QJEEM22Chd5c1: JEE Main 2022 Chemistry - Class 11 JEEM22Chd5c1 Solution: The number of N atoms in 681 g of C 7 H 5 N 3 O 6 is x × 10 21…

First, determine the molar mass of the compound C $_7$ H $_5$ N $_3$ O $_6$ :

Carbon (C): 12 g/mol, and there are 7 C atoms: $7 \times 12 = 84 \text{ g/mol}$

Hydrogen (H): 1 g/mol, and there are 5 H atoms: $5 \times 1 = 5 \text{ g/mol}$

Nitrogen (N): 14 g/mol, and there are 3 N atoms: $3 \times 14 = 42 \text{ g/mol}$

Oxygen (O): 16 g/mol, and there are 6 O atoms: $6 \times 16 = 96 \text{ g/mol}$

Calculate the total molar mass:

$\text{Molar mass of C}_7\text{H}_5\text{N}_3\text{O}_6 = 84 + 5 + 42 + 96 = 227 \text{ g/mol}$

Calculate the number of moles of C $_7$ H $_5$ N $_3$ O $_6$ in 681 grams:

$\text{Moles of C}_7\text{H}_5\text{N}_3\text{O}_6 = \frac{681 \text{ g}}{227 \text{ g/mol}} = 3 \text{ moles}$

Each molecule of C $_7$ H $_5$ N $_3$ O $_6$ contains 3 nitrogen atoms. Thus, in 3 moles, the total moles of nitrogen atoms is:

$\text{Moles of N atoms} = 3 \times 3 = 9 \text{ moles}$

Using Avogadro's number, calculate the number of nitrogen atoms:

$\text{Number of N atoms} = 9 \text{ moles} \times 6.02 \times 10^{23} \text{ atoms/mole}$

$= 54.18 \times 10^{23} = 5.418 \times 10^{24} \text{ N atoms}$

Convert this number to the form $x \times 10^{21}$ :

$5.418 \times 10^{24} = 5418 \times 10^{21}$

Thus, the value of $x$ is:

$x = 5418$