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QJEEM23Che285: JEE Main 2023 Chemistry - Class 11 JEEM23Che285 Solution: An athlete is given 100 g of glucose (C H O ) for energy. This…

An athlete is given 100 g of glucose (C 6_6 H 12_{12} O 6_6 ) for energy. This is equivalent to 1800kJ of energy. The 50% of this energy gained is utilized by the athlete for sports activities at the event. In order to avoid storage of energy, the weight of extra water he would need to perspire is ____________ g (Nearest integer)

Assume that there is no other way of consuming stored energy.

Given : The enthalpy of evaporation of water is 45 kJ mol −1^{-1}

Molar mass of C, H & O are 12, 1 and 16 g mol −1^{-1} .

Step-by-Step Verified Solution

C6H12O6( s)+6O2→6CO2( g)+6H2O(l)\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{~s})+6 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l})

Extra energy used to convert H2O(l)\mathrm{H}_2 \mathrm{O(l)} into H2O(l)\mathrm{H}_2 \mathrm{O}(\mathrm{l}) into H2O(g)\mathrm{H}_2 \mathrm{O}(\mathrm{g})

amp;=18002=900 kJamp;⇒900=nH2O×45amp;nH2O=90045=20 mole amp;WH2O=20×18=360 g\begin{aligned}& =\frac{1800}{2}=900 \mathrm{~kJ} \\\\& \Rightarrow 900=\mathrm{n}_{\mathrm{H}_2 \mathrm{O}} \times 45 \\\\& \mathrm{n}_{\mathrm{H}_2 \mathrm{O}}=\frac{900}{45}=20 \text { mole } \\\\& \mathrm{W}_{\mathrm{H}_2 \mathrm{O}}=20 \times 18=360 \mathrm{~g}\end{aligned}

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