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

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

Summary:

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 10²³. To 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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