Sigma Percentile
JEE Main 2017
LEVELJEE Main

Animated Solution for Chemistry - Basic Concepts in Chemistry: The most abundant elements by mass in the body of a healthy human adult are Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0 %); and Nitrogen (2.6%). The weight which a 75 kg person would gain if all atoms are replaced by atoms is

Select Answer:

Visualized Solution

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

Solution Diagram

The Human Body

A Chemical Masterpiece
Imagine for a moment that you are not just a person, but a walking, talking chemical reactor. The human body is an intricate masterpiece composed of various elements, each playing a critical role in sustaining life. When we break down a healthy human adult by mass, we find a fascinating distribution: Oxygen dominates at 61.4%, followed by Carbon at 22.9%, Hydrogen at 10.0%, and Nitrogen at 2.6%.
In this problem, we are presented with a fascinating thought experiment. We are asked to imagine a scenario where every single normal hydrogen atom () in a 75 kg person is magically replaced by its heavier isotope, deuterium (). Our mission is to calculate the exact weight this person would gain.

Analyzing the Initial State

Before we can replace anything, we must first understand what is already there. The problem states that the total mass of the person is . We are also given the mass percentage of hydrogen, which is exactly 10.0%.
This percentage is a powerful tool. It tells us that out of every 100 parts of body mass, 10 parts are purely hydrogen. To find the actual mass of hydrogen in this specific person, we simply need to calculate 10% of their total body weight.
Let's set up our master equation for the initial mass of hydrogen:
Now, we substitute the given total mass into our equation:
Calculating this is straightforward. Ten percent of 75 gives us exactly .
So, our 75 kg person carries around of normal hydrogen. The rest of the mass, , belongs to oxygen, carbon, nitrogen, and trace elements.

The Isotopic Swap

Now comes the thrilling part of our thought experiment. We are going to replace every single atom of normal hydrogen () with deuterium ().
To understand the impact of this swap, we must look at the atomic level. A normal hydrogen atom () has an atomic mass of approximately 1 atomic mass unit (amu), as its nucleus contains just one proton. Deuterium (), on the other hand, has a nucleus containing one proton and one neutron, giving it an atomic mass of approximately 2 amu.
This means that the mass of a deuterium atom is exactly twice the mass of a normal hydrogen atom:
Because we are replacing the atoms one-to-one, the total number of hydrogen atoms in the body remains completely unchanged. However, since every single atom is now twice as heavy, the total mass of hydrogen in the body must also double!

Calculating the Final Mass Gain

Let's calculate the new mass of the hydrogen content. Since the original mass was , and the mass has doubled, we multiply by 2:
The person now has of deuterium in their body. But we must be careful here. The question does not ask for the new mass of hydrogen, nor does it ask for the new total weight of the person. It specifically asks for the weight gained.
To find the weight gained, we must calculate the difference between the new mass of the hydrogen and the old mass of the hydrogen. The mass of all the other elements (oxygen, carbon, nitrogen) remains completely unaffected by this isotopic swap.
Let's set up our final calculation:
Substituting our values:
The difference is exactly 7.5 kg. This is the extra mass the person carries due to the heavier neutrons in the deuterium atoms.
This problem beautifully illustrates the connection between macroscopic mass percentages and microscopic atomic properties. By simply knowing the mass ratio of isotopes, we can predict large-scale changes in physical weight!

Similar Questions

LEVELJEE Main

Number of atoms in of Fe (atomic mass = ) is

(A)
twice that of N
(B)
half that of H
(C)
Both (a) and (b)
(D)
None of the above
JEE Main 2020
LEVELJEE Main

The mass percentage of nitrogen in histamine is ………

JEE Main 2020
LEVELJEE Main

The average molar mass of chlorine is . The ratio of to in naturally occurring chlorine is close to

(A)
(B)
(C)
(D)
JEE Main 2014
LEVELJEE Main

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is . The ratio of number of their molecule is

(A)
(B)
(C)
(D)
LEVELJEE Main

If we consider that , in place of , mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

(A)
be a function of the molecular mass of the substance
(B)
remain unchanged
(C)
increase two fold
(D)
decrease twice
LEVELJEE Main

In an organic compound of molar mass C, H and N atoms are present in by weight. Molecular formula can be

(A)
(B)
(C)
(D)
JEE Main 2020
LEVELJEE Main

Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve of iron in of wheat is ......... . Atomic weight : ; ;

JEE Advanced 2022
LEVELJEE Advanced

To check the principle of multiple proportions, a series of pure binary compounds () were analyzed and their composition is tabulated below. The correct option(s) is(are) \begin{array}{|c|c|c|} \hline \text{Compound} & \text{Weight \% of P} & \text{Weight \% of Q} \\ \hline 1 & 50 & 50 \\ \hline 2 & 44.4 & 55.6 \\ \hline 3 & 40 & 60 \\ \hline \end{array}

* Multiple Correct Options
(A)
If empirical formula of compound 3 is , then the empirical formula of compound 2 is .
(B)
If empirical formula of compound 3 is and atomic weight of element P is 20, then the atomic weight of Q is 45.
(C)
If empirical formula of compound 2 is PQ, then the empirical formula of the compound 1 is .
(D)
If atomic weight of P and Q are 70 and 35, respectively, then the empirical formula of compound 1 is .
JEE Main 2018
LEVELJEE Advanced

The ratio of mass per cent of C and H of an organic compound () is . If one molecule of the above compound () contains half as much oxygen as required to burn one molecule of compound completely to and . The empirical formula of compound is

(A)
(B)
(C)
(D)
JEE Main 2021
LEVELJEE Main

The mole fraction of a solute in a molal aqueous solution is ......... (Round off to the nearest integer). [Given, atomic masses , ]