Chemistry · 2025
JEE Main · 23 January 2025, Shift 1 · Q51
2.8 × 10^-3 mol of CO_2 is left after removing 10^21 molecules from its ' x ' mg sample. The mass of CO_2 taken initially is Given: N_A =6.02 × 10^23…
$\displaystyle 2.8 \times 10^{-3} \mathrm{~mol}$ of $\displaystyle \mathrm{CO}_2$ is left after removing $\displaystyle 10^{21}$ molecules from its ' $\displaystyle x$ ' mg sample. The mass of $\displaystyle \mathrm{CO}_2$ taken initially is
Given: $\displaystyle \mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}$
Official answer
From NTA’s final answer key for this paper.
(1)
196.$\displaystyle 2$ mg
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JEE Main 2025 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.