JEE Mains Chemistry Past PapershardNUMERICAL

An athlete is given 100 g of glucose (C6H12O6) for energy. This is equivalent to 1800 KJ of energy. JEE Mains Chemistry Past Papers Chemistry Question

Question

An athlete is given 100 g of glucose (C6H12O6) 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 & O are 12, 1 and 16 g mol–1

Answer: .

💡 Solution & Explanation

900 = O H2 n × 45 or O H2 n = 20 mol or 360 g | JEE(Main) 2023 | DATE : 25-01-2023 (SHIFT-1) | PAPER-1 | OFFICIAL PAPER| CHEMISTRY PAGE # 12

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