$2.8 \times 10^{-3} \mathrm{~mol}$ of $\mathrm{CO}_{2}$ is left after removing $10^{21}$ molecules from its ' x ' mg sample. The mass of $\mathrm{CO}_{2}$ taken initially is Given: $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}$

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Some Basic Concepts of Chemistry

2.8 \times 1 0^{- 3} mol

CO_{2}

is left after removing

1 0^{21}

molecules from its ' x ' mg sample. The mass of

CO_{2}

taken initially is
Given:

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

Answer: (A)

Let the initial mass of

C O_{2}

x

mg, giving initial moles

\frac{x \times 1 0 ^{- 3}}{44}

.

Moles of

C O_{2}

removed

= \frac{1 0 ^{21}}{6.02 \times 1 0 ^{23}} = 1.661 \times 1 0^{- 3} mol

.

Since

2.8 \times 1 0^{- 3} mol

is left after removal:

\frac{x \times 1 0 ^{- 3}}{44} = 2.8 \times 1 0^{- 3} + 1.661 \times 1 0^{- 3} = 4.461 \times 1 0^{- 3}

x = 4.461 \times 1 0^{- 3} \times 44 \times 1 0^{3} = 196.2 mg

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option