Solution

Theoretical density of a crystal is given by $$\rho = \cfrac {nM}{N_A a^3} \ g/cm^3$$

Given, $$M= 63.5 \ g \ ; \ \rho = 8.93 \ g/cm^3$$

For FCC, $$n=4$$

$$\Rightarrow a^3= \cfrac {4 \times 63.5}{6.022 \times 10^{23} \times 8.93}$$

$$\Rightarrow a = \left (\cfrac {4 \times 63.5}{6.022 \times 10^{23} \times 8.93}\right)^{1/3}= 3.6 \times 10^{-8} \ cm$$

Now, if $$r$$ is radius of copper atom, then in FCC,

$$r = \cfrac {a}{2 \sqrt 2}$$

$$r = \cfrac {3.6}{2 \sqrt 2} \times 10^{-8} \ cm= 127.8 \ pm$$

Solution

Friedel Crafts Acylation

SOLUTIONS

SOLUTION

only RCH₂NO₂ has α-hydrogen atoms - > will shows tautomerism

SOLUTION

Nylon 6,6 or polyamide is a substance obtained from the repeated condensation reaction of Hexamythyline di amine and adipic acid which when undergoes polymerisation gives rives polyamide and water molecules

Nylon 6,6 is a condensation product of adipic acid and hexamethylenediamine 

Nylon 6 is a condensation product of only one monomer caprolactum. 

Nylon 6,6 is stronger and more resistant to heat as compared to Nylon 6.
Nylon 6,6 is used for making carpets, tyre ropes etc. due to its strength and toughness.
Nylon 6 on the other hand is used for making clothes, nets, threads etc.

SOLUTION

NO⁺[BF4]⁻ 

⇒B.O.of NO⁺=3.0,, i.e, one sigma bond and two π bonds
∴ No. of π bonds = 2
No. of σ bond = 5
⇒B.O. of NO⁺= 3.0N
and B.O. of NO=2.5
Bond order is inversely proportional to bond length.
⇒NO+ is diamagnetic and BF4⁻ is also diamagnetic
⇒B−F bonds are longer in BF4⁻ than in BF₃ due to absence of pπ−pπ back bonding in [BF4⁻???????]

Solution

molality of aqueous solution = 5.2 mol kg^-1

Therefore, the number of moles of methyl alcohol in 1 kg of solvent (water) = 5.2 mol

And the number of moles of solvent (water)  = weight / molar mass = 1000 g / 18 g mol^-1 = 55.55 mol

Hence:

Mole fraction of methyl alcohol = n_MeOH /n_MeOH + n_water = 5.2 / 55.55 + 5.2 = 0.08559

SOLUTION

It undergoes auto reduction forming copper and sulphur dioxide.
$$2{ Cu }_{ 2 }O+{ Cu }_{ 2 }S\xrightarrow { heat } 6Cu+S{ O }_{ 2 }$$