When $100$ g of ethylene polymerises entirely to polyethene, the weight of polyethene formed as per the equation: $n (C_{2} H_{4}) \to {- (C H_{2} - C H_{2})}_{n} -$ is:

\frac{n}{2}

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100

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\frac{100}{n}

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100 n

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Solution

The correct option is B $100$ g
By applying $P O A C$ for $C$ atoms,
Moles of ethylene $\times$ $2$ $=$ Mole of polythene $\times$ $n$ $\times$ $2$
$\frac{100 g}{28} \times 2 = \frac{w t . o f p o l y e t h e n e}{28 \times n} \times n \times 2$
Weight of polyethene $= 100$ g

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Similar questions

n C H_{2} = C H_{2} ⟶ - (C H_{2} - C H_{2})_{n} -

n C H_{2} = C H_{2} \to (- C H_{2} - C H_{2} -)_{n}

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n C H_{2} = C H_{2} \to - (C H_{2} - C H_{2})_{n} -

C a C_{2} + 2 H_{2} O ⟶ C a {(O H)}_{2} + C_{2} H_{2}

C_{2} H_{2} + H_{2} ⟶ C_{2} H_{4}

n C_{2} H_{4} ⟶ {(- {C H}_{2} - {C H}_{2} -)}_{n}

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