SOLUTION

Molecular weight of CuSO₄.5H₂O = 249.5gmol⁻¹ 

The reaction for liberation of I₂;

2CuSO₄.5H₂O + KI = I₂ + Cu₂I₂ + 2K₂SO₄ + 5H₂O 

2.54g of I₂ = 2.54 /254 = 0.01 moles of I₂ 

Now, 1 mole of I₂ is liberated from 2 moles of CuSO₄.5H₂O 

∴0.01 mole of I₂ will be liberated from 2×0.01 mole i.e. 0.02 mole of CuSO₄.5H₂O. 

∴ Amount of CuSO₄.5H₂O=0.02 mole × 249.5gmol⁻¹=4.99gm

Solution

Because the oxidation state of oxygen in $${ H }_{ 2 }{ O }_{ 2 }$$ is $$-1$$ which can be oxidized to $$'O'$$ oxidation state by removing an $${ e }^{ - }$$ or else can be reduced by adding an $${ e }^{ - }$$ and oxidation state would be reduced to $$-2$$ oxidizing agent means oxidizing others itself getting reduced.

Solution

Because insects have the ability to digest silk and wool. Examples of insects that damage the clothes are-silverfish, cockroach, moth etc. 

Solution: 

CaF₂ ?Ca²⁺ + 2F⁻

Ksp=[Ca²⁺][F−]²=S(S +0.1)² = S × 0.12 = 5.3×10⁻⁹

Note: S<<0.1so, S+0.1≈0.1

⇒S=5.3×10⁻⁷ M

SOLUTION

 A cyclic bromonium ion is formed as an intermediate which is attacked from the opposite side to give anti addition product. 

SOLUTION

We know that Z=V(real)/ V(ideal)
Since Z > 1, it means V(real) > V(ideal)
At STP, V(ideal)=22.4 L=22.4 dm³
Hence, V(real) > 22.4 dm³

SOLUTION

 greater the number of alkyl groups attached to a positively charged carbon atom, the greater is the hyperconjugation interaction and stabilisation of the cation.

Hyperconjugation effect increases in the order :

(CH₃)₃C− < (CH₃)₂CH− < CH₃CH₂−

The σ electrons of C—H bond of the alkyl group enter into partial conjugation with the attached unsaturated system or with the unshared p orbital.

SOLUTION
 

k (Conductivity) =0.0248 Scm⁻¹

C (molarity) =0.20 M

T(temperature) =298 K=298 K

Λ_m=? (molar conductivity) 

Molar conductivity Λ_m = (1000 × k) /C

= (1000×0.0248) / 0.20

=124 S cm² mol⁻¹

The molar conductivity is 124 S cm² mol⁻¹

Solution

The solubility of silver bromide in hyposolution due to the formation of [Ag(S₂O₃)₂]^3-

Solution

The number of moles of organic compound $$\displaystyle = \frac {2.68}{134} = 0.2 $$ mol.
The number of moles of silver iodide $$\displaystyle  = \frac {14.08}{234.77} = 0.06 $$
Thus, 0.2 moles of organic compound gives 0.6 moles of $$AgI$$.
1 mole of organic compound will give 3 moles of $$AgI$$.
Hence, the number of methoxy groups present in the organic compound is 3.