Human body requires $2370 K . c a l$ of energy daily. If the heat of combustion of glucose is $- 790 K . c a l / m o l e$ the amount of glucose required for daily consumption is

650 g

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540 g

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327 g

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490 g

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Solution

The correct option is B $540 g$
Moles for $2370 K$ cal $= \frac{2370}{790}$ moles
$C_{6} H_{2} O_{16} =$ $72 + 12 + 96$
$=$ $84 + 96$
$=$ $180$
$\frac{A m o u n t o f g l u c o s e}{180 g} = m o l e s = \frac{237 ϕ}{79 ϕ}$
Amount of glucose $= \frac{237}{79}$ $\times 180 g$
$= 540 g$

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2870 K c a l

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C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O, △ H = - 72 k c a l

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