The number of N atoms is 681 g of $\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{6}$ is $\mathrm{x} \times 10^{21}$. The value of x is \_\_\_\_ ( $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}$ ) (Nearest Integer)

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Some Basic Concepts of Chemistry

The number of N atoms is 681 g of

C_{7} H_{5} N_{3} O_{6}

x \times 1 0^{21}

. The value of x is ____ (

N_{A} = 6.02 \times 1 0^{23} mol^{- 1}

) (Nearest Integer)

Answer: 5418

Solution

First, calculate the molar mass of C₇H₅N₃O₆: M = 7(12) + 5(1) + 3(14) + 6(16) = 84 + 5 + 42 + 96 = 227 g/mol.

Number of moles = 681/227 = 3 mol.

Since each molecule contains 3 nitrogen atoms, total moles of N = 3 × 3 = 9 mol.

Number of N atoms = 9 × 6.02 × 10²³ = 54.18 × 10²³ = 5418 × 10²¹.

Therefore, x = 5418.

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Integer answer

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