Polyethylene is obtained from calcium carbide.
$CaC_{2} + 2H_{2}O \to Ca(OH)_{2} + C_{2}H_{2}$$C_{2}H_{2} +H_{2} \to C_{2}H_{4}$$ nC_{2}H_{4} \to $
Therefore, the amount of polyethylene obtained for $64 \,kg \,CaC_{2}$ is
Updated On: Jul 6, 2022
$7\, kg$
$14\, kg$
$24\, kg$
$28\, kg$
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The Correct Option isD
Solution and Explanation
Moles of $CaC_{2} = \frac{64\times 10^{3}}{64} ?
$\therefore$ From the balanced chemical equation,
moles of $C_{2}H_{2}$ = moles of $ C_{2}H_{4}$
= moles of $CaC_{2}= 1 \times 10^{3}$$\therefore$ moles of polythene $ = \frac{1}{n} \times 1 \times 10^{3}$$\therefore$ weight of polythene $= \frac{1}{n} \times 1\times28n \, kg = 28\, kg$
It is described as the distribution rather than a specific number due to the occurrence of polymerization in such a way as to produce different chain lengths. Polymer MW is derived as follows:
\[M_{W} = \sum^{N}_{i=1} w_{i}MW_{i}.\]
Where,
wi = the weight fraction of polymer chains having a molecular weight of MWi.
The MW is typically measured by light dispersing experiments. The degree of dispersing arises from the molecule size and, thus, molecular weight dispensation can be mathematically set on the total scattering created by the sample.
