Combustion of glucose takes place according to the equation, |
\[{{C}_{6}}{{H}_{12}}{{O}_{6}}+6{{O}_{2}}\to 6C{{O}_{2}}+6{{H}_{2}}O\],\[\Delta H=-72\,kcal\]. |
How much energy will be required for the production of 1.6 g of glucose (Molecular mass of glucose = 180 g) |
A) 0.064 kcal
B) 0.64 kcal
C) 6.4 kcal
D) 64 kcal
Correct Answer: B
Solution :
[b] \[\Delta H\,per\,1.6g=\frac{72\times 1.6}{180}=0.64\,kcal\].You need to login to perform this action.
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