Q.

When 0.36 gm of glucose  (C6H12O6(s)) was burnt in a bomb calorimeter (Heat capacity 600 J/K), the temperature rise by 10 K. The standard molar enthalpy of combustion of glucose is  x×1016  J/mol. Find the value of  x.
[Given: Molecular mass of glucose = 180 g/mol]

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answer is 3.

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Detailed Solution

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qv=600×10=6000J

For 0.36 gm
For 1mole = 180 gm, ΔU=6000×1800.36=3×106J/mol

Δng=0ΔH=ΔU=3×106J/mole
x = 3

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