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Molecular Weight of Polymers

The formation...

Question

The formation of polyethylene from calcium carbide takes places as follows :
$C a C_{2} + 2 H_{2} O \to C a (O H)_{2} + C_{2} H_{2}$
$C_{2} H_{2} + H_{2} \to C_{2} H_{4}$
$n C_{2} H_{4} \to (C H_{2} - C H_{2})_{n}$
The amount of polyethylene obtained from $64$ kg of $C a C_{2}$ is :

14

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7

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21

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28

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Solution

The correct option is B $28$ kg
Mass of $C a C_{2}$ = $64$ Kg

Moles of $C a C_{2}$ = 64/64 = $1000$ mol

$C a C_{2} + 2 H_{2} O \to C a (O H)_{2} + C_{2} H_{2}$

For 1000 mole of $C a C_{2}$ , We get 1000 mole of $C_{2} H_{2}$

$C_{2} H_{2} + H_{2} \to C_{2} H_{4}$

For 100 mole of $C_{2} H_{2}$ , we get 1000 mole of $C_{2} H_{4}$

$n C_{2} H_{4} \to (C H_{2} - C H_{2})_{n}$

For n mole $C_{2} H_{2}$ , we get 1 mole of polyethylene.
For 1000 mole of $C_{2} H_{2}$ , we get $(\frac{1000}{n}) \times 1$
Number of moles of polyethylene formed are 1000/n.

Molecular weight of polyethylene is $n \times [(2 \times 12) + (2 \times 4)] = 28 n$

Mass of polyethylene produced $=$ Number of moles $\times$ Molecular weight
$= (\frac{1000}{n}) \times 28 n$

$= 28000$ gram

$= 28$ kg

Option D is correct.

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Similar questions

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

C_{2} H_{2} + H_{2} \to C_{2} H_{4}

n C_{2} H_{4} \to (C H_{2} - C H_{2}) n

C a C_{2}

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

C_{2} H_{2} + H_{2} \to C_{2} H_{4}

n C_{2} H_{4} \to {(- C H_{2} - C H_{2} -)}_{n}

64.1

C a C_{2}

C a C_{2} + H_{2} O \to C a (O H)_{2} + C_{2} H_{2} C_{2} H_{2} + H_{2} \to C_{2} H_{4}

n (C_{2} H_{4}) \to (- C H_{2} - C H_{2} -)_{n}

C a C_{2}

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}

C a C_{2}

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}

C a C_{2}

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