Chemistry

15.4

15.3

 Two monomers Hexamethylenediamine [H₂N- (CH₂)₆-NH₂] and adipic acid [HOOC- (CH₂)₄- COOH] combined to form the polymer and a water molecule is removed in this process. Hydrogen (H) is removed from both sides of Hexamethylenediamine and OH group is removed from both sides of adipic acid to give a water molecule.

The monomer is OC- (CH₂)₅-NH known as Caprolactam. The cyclic structure of the monomer changes to linear to form the polymer, as shown below:

The monomer is Tetrafluroethene (CF₂= CF₂), the double bond breaks to form the polymeric

15.2

Polymers are classified based on structure, into 3 types:

Linear Polymer: They have a long and straight chain of Ex: high-density Polythene, Polyvinyl chloride

Branched-chain Polymers: They have linear molecular chains along with some Ex: less density polythene.

Cross-linked or network Polymers: In these polymers, strong covalent bonds are between the linear chains. Generally, they contain 2 or 3 types of functional groups. Ex: Bakelite, melamine.

15.1

Polymer=Poly + Mer

Poly means “many” and “Mer” means unit or part. A polymer is a large molecule which is formed by linking repeating structural units. The structural units are generally simple molecules and they are linked by a covalent bond to form a polymer.

Given,

Volume of water, V = 450 mL = 0.45 L

Temperature, T= (37 + 273)K = 310 K

1.0 g of polymer of molar mass 185,000

Number of moles of polymer, n = 1 / 185,000 mol

We know that,

Osmotic pressure? = nRT/V

= 1 X 8.314 X 10³ X 310 / 185000 X 0.45

= 30.98 Pa

= 31 Pa (approx)

Given

Mass of acetic acid, w₁ = 75 g

Lowering of melting point? T_f = 1.5 K

K_f = 3.9 K kg/mol

Molar mass of ascorbic acid (C₆H₈O₆), M₂ 6 * 12 + 8 * 1 + 6 * 16 = 176 g/mol

We know that,

= 5.08 g

Hence,

5.08 g of ascorbic acid is needed to be dissolved.

16.32 

Alkyl groups are water hating groups that is they do not get assimilated in water easily whereas on the other hand functional groups like sulphonates, alcohol etc. are water liking groups that is these groups dissolve in water easily. Therefore hydrophobic part in the above detergents is the long chain alkyl group and the hydrophilic part is the alcohol group, sulphonates group etc.

Given,

Mass of water, w_l = 500 g

Boiling point of water = 99.63°C (at 750 mm Hg).

Molal elevation constant, K_b = 0.52 K kg/mol

Molar mass of sucrose (C₁₂H₂₂O₁₁), M₂   (11 * 12 + 22 * 1 + 11 * 16) = 342 g/mol

Elevation of boiling point ΔT_b = (100 + 273) - (99.63 + 273) = 0.37 K

We know that,

ΔT_b = K_b X 1000 X W₂ / M₂ X W₁

0.37 = 0.52 X 1000 X W₂ / 340 X 500

w₂ = 0.37 X 342 X 500 / 0.52 X 1000

w₂ = 121.67 g

Hence,

121.67 g (approx) Sucrose is added to 500g of water so that it boils at 100°C.

Given,

Vapour pressure of water, P_I^o = 23.8 mm of Hg

Weight of water, w₁ = 850 g

Weight of urea, w₂ = 50 g

Molecular weight of water, M₁ = 18 g/mol

Molecular weight of urea, M₂ = 60 g/mol

n₁ = w₁/M₁ = 850/18 = 47.22 mol

n₂ = w₂/M₂ = 50/60  = 0.83 mol

We have to calculate vapour pressure of water in the solution p₁

By using Raoult's therom,

P_I = 23.4 mm of Hg Hence,

The vapour pressure of water in the solution is 23.4mm of Hg and its relative lowering is 0.0173.

Given, P_A^o = 450 mm Hg

^P_Bo = 700 mm Hg

p_total = 600 mm of Hg

By using Rault's law,

p_total = P_A + P_B

p_total = P_A^ox_A + P_B^ox_B

p_total = P_A^ox_A + P_B^o ( 1 - xA )

p_{total = (}P_A^o- P_B^o)x_{A +} P_B^o

600 = (450 - 700) x_A + 700

-100 = -250 x_A

x_A = 0.4

∴ x_B = 1 - x_A

x_B = 1 – 0.4

x_B = 0.6

Now,

P_A = P_A^ox_A

P_A = 450 * 0.4

P_A = 180 mm of Hg and

P_B = P_B^ox

P_B = 700 * 0.6

P_B = 420 mm of Hg

Composition in vapour phase is calculated by

Mole fraction of liquid,

A =P_A / P_A + P_B

= 180/180+420

= 0.30

Mole fraction of liquid,

B =P_B / P_A + P_B

= 420 / 180+420

= 0.70