High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen. (a) Determine the concentration of Flourine (in wt%) that must be added if this substitution occurs for 12% of all the original hydrogen atoms.

Concentration of Flourine = 24.756%

Explanation:

Given that :

High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen.

the objective is to determine he concentration of Flourine (in wt%) that must be added if this substitution occurs for 12% of all the original hydrogen atoms.

At standard conditions , the atomic weight of the these compounds are as follows:

Carbon = 12.01 g/mol

Chlorine = 35.45 g/mol

Fluorine = 19.00 g/mol

Hydrogen = 1.008 g/mol

Oxygen = 16.00 g/mol

The chemical formula for polyethylene = (CH₂ - CH₂)ₙ

Therefore, for two carbons, there will be 4 hydrogens;

i.e

(CH₂ - CH₂)₂

( C₂H₄ - C₂H₄ )

Suppose the number of original hydrogen = 4moles

number of moles of Flourine F = 12% of 4

= 0.12 × 4

= 0.48 mol

∴ the number of remaining moles of Hydrogen is:

= 4 - 0.48

= 3.52 moles

number of moles of Carbon = 2 moles

∴ the mass of flourine F = number of moles of F × molar mass of F

= 0.48 × 19

= 9.12

The total mass of the compound now is = (0.48 × 19 ) + (3.52 × 1) + (2× 12)

= 9.12 + 3.52 + 24

= 36.64

Concentration of Flourine = (mass of flourine/total mass) × 100

Concentration of Flourine = (9.12/36.84 ) × 100

Concentration of Flourine = 0.24756 × 100

Concentration of Flourine = 24.756%

a

explanation:

the answer for sodium is 0.186

the answer for rubidium is 0.251