The Number of Atoms Present in 4.25 g of NH3 is Approximately

By BYJU'S Exam Prep

Updated on: September 13th, 2023

a) 1 X 1023

b) 1.5 X 1023

c) 0.25 X 1023

d) 6.02 X 1023

The number of atoms present is 6.02 X 1023. Hence, option D is the correct choice. Let us understand this chemical equation by breaking it down. We will go through the step-by-step process and understand how to reach the solution to this question. First, start by noting down the quantities mentioned in the question itself.

Find the Number of Atoms in 4.25 g of NH3

In the given question, we are supposed to find the number of atoms present in 4.25 grams of NH3. And to reach the solution, we will have to go through the steps mentioned below.

As per the question, it is given that:

Mass of ammonia = 4.25 g

Gram molecular mass of ammonia = 17 g

We know that,

Number of moles of ammonia = Mass of ammonia/ Gram molecular mass of ammonia

After substituting the values, we get:

= 4.25/17

= 0.25 mol

1 mole of ammonia has 6.022 x 1023 molecules of ammonia

So, the number of molecules present in 0.25 mol of ammonia = 0.25 x 6.022 x 1023

= 1.505 x 1023 molecules

There are 4 atoms in 1 molecule of ammonia (3 hydrogens and 1 nitrogen)

Number of atoms present in 1.505 x 1023 ammonia molecules = 4 x 1.505 x 1023

= 6.02 x 1023

Therefore, the number of atoms present is 1 x 1023


The Number of Atoms Present in 4.25 g of NH3 is Approximately?

The number of atoms present in 4.25 g of NH3 is approximately 6.02 x 1023 is 1 x 1023To reach this conclusion, candidates must go through all the steps mentioned above and understand the concepts that are applied to this equation. Such types of questions are common in competitive exams and it is extremely beneficial to solve as many numericals as possible before the exam.

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