JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
An athlete is given \(100\,g\) of glucose \(\left( C _6 H _{12} O _6\right)\) for energy. This is equivalent to \(1800\,kJ\) 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 \(C , H\) and \(O\) are \(12.1\) and \(16\,g\,mol ^{-1}\)
- A \(180\)
- B \(360\)
- C \(90\)
- D \(45\)
Answer & Solution
Correct Answer
(B) \(360\)
Step-by-step Solution
Detailed explanation
\(C _6 H _{12} O _6( s )+6 O _2 \rightarrow 6 CO _2( g )+6 H _2 O ( l )\) Extra energy used to convert \(H _2 O (1)\) into \(H _2 O (1)\) into \(H _2 O ( g )\) \(=\frac{1800}{2}=900\,kJ\) \(900= n _{ H _2 O } \times 45\) \(n _{ H _2 O }=\frac{900}{45}=20\,mole\)…
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