JEE MAIN - Chemistry (2023 - 25th January Morning Shift - No. 19)
An athlete is given 100 g of glucose (C$$_6$$H$$_{12}$$O$$_6$$) 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$$^{-1}$$
Molar mass of C, H & O are 12, 1 and 16 g mol$$^{-1}$$.
Answer
360
Explanation
$$
\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{~s})+6 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l})
$$
Extra energy used to convert $\mathrm{H}_2 \mathrm{O(l)}$ into $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ into $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$
$$ \begin{aligned} & =\frac{1800}{2}=900 \mathrm{~kJ} \\\\ & \Rightarrow 900=\mathrm{n}_{\mathrm{H}_2 \mathrm{O}} \times 45 \\\\ & \mathrm{n}_{\mathrm{H}_2 \mathrm{O}}=\frac{900}{45}=20 \text { mole } \\\\ & \mathrm{W}_{\mathrm{H}_2 \mathrm{O}}=20 \times 18=360 \mathrm{~g} \end{aligned} $$
Extra energy used to convert $\mathrm{H}_2 \mathrm{O(l)}$ into $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ into $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$
$$ \begin{aligned} & =\frac{1800}{2}=900 \mathrm{~kJ} \\\\ & \Rightarrow 900=\mathrm{n}_{\mathrm{H}_2 \mathrm{O}} \times 45 \\\\ & \mathrm{n}_{\mathrm{H}_2 \mathrm{O}}=\frac{900}{45}=20 \text { mole } \\\\ & \mathrm{W}_{\mathrm{H}_2 \mathrm{O}}=20 \times 18=360 \mathrm{~g} \end{aligned} $$
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