JEE Main 2025 Jan 23 Shift 1, Chemistry Q8: Mole Concept And Equivalent Concept

JEE Main2025Jan 23, Shift 1Chemistry

$2.8 \times 1 0^{- 3}$ mol of $C O_{2}$ is left after removing $1 0^{21}$ molecules from its 'x' mg sample. The mass of $C O_{2}$ taken initially is

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

Solution

Balance the moles: $n_{initial} = n_{removed} + n_{left}$ .

Moles removed $= \frac{1 0 ^{21}}{6.02 \times 1 0 ^{23}} \approx 1.66 \times 1 0^{- 3}$ mol.

Moles left $= 2.8 \times 1 0^{- 3}$ mol.

So $n_{initial} = (2.8 + 1.66) \times 1 0^{- 3} = 4.46 \times 1 0^{- 3}$ mol.

Initial mass $= n \times M = 4.46 \times 1 0^{- 3} \times 44 \approx 196.2$ mg.

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