite on the basis of their structures.

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Diamond has a crystalline lattice where each carbon atom undergoes sp³ hybridisation and linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion. The C–C bond length is 154 pm. The structure extends in space and produces a rigid three- dimensional network of carbon atoms. It is very difficult to break extended covalent bonding and, therefore, diamond is a the hardest substance on the earth.

Graphite has layered structure in which the layers are held by van there Waals forces and distance between two layers is 340 pm. Each layer is composed of planar hexagonal rings of carbon atoms. C—C bond length

...more

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1.4. Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

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