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How many atoms of N are in 137. 0 grams of N2O3? A. 1. 085 × 1023 B. 1. 802 × 1023 C. 5. 985 × 1023 D. 2. 171 × 1024 E. 3. 604 × 1024.

Sagot :

The number of atoms of nitrogen present in 137 grams of N₂O₃ in 2.1 × 10²⁴.

What is Avogadro's number?

Avogadro's number is that number which tells about the number of atoms or entities of any substance present in one mole of any substance, i.e. 1 mole = 6.022 × 10²³

First we have to convert mass of N₂O₃ to moles by using the below formula:

n = W/M, where

W = given mass = 137g

M = molar mass = 76g/mol

Moles of N₂O₃ = 137 / 76 = 1.8mol

From the stoichiometry of the mass and formula it is clear that:

In 76g of N₂O₃ = 2 moles of Nitrogens are present

In 137g of N₂O₃ = 2/76×137=3.6 moles of Nitrogens are present

No. of N atoms in 3.6 moles = 3.6 × 6.022 × 10²³ = 2.1 × 10²⁴

Hence option (D) is correct.

To know more about Avogadro's number, visit the below link:

https://brainly.com/question/10614569

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