What is the volume of water consumed during acid hydrolysis of $1.368 K g$ of sucrose?
(Given - molar masses of sucrose $= 342$ , water $= 18$ , density of water $= 1 g / {c m}^{3}$ )

0.072 {d m}^{3}

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0.720 {d m}^{3}

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0.18 {d m}^{3}

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0.018 {d m}^{3}

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Solution

The correct option is A $0.072 {d m}^{3}$
From the above reaction, we can conclude that one mole of sucrose during acid hydrolysis consumes one mole of water to form one mole each of glucose and fructose. Since we have,
$N u m b e r o f m o l e s o f s u c r o s e = \frac{m a s s o f s u c r o s e i n g r a m s}{g r a m m o l e c u l a r m a s s o f s u c r o s e} = \frac{1368}{342} = 4$

Thus, $4$ moles of sucrose will consume $4$ moles of water during acid hydrolysis which can be written as:

4 moles of water $= 4 \times G r a m m o l e c u l a r m a s s o f w a t e r = 4 \times 18 g r a m s = 72 g r a m s$

Since $V o l u m e = \frac{m a s s}{d e n s i t y}$ , we get

$V o l u m e o f w a t e r r e q u i r e d = \frac{72 g r a m s}{1 g / c m^{3}} = 72 c m^{3} = 0.072 d m^{3}$

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