
An athlete takes $\text{ 100 g }$ of glucose of energy equivalent to$\text{ }1560\text{ kJ }$. How much energy is taken up by $\text{ 1 g }$ a molecule of glucose?
(A) \[\text{ 15}\text{.6 kJ }\]
(B) \[\text{ 2808 kJ }\]
(C) \[\text{ 1560 kJ }\]
(D) \[\text{ 28}\text{.08 kJ }\]
Answer
225k+ views
Hint: Here, the energy produced by the $\text{ 1 g }$molecules of the glucose is determined by the amount of energy produced by the 1 mole of glucose. The mole is the ratio of the weight of the substance to the molecular weight. Use the concept of the mole to find out the amount of glucose in one gram of molecule and equate it to get the desired answer.
Complete step by step solution:
We have given that the athlete takes up $\text{ 100 g }$ of the glucose.
The $\text{ 100 g }$ of the glucose is equivalent to $\text{ }1560\text{ kJ }$ of the energy equivalent.
We have to find the amount of energy taken by the athlete when he consumes $\text{ 1 g }$ of glucose.
We can write the relation as,
$\text{ 100 g }=\text{ 1560 kJ}$
And we have to find the ‘X’ amount of energy.
$\text{ 1 g }=\text{ X kJ}$
Let's consider one gram of a glucose molecule. The 1 gram of glucose molecules would is considered as the 1 mole of glucose. We know that. 1 mole is equalled to the number of atoms present per molecule of glucose.
We know that the mass of 1 mole of glucose is equalled to the molecular weight of glucose which is$\text{ 180 g }$.
In other words, the 1 mole of glucose contains $\text{ 180 g }$of the glucose molecule.
$\text{ 1 mole of glucose = 180 g of Glucose }$
We know that one mole is equalled to one gram of molecules. Thus, we have,
$\text{ 1 g of glucose = 180 g of Glucose }$
Now, if $\text{ 100 g }$of glucose gives $\text{ }1560\text{ kJ }$energy equivalent then 1 g of glucose i.e. $\text{ 180 g }$Of glucose molecule would give,
\[\begin{matrix}
\text{100 g } & = & \text{1560 kJ} \\
180\text{ g} & = & \text{X Energy} \\
\end{matrix}\]
On solving further we get,
\[\text{Energy = }\left( \dfrac{1560}{100} \right)\text{ }\times \text{ 180 kJ = 2808 kJ}\]
Therefore, the energy taken when an athlete consumes the 1 gram of glucose is equalled to\[\text{ 2808 kJ }\].
Hence, (B) is the correct option.
Note: Note that, we have to consider the molecules of the glucose. Thus, we are considering that the 1 mole of glucose has $\text{ 180 g }$ of glucose in it. This is because 1 gram of a substance will be equal to the 1 mole of the substance only when it contains the weight which equals its molecular weight. In simpler words, the weight taken must be equal to the molecular weight.
$\begin{align}
&\text{ no}\text{.of moles = }\dfrac{\text{weight}}{\text{mol}\text{.wt}} \\
&\text{For 1 mole , weight = molecular weight} \\
\end{align}$
Complete step by step solution:
We have given that the athlete takes up $\text{ 100 g }$ of the glucose.
The $\text{ 100 g }$ of the glucose is equivalent to $\text{ }1560\text{ kJ }$ of the energy equivalent.
We have to find the amount of energy taken by the athlete when he consumes $\text{ 1 g }$ of glucose.
We can write the relation as,
$\text{ 100 g }=\text{ 1560 kJ}$
And we have to find the ‘X’ amount of energy.
$\text{ 1 g }=\text{ X kJ}$
Let's consider one gram of a glucose molecule. The 1 gram of glucose molecules would is considered as the 1 mole of glucose. We know that. 1 mole is equalled to the number of atoms present per molecule of glucose.
We know that the mass of 1 mole of glucose is equalled to the molecular weight of glucose which is$\text{ 180 g }$.
In other words, the 1 mole of glucose contains $\text{ 180 g }$of the glucose molecule.
$\text{ 1 mole of glucose = 180 g of Glucose }$
We know that one mole is equalled to one gram of molecules. Thus, we have,
$\text{ 1 g of glucose = 180 g of Glucose }$
Now, if $\text{ 100 g }$of glucose gives $\text{ }1560\text{ kJ }$energy equivalent then 1 g of glucose i.e. $\text{ 180 g }$Of glucose molecule would give,
\[\begin{matrix}
\text{100 g } & = & \text{1560 kJ} \\
180\text{ g} & = & \text{X Energy} \\
\end{matrix}\]
On solving further we get,
\[\text{Energy = }\left( \dfrac{1560}{100} \right)\text{ }\times \text{ 180 kJ = 2808 kJ}\]
Therefore, the energy taken when an athlete consumes the 1 gram of glucose is equalled to\[\text{ 2808 kJ }\].
Hence, (B) is the correct option.
Note: Note that, we have to consider the molecules of the glucose. Thus, we are considering that the 1 mole of glucose has $\text{ 180 g }$ of glucose in it. This is because 1 gram of a substance will be equal to the 1 mole of the substance only when it contains the weight which equals its molecular weight. In simpler words, the weight taken must be equal to the molecular weight.
$\begin{align}
&\text{ no}\text{.of moles = }\dfrac{\text{weight}}{\text{mol}\text{.wt}} \\
&\text{For 1 mole , weight = molecular weight} \\
\end{align}$
Recently Updated Pages
JEE Main 2026 Session 1 Correction Window Started: Check Dates, Edit Link & Fees

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Trending doubts
JEE Main 2026: City Intimation Slip and Exam Dates Released, Application Form Closed, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

Organic Chemistry Some Basic Principles And Techniques Class 11 Chemistry Chapter 8 CBSE Notes - 2025-26

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

