Combustion of glucose takes place according to the equation, C_6 H_ 12 O_6+6 O_2 6 CO_2+6 H_2 O Δ H=-72 k-cal The energy required for combustion of 1.6 g of glucose is

Correct answer: B. 0.64 k-cal gathered Molar mass of C_6 H_ 12 O_6=180 g mol^ -1 180 g of glucose require energy =72 kcal 1.6 g of glucose require energy =(72 × 1.6)/(180) gathered =0.64 k cal

JIPMER 2017

Exam: JIPMER
Year: 2017
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

JIPMER 2017 · Chemistry · Some Basic Concepts of Chemistry

Combustion of glucose takes place according to the equation, C_6 H_...

Combustion of glucose takes place according to the equation,

\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+6 \mathrm{O}_2 \longrightarrow 6 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O}

\Delta H=-72 \mathrm{k}-\mathrm{cal}

The energy required for combustion of 1.6 g of glucose is

Options

A0.064 k-cal
B0.64 k-cal
C $6.4 \mathrm{k}-\mathrm{cal}$
D64 k-cal

Correct answer

B. 0.64 k-cal

Step-by-step solution

\begin{gathered}\text{Molar mass of } \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6=180 \mathrm{ g} \mathrm{ mol}^{-1} \\ \because 180 \mathrm{ g} \text{ of glucose require energy }=72 \mathrm{kcal} \\ \therefore 1.6 \mathrm{ g} \text{ of glucose require energy }=\frac{72 \times 1.6}{180}\end{gathered} =0.64 \mathrm{k}

\mathrm{cal}

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