Question

Human body requires $2370K.cal$ of energy daily. If the heat of combustion of glucose is $-790K.cal/mole$ the amount of glucose required for daily consumption is

• A. $650g$
• B. $540g$
• C. $327g$
• D. $490g$

Answer 1

The correct option is B $540g$

Moles for $2370K$ cal $=\frac{2370}{790}$ moles
$C_6{H}_2{O}_{16}=$ $72+12+96$
$=$ $84+96$
$=$ $180$
$\frac{Amountofglucose}{180g}=moles=\frac{237\varphi }{79\varphi }$
Amount of glucose $=\frac{237}{79}$ $\times 180g$
$=540g$