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Derivation of Kp and Kc

Equivalents o...

Question

Equivalents of $C_{6} H_{12} O_{6}$ per mol of $H_{2} O$ used :

2

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6

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4

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none of these

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Solution

The correct option is A $2$
$12 m o l H_{2} O u s e d = 1 m o l C_{6} H_{12} O_{6} \equiv 24 e q u i v a l e n t s C_{6} H_{12} O_{6}$
$1 m o l H_{2} O u s e d \equiv 2 e q u i v a l e n t s C_{6} H_{12} O_{6}$
$6 C O_{2} O . N = 24 ⟶ C_{6} H_{12} O_{6} = 0$
$c h a n g e = 24 u n i t s$
$E q u i v a l e n t w e i g h t o f C O_{2} = \frac{6 m o l}{24} = \frac{m o l . w t .}{4}$
$∴ 1 m o l C O_{2} = 4 e q u i v a l e n t s o f C O_{2}$
$E q u i v a l e n t w e i g h t o f C_{6} H_{12} O_{6} = \frac{m o l . w t .}{24}$
$∴ 1 m o l C_{6} H_{12} O_{6} = 24 e q u i v a l e n t s o f C_{6} H_{12} O_{6}$

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