Polymers

Calculation of water molecules lost

Each monomer of $X$ and $Y$ combines by losing $2$ moles of water per linkage. When $500$ moles of each react, total possible water molecules lost will be:

$ 500 \times 2 - 1 = 999 $

Mass of water lost

Mass of water lost = (number of moles of water) × (molar mass of water)

$ = 999 \times 18 = 17982 \text{ g} $

Mass of polymer formed

Total mass of reactants = (mass of 500 mol of X) + (mass of 500 mol of Y)

$ = 75 \times 500 + 131 \times 500 = 37500 + 65500 = 103000 \text{ g} $

Therefore, mass of polymer formed, $Z$, after losing water is:

$ 103000 - 17982 = 85018 \text{ g} $

Hence, the mass of polymer $Z$ formed is 85,018 g.

Monomer $(X) \rightarrow$ positive carbylamine test The monomer used in the synthesis of Nylon -6,6 are hexamethylene diamine and adipic acid. Positive carbylamine test is given by primary amine compounds. Hexamethylene diamine is a primary amine compound and can give positive carbylamine test. So, the monomer $X$ is hoxamethylene diamine. $ \begin{aligned} &\begin{aligned} x \rightarrow \mathrm{H_2N}-\left(\mathrm{CH}_2\right)_6-\mathrm{NH}_2 \quad & \text { Molarmars } & \\ = & 6 \times 12 \mathrm{~g} / \mathrm{mol}+2 \times 14 \mathrm{~g} / \mathrm{mol} +16 \times 1 \mathrm{~g} / \mathrm{mol} & \\ = & 116 \mathrm{~g} / \mathrm{mol} & \end{aligned}\\ &\text { Moles of } X \text { analyzed using Dumas method }=10 \mathrm{~mol} \end{aligned} $According to Dumas method, the nitrogen present in the compound is released as nitrogen gas $\left(\mathrm{N}_2\right)$, The general chemical equation fol Dumas method can be written as$ \mathrm{C}_x \mathrm{H}_y \mathrm{~N}_z+\left(2 x+\frac{y}{2}\right) \mathrm{CuO}(\mathrm{~s}) \xrightarrow{} $(organic Compound)$ x \mathrm{CO}_2(g)+\frac{y}{2} H_2 \mathrm{O}(gold )+\frac{z}{2} \mathrm{~N}_2(g)+\left(2 x+\frac{y}{2}\right) \mathrm{Cu}(s) $The compound used in Dumas method is hexamethylene diamine. Its formula is $\mathrm{C}_6 \mathrm{H}_{16} \mathrm{~N}_2$$ \text { So, the equation can be written as, } $$ \mathrm{C}_6 \mathrm{H}_{16} \mathrm{~N}_2+20 \mathrm{CuO} \rightarrow $$ 6 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}+1 \mathrm{~N}_2+20 \mathrm{Cu} $$ \text { So, } 1 \text { mole of } \mathrm{C}_6 \mathrm{H}_{16} \mathrm{~N}_2 \text { gives. } 1 \text { mole of } \mathrm{N}_2 \text {. } $Given 10, moles of $\mathrm{C}_{6}\mathrm{H}_{16} \mathrm{~N}_2$ (hexamethylene diamine) gives 10 moles of $\mathrm{N}_2$ gas.Moles of $\mathrm{N}_2$ gas evolved $=10$ mol Mass can be calculated as, \begin{aligned} &\text { Moles }=\frac{\text { mass }}{\text { molarmass }}\\ &\text { Moloumass of } \mathrm{N}_2=28 \mathrm{~g} / \mathrm{mol} \end{aligned}mass $=$ moles $\times$ molar mass$ \begin{aligned} &\begin{aligned} & =10 \text { mol } x 28 \mathrm{~g} / \mathrm{mol} \\ & =280 \mathrm{~g} \end{aligned}\\ &\text { Answer: } 280 \% \end{aligned} $

Novolac is


Molar mass of monomer is $107 \mathrm{~g} / \mathrm{mol}$

$ \mathrm{n}=\frac{963}{107}=9 $

The number of monomer units present in it is 9.

(A) The monomer A on ozonolysis yields two molecules of HCHO and a molecule of CH$_3$COCHO. During ozonolysis, the double or triple bonds in the molecule are replaced by carbonyl groups.

Therefore, if we remove the oxygen atoms from 2 molecules of HCHO and a molecule of CH$_3$COCHO and replaced them with double bond, we can deduce the structure of monomer A.

The structure of monomer A we got is the structure of isoprene. Therefore, monomer A is isoprene.

(B) The isoprene is a monomer for natural rubber.

To write the “all $cis$” form of the polymer of monomer A that is isoprene, we need to write the structure of polymer such that the identical groups present across double bonds should lie in the same plane.

The “all $cis$” for of the polymer is shown below.

Final Answer

Hints:

The ozonolysis eliminates the double or triple bond from the molecule and replace it with carbonyl groups. The removal of oxygen atoms and replacing it with double bonds will give the structure of monomer A.